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69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,621,420,675 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 154 | 0 | n=int(input())
a1=[int(x) for x in input().split()]
a2=[int(y) for y in input().split()]
a3=[int(z) for z in input().split()]
s1=sum(a1)
s2=sum(a2)
s3=sum(a3)
s4=s1+s2+s3
if(s4==0):
print("YES")
else:
print("NO")
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
n=int(input())
a1=[int(x) for x in input().split()]
a2=[int(y) for y in input().split()]
a3=[int(z) for z in input().split()]
s1=sum(a1)
s2=sum(a2)
s3=sum(a3)
s4=s1+s2+s3
if(s4==0):
print("YES")
else:
print("NO")
``` | 0 |
777 | C | Alyona and Spreadsheet | PROGRAMMING | 1,600 | [
"binary search",
"data structures",
"dp",
"greedy",
"implementation",
"two pointers"
] | null | null | During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit tables.
Now she has a table filled with integers. The table consists of *n* rows and *m* columns. By *a**i*,<=*j* we will denote the integer located at the *i*-th row and the *j*-th column. We say that the table is sorted in non-decreasing order in the column *j* if *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from 1 to *n*<=-<=1.
Teacher gave Alyona *k* tasks. For each of the tasks two integers *l* and *r* are given and Alyona has to answer the following question: if one keeps the rows from *l* to *r* inclusive and deletes all others, will the table be sorted in non-decreasing order in at least one column? Formally, does there exist such *j* that *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from *l* to *r*<=-<=1 inclusive.
Alyona is too small to deal with this task and asks you to help! | The first line of the input contains two positive integers *n* and *m* (1<=≤<=*n*·*m*<=≤<=100<=000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table.
Each of the following *n* lines contains *m* integers. The *j*-th integers in the *i* of these lines stands for *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=109).
The next line of the input contains an integer *k* (1<=≤<=*k*<=≤<=100<=000) — the number of task that teacher gave to Alyona.
The *i*-th of the next *k* lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). | Print "Yes" to the *i*-th line of the output if the table consisting of rows from *l**i* to *r**i* inclusive is sorted in non-decreasing order in at least one column. Otherwise, print "No". | [
"5 4\n1 2 3 5\n3 1 3 2\n4 5 2 3\n5 5 3 2\n4 4 3 4\n6\n1 1\n2 5\n4 5\n3 5\n1 3\n1 5\n"
] | [
"Yes\nNo\nYes\nYes\nYes\nNo\n"
] | In the sample, the whole table is not sorted in any column. However, rows 1–3 are sorted in column 1, while rows 4–5 are sorted in column 3. | 1,500 | [
{
"input": "5 4\n1 2 3 5\n3 1 3 2\n4 5 2 3\n5 5 3 2\n4 4 3 4\n6\n1 1\n2 5\n4 5\n3 5\n1 3\n1 5",
"output": "Yes\nNo\nYes\nYes\nYes\nNo"
},
{
"input": "1 1\n1\n1\n1 1",
"output": "Yes"
},
{
"input": "10 1\n523130301\n127101624\n15573616\n703140639\n628818570\n957494759\n161270109\n386865653\n67832626\n53360557\n17\n4 5\n4 7\n8 8\n9 9\n3 9\n8 10\n8 9\n7 9\n4 5\n2 9\n4 6\n2 4\n2 6\n4 6\n7 9\n2 4\n8 10",
"output": "No\nNo\nYes\nYes\nNo\nNo\nNo\nNo\nNo\nNo\nNo\nNo\nNo\nNo\nNo\nNo\nNo"
},
{
"input": "15 1\n556231456\n573340933\n626155933\n397229387\n10255952\n376567394\n906742013\n269437009\n31298788\n712285290\n620239975\n379221898\n229140718\n95080095\n997123854\n18\n5 15\n1 12\n4 10\n2 15\n12 15\n15 15\n2 2\n15 15\n15 15\n13 13\n10 14\n3 6\n14 15\n3 6\n4 4\n14 15\n12 14\n1 9",
"output": "No\nNo\nNo\nNo\nNo\nYes\nYes\nYes\nYes\nYes\nNo\nNo\nYes\nNo\nYes\nYes\nNo\nNo"
},
{
"input": "11 1\n501465490\n366941771\n415080944\n385243536\n445132523\n697044413\n894369800\n812743722\n23684788\n466526046\n953916313\n45\n2 4\n8 9\n7 7\n4 9\n2 9\n2 11\n4 4\n5 7\n1 2\n5 10\n4 6\n1 7\n4 4\n1 6\n4 7\n10 11\n1 8\n6 11\n8 8\n8 10\n1 1\n5 10\n9 10\n6 9\n6 11\n1 1\n9 9\n5 11\n1 2\n9 11\n2 6\n3 7\n11 11\n6 7\n11 11\n7 8\n5 8\n11 11\n5 6\n4 5\n2 6\n5 10\n9 9\n1 1\n1 1",
"output": "No\nNo\nYes\nNo\nNo\nNo\nYes\nYes\nNo\nNo\nYes\nNo\nYes\nNo\nYes\nYes\nNo\nNo\nYes\nNo\nYes\nNo\nYes\nNo\nNo\nYes\nYes\nNo\nNo\nYes\nNo\nNo\nYes\nYes\nYes\nNo\nNo\nYes\nYes\nYes\nNo\nNo\nYes\nYes\nYes"
}
] | 1,623,138,393 | 2,147,483,647 | PyPy 3 | OK | TESTS | 114 | 498 | 16,384,000 | ###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
# from itertools import permutations
# from functools import cmp_to_key # for adding custom comparator
# from fractions import Fraction
from collections import *
from sys import stdin
# from bisect import *
from heapq import *
from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
rr = lambda x : reversed(range(x))
mod = int(1e9)+7
inf = float("inf")
n, m = gil()
mat = [gil() for _ in range(n)]
p = set()
for j in range(m):
l, r = 0, 0
while r+1 < n :
if mat[r+1][j] < mat[r][j]:
if l != r:p.add((l+1, r+1))
l = r+1
r += 1
if l != r:p.add((l+1, r+1))
p = list(p)
p.sort(reverse=True)
endMax = [i for i in range(n+1)]
h = []
for start in range(1, n+1):
while p and p[-1][0] == start:
heappush(h, -p.pop()[1])
while h and -h[0] <= start:
heappop(h)
if h :
endMax[start] = -h[0]
ans = []
for _ in range(gil()[0]):
l, r = gil()
ans.append('Yes' if endMax[l] >= r else 'No')
print('\n'.join(ans)) | Title: Alyona and Spreadsheet
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit tables.
Now she has a table filled with integers. The table consists of *n* rows and *m* columns. By *a**i*,<=*j* we will denote the integer located at the *i*-th row and the *j*-th column. We say that the table is sorted in non-decreasing order in the column *j* if *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from 1 to *n*<=-<=1.
Teacher gave Alyona *k* tasks. For each of the tasks two integers *l* and *r* are given and Alyona has to answer the following question: if one keeps the rows from *l* to *r* inclusive and deletes all others, will the table be sorted in non-decreasing order in at least one column? Formally, does there exist such *j* that *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from *l* to *r*<=-<=1 inclusive.
Alyona is too small to deal with this task and asks you to help!
Input Specification:
The first line of the input contains two positive integers *n* and *m* (1<=≤<=*n*·*m*<=≤<=100<=000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table.
Each of the following *n* lines contains *m* integers. The *j*-th integers in the *i* of these lines stands for *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=109).
The next line of the input contains an integer *k* (1<=≤<=*k*<=≤<=100<=000) — the number of task that teacher gave to Alyona.
The *i*-th of the next *k* lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
Output Specification:
Print "Yes" to the *i*-th line of the output if the table consisting of rows from *l**i* to *r**i* inclusive is sorted in non-decreasing order in at least one column. Otherwise, print "No".
Demo Input:
['5 4\n1 2 3 5\n3 1 3 2\n4 5 2 3\n5 5 3 2\n4 4 3 4\n6\n1 1\n2 5\n4 5\n3 5\n1 3\n1 5\n']
Demo Output:
['Yes\nNo\nYes\nYes\nYes\nNo\n']
Note:
In the sample, the whole table is not sorted in any column. However, rows 1–3 are sorted in column 1, while rows 4–5 are sorted in column 3. | ```python
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
# from itertools import permutations
# from functools import cmp_to_key # for adding custom comparator
# from fractions import Fraction
from collections import *
from sys import stdin
# from bisect import *
from heapq import *
from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
rr = lambda x : reversed(range(x))
mod = int(1e9)+7
inf = float("inf")
n, m = gil()
mat = [gil() for _ in range(n)]
p = set()
for j in range(m):
l, r = 0, 0
while r+1 < n :
if mat[r+1][j] < mat[r][j]:
if l != r:p.add((l+1, r+1))
l = r+1
r += 1
if l != r:p.add((l+1, r+1))
p = list(p)
p.sort(reverse=True)
endMax = [i for i in range(n+1)]
h = []
for start in range(1, n+1):
while p and p[-1][0] == start:
heappush(h, -p.pop()[1])
while h and -h[0] <= start:
heappop(h)
if h :
endMax[start] = -h[0]
ans = []
for _ in range(gil()[0]):
l, r = gil()
ans.append('Yes' if endMax[l] >= r else 'No')
print('\n'.join(ans))
``` | 3 |
|
271 | A | Beautiful Year | PROGRAMMING | 800 | [
"brute force"
] | null | null | It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits. | The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number. | Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists. | [
"1987\n",
"2013\n"
] | [
"2013\n",
"2014\n"
] | none | 500 | [
{
"input": "1987",
"output": "2013"
},
{
"input": "2013",
"output": "2014"
},
{
"input": "1000",
"output": "1023"
},
{
"input": "1001",
"output": "1023"
},
{
"input": "1234",
"output": "1235"
},
{
"input": "5555",
"output": "5601"
},
{
"input": "9000",
"output": "9012"
},
{
"input": "1111",
"output": "1203"
},
{
"input": "8999",
"output": "9012"
},
{
"input": "4572",
"output": "4573"
},
{
"input": "6666",
"output": "6701"
},
{
"input": "2001",
"output": "2013"
},
{
"input": "3000",
"output": "3012"
},
{
"input": "7712",
"output": "7801"
},
{
"input": "8088",
"output": "8091"
},
{
"input": "1594",
"output": "1596"
},
{
"input": "6016",
"output": "6017"
},
{
"input": "8800",
"output": "8901"
},
{
"input": "3331",
"output": "3401"
},
{
"input": "5090",
"output": "5091"
},
{
"input": "2342",
"output": "2345"
},
{
"input": "2334",
"output": "2340"
},
{
"input": "1123",
"output": "1203"
},
{
"input": "8989",
"output": "9012"
},
{
"input": "8977",
"output": "9012"
},
{
"input": "6869",
"output": "6870"
},
{
"input": "8999",
"output": "9012"
}
] | 1,690,478,980 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 92 | 0 | s = int(input()) + 1
while len(set(str(s)))<4:s+=1
print(s) | Title: Beautiful Year
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
Input Specification:
The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.
Output Specification:
Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.
Demo Input:
['1987\n', '2013\n']
Demo Output:
['2013\n', '2014\n']
Note:
none | ```python
s = int(input()) + 1
while len(set(str(s)))<4:s+=1
print(s)
``` | 3 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | You are given two lists of non-zero digits.
Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer? | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=9) — the lengths of the first and the second lists, respectively.
The second line contains *n* distinct digits *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=9) — the elements of the first list.
The third line contains *m* distinct digits *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=9) — the elements of the second list. | Print the smallest pretty integer. | [
"2 3\n4 2\n5 7 6\n",
"8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1\n"
] | [
"25\n",
"1\n"
] | In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.
In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer. | 0 | [
{
"input": "2 3\n4 2\n5 7 6",
"output": "25"
},
{
"input": "8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "1 1\n9\n1",
"output": "19"
},
{
"input": "9 1\n5 4 2 3 6 1 7 9 8\n9",
"output": "9"
},
{
"input": "5 3\n7 2 5 8 6\n3 1 9",
"output": "12"
},
{
"input": "4 5\n5 2 6 4\n8 9 1 3 7",
"output": "12"
},
{
"input": "5 9\n4 2 1 6 7\n2 3 4 5 6 7 8 9 1",
"output": "1"
},
{
"input": "9 9\n5 4 3 2 1 6 7 8 9\n3 2 1 5 4 7 8 9 6",
"output": "1"
},
{
"input": "9 5\n2 3 4 5 6 7 8 9 1\n4 2 1 6 7",
"output": "1"
},
{
"input": "9 9\n1 2 3 4 5 6 7 8 9\n1 2 3 4 5 6 7 8 9",
"output": "1"
},
{
"input": "9 9\n1 2 3 4 5 6 7 8 9\n9 8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "9 9\n9 8 7 6 5 4 3 2 1\n1 2 3 4 5 6 7 8 9",
"output": "1"
},
{
"input": "9 9\n9 8 7 6 5 4 3 2 1\n9 8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "1 1\n8\n9",
"output": "89"
},
{
"input": "1 1\n9\n8",
"output": "89"
},
{
"input": "1 1\n1\n2",
"output": "12"
},
{
"input": "1 1\n2\n1",
"output": "12"
},
{
"input": "1 1\n9\n9",
"output": "9"
},
{
"input": "1 1\n1\n1",
"output": "1"
},
{
"input": "4 5\n3 2 4 5\n1 6 5 9 8",
"output": "5"
},
{
"input": "3 2\n4 5 6\n1 5",
"output": "5"
},
{
"input": "5 4\n1 3 5 6 7\n2 4 3 9",
"output": "3"
},
{
"input": "5 5\n1 3 5 7 9\n2 4 6 8 9",
"output": "9"
},
{
"input": "2 2\n1 8\n2 8",
"output": "8"
},
{
"input": "5 5\n5 6 7 8 9\n1 2 3 4 5",
"output": "5"
},
{
"input": "5 5\n1 2 3 4 5\n1 2 3 4 5",
"output": "1"
},
{
"input": "5 5\n1 2 3 4 5\n2 3 4 5 6",
"output": "2"
},
{
"input": "2 2\n1 5\n2 5",
"output": "5"
},
{
"input": "4 4\n1 3 5 8\n2 4 6 8",
"output": "8"
},
{
"input": "3 3\n1 5 3\n2 5 7",
"output": "5"
},
{
"input": "3 3\n3 6 8\n2 6 9",
"output": "6"
},
{
"input": "2 2\n1 4\n2 4",
"output": "4"
},
{
"input": "5 3\n3 4 5 6 7\n1 5 9",
"output": "5"
},
{
"input": "4 4\n1 2 3 4\n2 5 6 7",
"output": "2"
},
{
"input": "5 5\n1 2 3 4 5\n9 2 1 7 5",
"output": "1"
},
{
"input": "2 2\n1 3\n2 3",
"output": "3"
},
{
"input": "3 3\n3 2 1\n3 2 1",
"output": "1"
},
{
"input": "3 3\n1 3 5\n2 3 6",
"output": "3"
},
{
"input": "3 3\n5 6 7\n5 6 7",
"output": "5"
},
{
"input": "1 2\n5\n2 5",
"output": "5"
},
{
"input": "3 3\n2 4 9\n7 8 9",
"output": "9"
},
{
"input": "3 3\n1 2 4\n3 4 5",
"output": "4"
},
{
"input": "3 2\n1 4 9\n2 4",
"output": "4"
},
{
"input": "3 3\n3 5 6\n1 5 9",
"output": "5"
},
{
"input": "3 2\n1 2 4\n3 4",
"output": "4"
},
{
"input": "2 4\n8 9\n1 2 3 9",
"output": "9"
},
{
"input": "1 2\n9\n8 9",
"output": "9"
},
{
"input": "3 2\n1 2 4\n4 2",
"output": "2"
},
{
"input": "2 3\n4 5\n1 3 5",
"output": "5"
},
{
"input": "3 2\n1 2 3\n2 3",
"output": "2"
},
{
"input": "4 3\n1 3 5 9\n2 8 9",
"output": "9"
},
{
"input": "2 2\n1 9\n9 2",
"output": "9"
}
] | 1,508,056,793 | 2,093 | Python 3 | OK | TESTS | 52 | 62 | 5,529,600 | x,y = map(int,input().split())
lst1 = []
lst2 = []
num1 = map(int,input().split())
num2 = map(int,input().split())
for a in num1:
lst1.append(a)
for b in num2:
lst2.append(b)
lst1.sort()
lst2.sort()
intersec = []
for x in range(0,len(lst1)):
for y in range(0,len(lst2)):
if lst1[x] == lst2[y]:
intersec.append(lst1[x])
if len(intersec) == 0:
if lst1[0] > lst2[0]:
print(str(lst2[0])+str(lst1[0]))
else:
print(str(lst1[0])+str(lst2[0]))
else:
print(intersec[0])
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two lists of non-zero digits.
Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=9) — the lengths of the first and the second lists, respectively.
The second line contains *n* distinct digits *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=9) — the elements of the first list.
The third line contains *m* distinct digits *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=9) — the elements of the second list.
Output Specification:
Print the smallest pretty integer.
Demo Input:
['2 3\n4 2\n5 7 6\n', '8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1\n']
Demo Output:
['25\n', '1\n']
Note:
In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.
In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer. | ```python
x,y = map(int,input().split())
lst1 = []
lst2 = []
num1 = map(int,input().split())
num2 = map(int,input().split())
for a in num1:
lst1.append(a)
for b in num2:
lst2.append(b)
lst1.sort()
lst2.sort()
intersec = []
for x in range(0,len(lst1)):
for y in range(0,len(lst2)):
if lst1[x] == lst2[y]:
intersec.append(lst1[x])
if len(intersec) == 0:
if lst1[0] > lst2[0]:
print(str(lst2[0])+str(lst1[0]))
else:
print(str(lst1[0])+str(lst2[0]))
else:
print(intersec[0])
``` | 3 |
|
911 | B | Two Cakes | PROGRAMMING | 1,200 | [
"binary search",
"brute force",
"implementation"
] | null | null | It's New Year's Eve soon, so Ivan decided it's high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into *a* pieces, and the second one — into *b* pieces.
Ivan knows that there will be *n* people at the celebration (including himself), so Ivan has set *n* plates for the cakes. Now he is thinking about how to distribute the cakes between the plates. Ivan wants to do it in such a way that all following conditions are met:
1. Each piece of each cake is put on some plate; 1. Each plate contains at least one piece of cake; 1. No plate contains pieces of both cakes.
To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number *x* such that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least *x* pieces of cake.
Help Ivan to calculate this number *x*! | The first line contains three integers *n*, *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100, 2<=≤<=*n*<=≤<=*a*<=+<=*b*) — the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively. | Print the maximum possible number *x* such that Ivan can distribute the cake in such a way that each plate will contain at least *x* pieces of cake. | [
"5 2 3\n",
"4 7 10\n"
] | [
"1\n",
"3\n"
] | In the first example there is only one way to distribute cakes to plates, all of them will have 1 cake on it.
In the second example you can have two plates with 3 and 4 pieces of the first cake and two plates both with 5 pieces of the second cake. Minimal number of pieces is 3. | 0 | [
{
"input": "5 2 3",
"output": "1"
},
{
"input": "4 7 10",
"output": "3"
},
{
"input": "100 100 100",
"output": "2"
},
{
"input": "10 100 3",
"output": "3"
},
{
"input": "2 9 29",
"output": "9"
},
{
"input": "4 6 10",
"output": "3"
},
{
"input": "3 70 58",
"output": "35"
},
{
"input": "5 7 10",
"output": "3"
},
{
"input": "5 30 22",
"output": "10"
},
{
"input": "5 5 6",
"output": "2"
},
{
"input": "2 4 3",
"output": "3"
},
{
"input": "10 10 31",
"output": "3"
},
{
"input": "2 1 1",
"output": "1"
},
{
"input": "10 98 99",
"output": "19"
},
{
"input": "4 10 16",
"output": "5"
},
{
"input": "11 4 8",
"output": "1"
},
{
"input": "5 10 14",
"output": "4"
},
{
"input": "6 7 35",
"output": "7"
},
{
"input": "5 6 7",
"output": "2"
},
{
"input": "4 15 3",
"output": "3"
},
{
"input": "7 48 77",
"output": "16"
},
{
"input": "4 4 10",
"output": "3"
},
{
"input": "4 7 20",
"output": "6"
},
{
"input": "5 2 8",
"output": "2"
},
{
"input": "3 2 3",
"output": "1"
},
{
"input": "14 95 1",
"output": "1"
},
{
"input": "99 82 53",
"output": "1"
},
{
"input": "10 71 27",
"output": "9"
},
{
"input": "5 7 8",
"output": "2"
},
{
"input": "11 77 77",
"output": "12"
},
{
"input": "10 5 28",
"output": "3"
},
{
"input": "7 3 12",
"output": "2"
},
{
"input": "10 15 17",
"output": "3"
},
{
"input": "7 7 7",
"output": "1"
},
{
"input": "4 11 18",
"output": "6"
},
{
"input": "3 3 4",
"output": "2"
},
{
"input": "9 2 10",
"output": "1"
},
{
"input": "100 90 20",
"output": "1"
},
{
"input": "3 2 2",
"output": "1"
},
{
"input": "12 45 60",
"output": "8"
},
{
"input": "3 94 79",
"output": "47"
},
{
"input": "41 67 34",
"output": "2"
},
{
"input": "9 3 23",
"output": "2"
},
{
"input": "10 20 57",
"output": "7"
},
{
"input": "55 27 30",
"output": "1"
},
{
"input": "100 100 10",
"output": "1"
},
{
"input": "20 8 70",
"output": "3"
},
{
"input": "3 3 3",
"output": "1"
},
{
"input": "4 9 15",
"output": "5"
},
{
"input": "3 1 3",
"output": "1"
},
{
"input": "2 94 94",
"output": "94"
},
{
"input": "5 3 11",
"output": "2"
},
{
"input": "4 3 2",
"output": "1"
},
{
"input": "12 12 100",
"output": "9"
},
{
"input": "6 75 91",
"output": "25"
},
{
"input": "3 4 3",
"output": "2"
},
{
"input": "3 2 5",
"output": "2"
},
{
"input": "6 5 15",
"output": "3"
},
{
"input": "4 3 6",
"output": "2"
},
{
"input": "3 9 9",
"output": "4"
},
{
"input": "26 93 76",
"output": "6"
},
{
"input": "41 34 67",
"output": "2"
},
{
"input": "6 12 6",
"output": "3"
},
{
"input": "5 20 8",
"output": "5"
},
{
"input": "2 1 3",
"output": "1"
},
{
"input": "35 66 99",
"output": "4"
},
{
"input": "30 7 91",
"output": "3"
},
{
"input": "5 22 30",
"output": "10"
},
{
"input": "8 19 71",
"output": "10"
},
{
"input": "3 5 6",
"output": "3"
},
{
"input": "5 3 8",
"output": "2"
},
{
"input": "2 4 2",
"output": "2"
},
{
"input": "4 3 7",
"output": "2"
},
{
"input": "5 20 10",
"output": "5"
},
{
"input": "5 100 50",
"output": "25"
},
{
"input": "6 3 10",
"output": "2"
},
{
"input": "2 90 95",
"output": "90"
},
{
"input": "4 8 6",
"output": "3"
},
{
"input": "6 10 3",
"output": "2"
},
{
"input": "3 3 5",
"output": "2"
},
{
"input": "5 33 33",
"output": "11"
},
{
"input": "5 5 8",
"output": "2"
},
{
"input": "19 24 34",
"output": "3"
},
{
"input": "5 5 12",
"output": "3"
},
{
"input": "8 7 10",
"output": "2"
},
{
"input": "5 56 35",
"output": "17"
},
{
"input": "4 3 5",
"output": "1"
},
{
"input": "18 100 50",
"output": "8"
},
{
"input": "5 6 8",
"output": "2"
},
{
"input": "5 98 100",
"output": "33"
},
{
"input": "6 5 8",
"output": "2"
},
{
"input": "3 40 80",
"output": "40"
},
{
"input": "4 8 11",
"output": "4"
},
{
"input": "66 100 99",
"output": "3"
},
{
"input": "17 100 79",
"output": "10"
},
{
"input": "3 2 10",
"output": "2"
},
{
"input": "99 100 99",
"output": "2"
},
{
"input": "21 100 5",
"output": "5"
},
{
"input": "3 10 2",
"output": "2"
},
{
"input": "4 100 63",
"output": "33"
},
{
"input": "2 2 10",
"output": "2"
},
{
"input": "5 94 79",
"output": "31"
},
{
"input": "4 12 5",
"output": "4"
},
{
"input": "5 5 40",
"output": "5"
},
{
"input": "99 99 99",
"output": "1"
},
{
"input": "8 97 44",
"output": "16"
},
{
"input": "11 4 10",
"output": "1"
},
{
"input": "6 3 3",
"output": "1"
},
{
"input": "7 3 4",
"output": "1"
},
{
"input": "8 4 4",
"output": "1"
},
{
"input": "9 4 5",
"output": "1"
},
{
"input": "12 6 6",
"output": "1"
},
{
"input": "4 48 89",
"output": "29"
},
{
"input": "8 3 6",
"output": "1"
},
{
"input": "4 6 3",
"output": "2"
},
{
"input": "5 5 1",
"output": "1"
},
{
"input": "11 6 5",
"output": "1"
},
{
"input": "4 5 4",
"output": "2"
},
{
"input": "6 6 4",
"output": "1"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "4 1 3",
"output": "1"
},
{
"input": "3 3 1",
"output": "1"
},
{
"input": "9 4 6",
"output": "1"
},
{
"input": "6 5 6",
"output": "1"
},
{
"input": "2 2 3",
"output": "2"
},
{
"input": "4 5 1",
"output": "1"
},
{
"input": "13 6 7",
"output": "1"
},
{
"input": "14 7 7",
"output": "1"
},
{
"input": "12 97 13",
"output": "8"
},
{
"input": "4 2 9",
"output": "2"
},
{
"input": "10 20 59",
"output": "7"
},
{
"input": "12 34 56",
"output": "7"
},
{
"input": "4 5 9",
"output": "3"
},
{
"input": "2 2 2",
"output": "2"
},
{
"input": "4 66 41",
"output": "22"
}
] | 1,580,570,076 | 2,147,483,647 | PyPy 3 | OK | TESTS | 135 | 156 | 0 | n,a,b = [int(i) for i in input().split()]
maxi = -1
for i in range(1,n):
if min(a//i,b//(n-i))>maxi:
maxi = int(min(a//i,b//(n-i)))
print(maxi)
| Title: Two Cakes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's New Year's Eve soon, so Ivan decided it's high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into *a* pieces, and the second one — into *b* pieces.
Ivan knows that there will be *n* people at the celebration (including himself), so Ivan has set *n* plates for the cakes. Now he is thinking about how to distribute the cakes between the plates. Ivan wants to do it in such a way that all following conditions are met:
1. Each piece of each cake is put on some plate; 1. Each plate contains at least one piece of cake; 1. No plate contains pieces of both cakes.
To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number *x* such that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least *x* pieces of cake.
Help Ivan to calculate this number *x*!
Input Specification:
The first line contains three integers *n*, *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100, 2<=≤<=*n*<=≤<=*a*<=+<=*b*) — the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively.
Output Specification:
Print the maximum possible number *x* such that Ivan can distribute the cake in such a way that each plate will contain at least *x* pieces of cake.
Demo Input:
['5 2 3\n', '4 7 10\n']
Demo Output:
['1\n', '3\n']
Note:
In the first example there is only one way to distribute cakes to plates, all of them will have 1 cake on it.
In the second example you can have two plates with 3 and 4 pieces of the first cake and two plates both with 5 pieces of the second cake. Minimal number of pieces is 3. | ```python
n,a,b = [int(i) for i in input().split()]
maxi = -1
for i in range(1,n):
if min(a//i,b//(n-i))>maxi:
maxi = int(min(a//i,b//(n-i)))
print(maxi)
``` | 3 |
|
1,009 | B | Minimum Ternary String | PROGRAMMING | 1,400 | [
"greedy",
"implementation"
] | null | null | You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$. | The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive). | Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero). | [
"100210\n",
"11222121\n",
"20\n"
] | [
"001120\n",
"11112222\n",
"20\n"
] | none | 0 | [
{
"input": "100210",
"output": "001120"
},
{
"input": "11222121",
"output": "11112222"
},
{
"input": "20",
"output": "20"
},
{
"input": "1002",
"output": "0012"
},
{
"input": "10",
"output": "01"
},
{
"input": "000021",
"output": "000012"
},
{
"input": "021",
"output": "012"
},
{
"input": "2",
"output": "2"
},
{
"input": "201",
"output": "120"
},
{
"input": "2112120",
"output": "1112220"
},
{
"input": "102",
"output": "012"
},
{
"input": "202",
"output": "202"
},
{
"input": "220201",
"output": "122020"
},
{
"input": "12",
"output": "12"
},
{
"input": "100022202",
"output": "000122202"
},
{
"input": "01",
"output": "01"
},
{
"input": "1",
"output": "1"
}
] | 1,664,019,446 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 77 | 2,048,000 |
for i in range(int(input())):
n = int(input())
a=list(map(int,input().split()))
a1=[0]*10000
for i in a:
a1[i]+=1
temp=None
best1=None
best2=None
r=999999999
for i in range(len(a1)):
if a1[i]>=4:
best2=i
best1=i
break
elif a1[i]>=2:
if temp != None:
if r > abs(temp - i):
best1 = temp
r = abs(temp - i)
best2 = i
temp = i
print(best1,best1,best2,best2)
| Title: Minimum Ternary String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$.
Input Specification:
The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive).
Output Specification:
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
Demo Input:
['100210\n', '11222121\n', '20\n']
Demo Output:
['001120\n', '11112222\n', '20\n']
Note:
none | ```python
for i in range(int(input())):
n = int(input())
a=list(map(int,input().split()))
a1=[0]*10000
for i in a:
a1[i]+=1
temp=None
best1=None
best2=None
r=999999999
for i in range(len(a1)):
if a1[i]>=4:
best2=i
best1=i
break
elif a1[i]>=2:
if temp != None:
if r > abs(temp - i):
best1 = temp
r = abs(temp - i)
best2 = i
temp = i
print(best1,best1,best2,best2)
``` | -1 |
|
490 | C | Hacking Cypher | PROGRAMMING | 1,700 | [
"brute force",
"math",
"number theory",
"strings"
] | null | null | Polycarpus participates in a competition for hacking into a new secure messenger. He's almost won.
Having carefully studied the interaction protocol, Polycarpus came to the conclusion that the secret key can be obtained if he properly cuts the public key of the application into two parts. The public key is a long integer which may consist of even a million digits!
Polycarpus needs to find such a way to cut the public key into two nonempty parts, that the first (left) part is divisible by *a* as a separate number, and the second (right) part is divisible by *b* as a separate number. Both parts should be positive integers that have no leading zeros. Polycarpus knows values *a* and *b*.
Help Polycarpus and find any suitable method to cut the public key. | The first line of the input contains the public key of the messenger — an integer without leading zeroes, its length is in range from 1 to 106 digits. The second line contains a pair of space-separated positive integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=108). | In the first line print "YES" (without the quotes), if the method satisfying conditions above exists. In this case, next print two lines — the left and right parts after the cut. These two parts, being concatenated, must be exactly identical to the public key. The left part must be divisible by *a*, and the right part must be divisible by *b*. The two parts must be positive integers having no leading zeros. If there are several answers, print any of them.
If there is no answer, print in a single line "NO" (without the quotes). | [
"116401024\n97 1024\n",
"284254589153928171911281811000\n1009 1000\n",
"120\n12 1\n"
] | [
"YES\n11640\n1024\n",
"YES\n2842545891539\n28171911281811000\n",
"NO\n"
] | none | 1,500 | [
{
"input": "116401024\n97 1024",
"output": "YES\n11640\n1024"
},
{
"input": "284254589153928171911281811000\n1009 1000",
"output": "YES\n2842545891539\n28171911281811000"
},
{
"input": "120\n12 1",
"output": "NO"
},
{
"input": "604\n6 4",
"output": "YES\n60\n4"
},
{
"input": "2108\n7 8",
"output": "YES\n210\n8"
},
{
"input": "7208\n10 1",
"output": "YES\n720\n8"
},
{
"input": "97502821\n25 91",
"output": "YES\n9750\n2821"
},
{
"input": "803405634\n309 313",
"output": "YES\n80340\n5634"
},
{
"input": "15203400\n38 129",
"output": "NO"
},
{
"input": "8552104774\n973 76",
"output": "NO"
},
{
"input": "2368009434\n320 106",
"output": "YES\n236800\n9434"
},
{
"input": "425392502895812\n4363 2452",
"output": "YES\n42539250\n2895812"
},
{
"input": "142222201649130\n4854 7853",
"output": "YES\n14222220\n1649130"
},
{
"input": "137871307228140\n9375 9092",
"output": "NO"
},
{
"input": "8784054131798916\n9 61794291",
"output": "YES\n87840\n54131798916"
},
{
"input": "24450015102786098\n75 55729838",
"output": "YES\n244500\n15102786098"
},
{
"input": "100890056766780885\n177 88010513",
"output": "YES\n1008900\n56766780885"
},
{
"input": "2460708054301924950\n9428 85246350",
"output": "YES\n24607080\n54301924950"
},
{
"input": "39915186055525904358\n90102 63169402",
"output": "YES\n399151860\n55525904358"
},
{
"input": "199510140021146591389\n458644 28692797",
"output": "YES\n1995101400\n21146591389"
},
{
"input": "4802711808015050898224\n8381696 51544172",
"output": "YES\n48027118080\n15050898224"
},
{
"input": "6450225349035040017740\n8872387 56607460",
"output": "YES\n64502253490\n35040017740"
},
{
"input": "4530228043401488\n71454701 8",
"output": "YES\n453022804340\n1488"
},
{
"input": "18769213650033200\n56876405 100",
"output": "YES\n187692136500\n33200"
},
{
"input": "389744672208415\n17019418 765",
"output": "YES\n38974467220\n8415"
},
{
"input": "1256363256202133560\n26228878 7460",
"output": "YES\n125636325620\n2133560"
},
{
"input": "10213094404080691512\n64639838 83359",
"output": "YES\n102130944040\n80691512"
},
{
"input": "14525757302059286788\n44151238 152801",
"output": "YES\n145257573020\n59286788"
},
{
"input": "443852406270256089240\n54194433 423288",
"output": "YES\n443852406270\n256089240"
},
{
"input": "6450225349035040017740\n8872387 56607460",
"output": "YES\n64502253490\n35040017740"
},
{
"input": "16375289070073689\n33903290 216",
"output": "NO"
},
{
"input": "3415280033041307294\n15179 79809921",
"output": "NO"
},
{
"input": "4261508098904115227\n52546339 6430",
"output": "NO"
},
{
"input": "15016\n15 16",
"output": "YES\n150\n16"
},
{
"input": "120007\n120 7",
"output": "YES\n12000\n7"
},
{
"input": "23\n2 3",
"output": "YES\n2\n3"
}
] | 1,688,248,183 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 35 | 1,000 | 3,072,000 | s=input()
a,b=input().split()
l1=[]
l2=[]
ss=s
ans=s
ans1,ans2="",""
flag=False
while s:
if(int(s)%int(a)==0 and s==s.lstrip('0')
and len(ss[len(s):])>0
and int(ss[len(s):])%int(b)==0
and ss[len(s):]==ss[len(s):].lstrip('0')
and s+ss[len(s):]==ans):
ans1=s
ans2=ss[len(s):]
flag=True
break
s=s[:-1]
else:
s=s[:-1]
if(flag):
print("YES")
print(ans1)
print(ans2)
else:
print("NO")
| Title: Hacking Cypher
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus participates in a competition for hacking into a new secure messenger. He's almost won.
Having carefully studied the interaction protocol, Polycarpus came to the conclusion that the secret key can be obtained if he properly cuts the public key of the application into two parts. The public key is a long integer which may consist of even a million digits!
Polycarpus needs to find such a way to cut the public key into two nonempty parts, that the first (left) part is divisible by *a* as a separate number, and the second (right) part is divisible by *b* as a separate number. Both parts should be positive integers that have no leading zeros. Polycarpus knows values *a* and *b*.
Help Polycarpus and find any suitable method to cut the public key.
Input Specification:
The first line of the input contains the public key of the messenger — an integer without leading zeroes, its length is in range from 1 to 106 digits. The second line contains a pair of space-separated positive integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=108).
Output Specification:
In the first line print "YES" (without the quotes), if the method satisfying conditions above exists. In this case, next print two lines — the left and right parts after the cut. These two parts, being concatenated, must be exactly identical to the public key. The left part must be divisible by *a*, and the right part must be divisible by *b*. The two parts must be positive integers having no leading zeros. If there are several answers, print any of them.
If there is no answer, print in a single line "NO" (without the quotes).
Demo Input:
['116401024\n97 1024\n', '284254589153928171911281811000\n1009 1000\n', '120\n12 1\n']
Demo Output:
['YES\n11640\n1024\n', 'YES\n2842545891539\n28171911281811000\n', 'NO\n']
Note:
none | ```python
s=input()
a,b=input().split()
l1=[]
l2=[]
ss=s
ans=s
ans1,ans2="",""
flag=False
while s:
if(int(s)%int(a)==0 and s==s.lstrip('0')
and len(ss[len(s):])>0
and int(ss[len(s):])%int(b)==0
and ss[len(s):]==ss[len(s):].lstrip('0')
and s+ss[len(s):]==ans):
ans1=s
ans2=ss[len(s):]
flag=True
break
s=s[:-1]
else:
s=s[:-1]
if(flag):
print("YES")
print(ans1)
print(ans2)
else:
print("NO")
``` | 0 |
|
845 | C | Two TVs | PROGRAMMING | 1,500 | [
"data structures",
"greedy",
"sortings"
] | null | null | Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains *n* shows, *i*-th of them starts at moment *l**i* and ends at moment *r**i*.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all *n* shows. Are two TVs enough to do so? | The first line contains one integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of shows.
Each of the next *n* lines contains two integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=109) — starting and ending time of *i*-th show. | If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes). | [
"3\n1 2\n2 3\n4 5\n",
"4\n1 2\n2 3\n2 3\n1 2\n"
] | [
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "3\n1 2\n2 3\n4 5",
"output": "YES"
},
{
"input": "4\n1 2\n2 3\n2 3\n1 2",
"output": "NO"
},
{
"input": "4\n0 1\n1 2\n2 3\n3 4",
"output": "YES"
},
{
"input": "3\n1 2\n2 3\n2 4",
"output": "NO"
},
{
"input": "3\n0 100\n0 100\n0 100",
"output": "NO"
},
{
"input": "1\n0 1000000000",
"output": "YES"
},
{
"input": "2\n0 1\n0 1",
"output": "YES"
},
{
"input": "3\n2 3\n4 5\n1 6",
"output": "YES"
},
{
"input": "5\n1 3\n1 4\n4 10\n5 8\n9 11",
"output": "YES"
},
{
"input": "3\n1 2\n1 2\n2 3",
"output": "NO"
},
{
"input": "4\n1 100\n10 15\n20 25\n30 35",
"output": "YES"
},
{
"input": "3\n1 8\n6 7\n8 11",
"output": "YES"
},
{
"input": "5\n1 2\n3 5\n4 7\n8 9\n5 10",
"output": "NO"
},
{
"input": "4\n1 7\n2 3\n4 5\n6 7",
"output": "YES"
},
{
"input": "4\n1 100\n50 51\n60 90\n51 52",
"output": "NO"
},
{
"input": "3\n1 10\n2 9\n3 8",
"output": "NO"
},
{
"input": "2\n0 4\n0 4",
"output": "YES"
},
{
"input": "2\n0 2\n0 6",
"output": "YES"
},
{
"input": "5\n3 4\n21 26\n12 17\n9 14\n15 16",
"output": "YES"
},
{
"input": "5\n1 4\n13 15\n11 12\n9 15\n2 5",
"output": "YES"
},
{
"input": "4\n16 19\n9 14\n14 15\n15 19",
"output": "YES"
},
{
"input": "5\n16 19\n23 29\n3 8\n23 26\n22 23",
"output": "NO"
},
{
"input": "5\n19 23\n12 17\n16 21\n20 23\n8 10",
"output": "NO"
},
{
"input": "5\n8 10\n4 10\n3 4\n14 15\n17 19",
"output": "YES"
},
{
"input": "3\n2 8\n5 7\n6 7",
"output": "NO"
},
{
"input": "5\n10 12\n4 6\n21 24\n9 12\n7 13",
"output": "NO"
},
{
"input": "5\n0 3\n14 16\n6 8\n5 9\n9 15",
"output": "YES"
},
{
"input": "5\n6 12\n23 25\n6 7\n19 25\n10 11",
"output": "YES"
},
{
"input": "5\n15 18\n23 24\n23 28\n22 24\n15 19",
"output": "NO"
},
{
"input": "4\n1 8\n8 9\n5 7\n1 4",
"output": "YES"
},
{
"input": "3\n6 10\n1 9\n2 5",
"output": "YES"
},
{
"input": "3\n1 8\n5 6\n6 9",
"output": "NO"
},
{
"input": "4\n2 3\n5 9\n8 10\n9 10",
"output": "NO"
},
{
"input": "4\n0 8\n6 7\n5 9\n1 4",
"output": "NO"
},
{
"input": "3\n6 9\n0 1\n0 2",
"output": "YES"
},
{
"input": "5\n0 6\n21 25\n18 19\n0 3\n6 12",
"output": "YES"
},
{
"input": "4\n1 5\n6 9\n4 8\n1 3",
"output": "YES"
},
{
"input": "2\n2 5\n0 5",
"output": "YES"
},
{
"input": "4\n5 8\n11 15\n3 7\n10 14",
"output": "YES"
},
{
"input": "3\n12 14\n0 4\n2 3",
"output": "YES"
},
{
"input": "4\n4 10\n0 1\n2 10\n0 5",
"output": "NO"
},
{
"input": "4\n0 3\n0 1\n2 4\n2 5",
"output": "NO"
}
] | 1,544,719,862 | 2,147,483,647 | PyPy 3 | OK | TESTS | 58 | 1,028 | 20,377,600 | from sys import stdin
n=int(stdin.readline().strip())
s=[tuple(map(int,stdin.readline().strip().split())) for i in range(n)]
s.sort()
t1=-1,-1
t2=-1,-1
for i in s:
if i[0]>t1[1]:
t1=i
elif i[0]>t2[1]:
t2=i
else:
print("NO")
exit(0)
print("YES")
| Title: Two TVs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains *n* shows, *i*-th of them starts at moment *l**i* and ends at moment *r**i*.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all *n* shows. Are two TVs enough to do so?
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of shows.
Each of the next *n* lines contains two integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=109) — starting and ending time of *i*-th show.
Output Specification:
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
Demo Input:
['3\n1 2\n2 3\n4 5\n', '4\n1 2\n2 3\n2 3\n1 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
from sys import stdin
n=int(stdin.readline().strip())
s=[tuple(map(int,stdin.readline().strip().split())) for i in range(n)]
s.sort()
t1=-1,-1
t2=-1,-1
for i in s:
if i[0]>t1[1]:
t1=i
elif i[0]>t2[1]:
t2=i
else:
print("NO")
exit(0)
print("YES")
``` | 3 |
|
701 | C | They Are Everywhere | PROGRAMMING | 1,500 | [
"binary search",
"strings",
"two pointers"
] | null | null | Sergei B., the young coach of Pokemons, has found the big house which consists of *n* flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number *n* is only connected with the flat number *n*<=-<=1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit. | The first line contains the integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of flats in the house.
The second line contains the row *s* with the length *n*, it consists of uppercase and lowercase letters of English alphabet, the *i*-th letter equals the type of Pokemon, which is in the flat number *i*. | Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house. | [
"3\nAaA\n",
"7\nbcAAcbc\n",
"6\naaBCCe\n"
] | [
"2\n",
"3\n",
"5\n"
] | In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.
In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.
In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6. | 1,000 | [
{
"input": "3\nAaA",
"output": "2"
},
{
"input": "7\nbcAAcbc",
"output": "3"
},
{
"input": "6\naaBCCe",
"output": "5"
},
{
"input": "1\nA",
"output": "1"
},
{
"input": "1\ng",
"output": "1"
},
{
"input": "52\nabcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "52"
},
{
"input": "2\nAA",
"output": "1"
},
{
"input": "4\nqqqE",
"output": "2"
},
{
"input": "10\nrrrrroooro",
"output": "2"
},
{
"input": "15\nOCOCCCCiCOCCCOi",
"output": "3"
},
{
"input": "20\nVEVnVVnWnVEVVnEVBEWn",
"output": "5"
},
{
"input": "25\ncpcyPPjPPcPPPPcppPcPpppcP",
"output": "6"
},
{
"input": "30\nsssssAsesssssssssssssessssssss",
"output": "3"
},
{
"input": "35\ngdXdddgddddddddggggXdbgdggdgddddddb",
"output": "4"
},
{
"input": "40\nIgsggIiIggzgigIIiiIIIiIgIggIzgIiiiggggIi",
"output": "9"
},
{
"input": "45\neteeeeeteaattaeetaetteeettoetettteyeteeeotaae",
"output": "9"
},
{
"input": "50\nlUlUllUlUllllUllllUllllUlUlllUlllUlllllUUlllUUlkUl",
"output": "3"
},
{
"input": "55\nAAAAASAAAASAASAAAAAAAAAAAAASAAAAAAAAAAAAAAAASAAAAAAAAAA",
"output": "2"
},
{
"input": "60\nRRRrSRRRRRRRRRSSRRRSRRRRRRRRrRSRRRRRRRRRRRRRRSRRRRRSSRSRrRRR",
"output": "3"
},
{
"input": "65\nhhMhMhhhhhhhhhhhMhhMMMhhhhBhhhhMhhhhMhhhhhMhhhBhhhhhhhhhhBhhhhhhh",
"output": "5"
},
{
"input": "70\nwAwwwAwwwwwwwwwwwwwwAwAAwwAwwwwwwwwAwAAAwAAwwwwwwwwwAwwwwwwwwwwwwAAwww",
"output": "2"
},
{
"input": "75\niiiXXiiyiiiXyXiiyXiiXiiiiiiXXyiiiiXXiiXiiXifiXiXXiifiiiiiiXfXiyiXXiXiiiiXiX",
"output": "4"
},
{
"input": "80\nSrSrrrrrrrrrrrrrrSSSrrrrrrSrrrrSrrrrrrrrrrSSrrrrrrrrrrrSrrrSrrrrSrrrrSrrrrSSrSSr",
"output": "2"
},
{
"input": "85\nwkMMMwMMkMMMMMMMkkkkMMMMzkkMMwMMkkwMkMwkMMkMMwwMzMMMkkMwwMMMMMMkMMkMzMMMkMMkwMkMMMkMM",
"output": "6"
},
{
"input": "90\nZllZZZyZlZlllZlylllZlllZZllllllllZZllllllllllyylZZyvZvZlllZZlZllZlZlllZyllZyZlllZlllllllZl",
"output": "5"
},
{
"input": "95\nEmuBuEBmmEBBElBlElmmBEmmmEmmEuBEEmummmEmBBBBEWBBBmEEBmmummBBmmlluBBmElmEBEmBmBBmBmuLmEBBmlEBmBu",
"output": "39"
},
{
"input": "100\nfAfhfNNNhfffAfNffNANhffNffffffNfffffhNfffNfffNfffNfNNNffNfffNfhNNNffffNfNfffAffNffNffNNfhfffNhNffNNN",
"output": "4"
},
{
"input": "7\nabcbbab",
"output": "3"
},
{
"input": "65\nkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkjjjjkkkkkkkkkkkkkkkkkkkllllllljk",
"output": "3"
},
{
"input": "9\nbcdddbddc",
"output": "3"
},
{
"input": "71\nlqcrdopdpfagcilpcmheqhjbmnciqmeoabflrenbkrsaabaanpehdmpqcdqnljcrsplnlco",
"output": "36"
},
{
"input": "8\nabbbbbbc",
"output": "8"
}
] | 1,679,629,090 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 11 | 342 | 409,600 | # Implementé el hint, y si lo logro es gracias al análisis de MF
num_flats = int(input())
flats = input()
distinct_pokemon = set(flats)
num_pokemon = len(distinct_pokemon)
'''
El siguiente código es una implementación de:
Hay varias formas de hacerlo, una de las más simple consiste en mantener un rango
en que están presentes todos los pokemons para cada posición final.
Empezando de la primera posición en que estén todos los pokemons hacia la izquierda,
si avanzamos el borde derecho en una posición podemos mover el izquierdo
mientras la ocurrencia del pokemon en el borde izquierdo dentro del rango sea mayor a 1.
La respuesta final será el mínimo de estos rangos.
- Página del curso.'''
left = 0
initial_right = 0
for pokemon in distinct_pokemon:
initial_right = max(initial_right, flats.index(pokemon))
initial_right += 1
step_size = initial_right//num_pokemon
# If this doesn't work I'm going to be very sad :(
while True:
pokemon_in_range = set(flats[left + step_size:initial_right])
if len(pokemon_in_range) == num_pokemon:
left += step_size
else:
step_size = step_size // 2
if step_size == 0:
break
#print(f"Left: {left}, Step size: {step_size}, Right: {initial_right}")
min_range_length = initial_right - left
# All distinct pokemon are included in the first range
for right in range(initial_right, num_flats):
# More "efficient" uglier code
while flats[left] in flats[left+1:right]:
left += 1
min_range_length = min(min_range_length, right - left)
if min_range_length == num_pokemon:
break
print(min_range_length)
| Title: They Are Everywhere
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sergei B., the young coach of Pokemons, has found the big house which consists of *n* flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number *n* is only connected with the flat number *n*<=-<=1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of flats in the house.
The second line contains the row *s* with the length *n*, it consists of uppercase and lowercase letters of English alphabet, the *i*-th letter equals the type of Pokemon, which is in the flat number *i*.
Output Specification:
Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.
Demo Input:
['3\nAaA\n', '7\nbcAAcbc\n', '6\naaBCCe\n']
Demo Output:
['2\n', '3\n', '5\n']
Note:
In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.
In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.
In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6. | ```python
# Implementé el hint, y si lo logro es gracias al análisis de MF
num_flats = int(input())
flats = input()
distinct_pokemon = set(flats)
num_pokemon = len(distinct_pokemon)
'''
El siguiente código es una implementación de:
Hay varias formas de hacerlo, una de las más simple consiste en mantener un rango
en que están presentes todos los pokemons para cada posición final.
Empezando de la primera posición en que estén todos los pokemons hacia la izquierda,
si avanzamos el borde derecho en una posición podemos mover el izquierdo
mientras la ocurrencia del pokemon en el borde izquierdo dentro del rango sea mayor a 1.
La respuesta final será el mínimo de estos rangos.
- Página del curso.'''
left = 0
initial_right = 0
for pokemon in distinct_pokemon:
initial_right = max(initial_right, flats.index(pokemon))
initial_right += 1
step_size = initial_right//num_pokemon
# If this doesn't work I'm going to be very sad :(
while True:
pokemon_in_range = set(flats[left + step_size:initial_right])
if len(pokemon_in_range) == num_pokemon:
left += step_size
else:
step_size = step_size // 2
if step_size == 0:
break
#print(f"Left: {left}, Step size: {step_size}, Right: {initial_right}")
min_range_length = initial_right - left
# All distinct pokemon are included in the first range
for right in range(initial_right, num_flats):
# More "efficient" uglier code
while flats[left] in flats[left+1:right]:
left += 1
min_range_length = min(min_range_length, right - left)
if min_range_length == num_pokemon:
break
print(min_range_length)
``` | 0 |
|
818 | A | Diplomas and Certificates | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners. | The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas. | Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners. | [
"18 2\n",
"9 10\n",
"1000000000000 5\n",
"1000000000000 499999999999\n"
] | [
"3 6 9\n",
"0 0 9\n",
"83333333333 416666666665 500000000002\n",
"1 499999999999 500000000000\n"
] | none | 0 | [
{
"input": "18 2",
"output": "3 6 9"
},
{
"input": "9 10",
"output": "0 0 9"
},
{
"input": "1000000000000 5",
"output": "83333333333 416666666665 500000000002"
},
{
"input": "1000000000000 499999999999",
"output": "1 499999999999 500000000000"
},
{
"input": "1 1",
"output": "0 0 1"
},
{
"input": "5 3",
"output": "0 0 5"
},
{
"input": "42 6",
"output": "3 18 21"
},
{
"input": "1000000000000 1000",
"output": "499500499 499500499000 500000000501"
},
{
"input": "999999999999 999999",
"output": "499999 499998500001 500000999999"
},
{
"input": "732577309725 132613",
"output": "2762066 366285858458 366288689201"
},
{
"input": "152326362626 15",
"output": "4760198832 71402982480 76163181314"
},
{
"input": "2 1",
"output": "0 0 2"
},
{
"input": "1000000000000 500000000000",
"output": "0 0 1000000000000"
},
{
"input": "100000000000 50000000011",
"output": "0 0 100000000000"
},
{
"input": "1000000000000 32416187567",
"output": "15 486242813505 513757186480"
},
{
"input": "1000000000000 7777777777",
"output": "64 497777777728 502222222208"
},
{
"input": "1000000000000 77777777777",
"output": "6 466666666662 533333333332"
},
{
"input": "100000000000 578485652",
"output": "86 49749766072 50250233842"
},
{
"input": "999999999999 10000000000",
"output": "49 490000000000 509999999950"
},
{
"input": "7 2",
"output": "1 2 4"
},
{
"input": "420506530901 752346673804",
"output": "0 0 420506530901"
},
{
"input": "960375521135 321688347872",
"output": "1 321688347872 638687173262"
},
{
"input": "1000000000000 1000000000000",
"output": "0 0 1000000000000"
},
{
"input": "99999999999 15253636363",
"output": "3 45760909089 54239090907"
},
{
"input": "19 2",
"output": "3 6 10"
},
{
"input": "999999999999 1000000000000",
"output": "0 0 999999999999"
},
{
"input": "1000000000000 5915587276",
"output": "84 496909331184 503090668732"
},
{
"input": "1000000000000 1000000006",
"output": "499 499000002994 500999996507"
},
{
"input": "549755813888 134217728",
"output": "2047 274743689216 275012122625"
},
{
"input": "99999999999 3333333",
"output": "14999 49996661667 50003323333"
},
{
"input": "9 1",
"output": "2 2 5"
},
{
"input": "1000000000000 250000000001",
"output": "1 250000000001 749999999998"
},
{
"input": "5 1",
"output": "1 1 3"
},
{
"input": "3107038133 596040207",
"output": "2 1192080414 1914957717"
},
{
"input": "1000000000000 73786977",
"output": "6776 499980556152 500019437072"
},
{
"input": "1000000000000 73786976",
"output": "6776 499980549376 500019443848"
},
{
"input": "1000000000000 25000000000",
"output": "19 475000000000 524999999981"
},
{
"input": "216929598879 768233755932",
"output": "0 0 216929598879"
},
{
"input": "1000000000000 250000000000",
"output": "1 250000000000 749999999999"
},
{
"input": "1000000000000 100000000001",
"output": "4 400000000004 599999999992"
},
{
"input": "100000000000 100000000001",
"output": "0 0 100000000000"
},
{
"input": "900000000000 100281800001",
"output": "4 401127200004 498872799992"
},
{
"input": "906028900004 109123020071",
"output": "4 436492080284 469536819716"
},
{
"input": "1000000000000 1",
"output": "250000000000 250000000000 500000000000"
}
] | 1,662,705,238 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 46 | 0 | def solution(n, k):
d = (n//2)//(k+1)
if d > 0:
return [d, k*d, n - (k+1)*d]
else:
return [0, 0, n]
n, k = [int(x) for x in input().strip().split(" ")]
ans = solution(n, k)
print(ans[0], ans[1], ans[2]) | Title: Diplomas and Certificates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners.
Input Specification:
The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas.
Output Specification:
Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners.
Demo Input:
['18 2\n', '9 10\n', '1000000000000 5\n', '1000000000000 499999999999\n']
Demo Output:
['3 6 9\n', '0 0 9\n', '83333333333 416666666665 500000000002\n', '1 499999999999 500000000000\n']
Note:
none | ```python
def solution(n, k):
d = (n//2)//(k+1)
if d > 0:
return [d, k*d, n - (k+1)*d]
else:
return [0, 0, n]
n, k = [int(x) for x in input().strip().split(" ")]
ans = solution(n, k)
print(ans[0], ans[1], ans[2])
``` | 3 |
|
796 | A | Buying A House | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.
The girl lives in house *m* of a village. There are *n* houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house *n*. The village is also well-structured: house *i* and house *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.
You will be given *n* integers *a*1,<=*a*2,<=...,<=*a**n* that denote the availability and the prices of the houses. If house *i* is occupied, and therefore cannot be bought, then *a**i* equals 0. Otherwise, house *i* can be bought, and *a**i* represents the money required to buy it, in dollars.
As Zane has only *k* dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love. | The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — denoting the availability and the prices of the houses.
It is guaranteed that *a**m*<==<=0 and that it is possible to purchase some house with no more than *k* dollars. | Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy. | [
"5 1 20\n0 27 32 21 19\n",
"7 3 50\n62 0 0 0 99 33 22\n",
"10 5 100\n1 0 1 0 0 0 0 0 1 1\n"
] | [
"40",
"30",
"20"
] | In the first sample, with *k* = 20 dollars, Zane can buy only house 5. The distance from house *m* = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.
In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house *m* = 3 and house 6 are only 30 meters away, while house *m* = 3 and house 7 are 40 meters away. | 500 | [
{
"input": "5 1 20\n0 27 32 21 19",
"output": "40"
},
{
"input": "7 3 50\n62 0 0 0 99 33 22",
"output": "30"
},
{
"input": "10 5 100\n1 0 1 0 0 0 0 0 1 1",
"output": "20"
},
{
"input": "5 3 1\n1 1 0 0 1",
"output": "10"
},
{
"input": "5 5 5\n1 0 5 6 0",
"output": "20"
},
{
"input": "15 10 50\n20 0 49 50 50 50 50 50 50 0 50 50 49 0 20",
"output": "10"
},
{
"input": "7 5 1\n0 100 2 2 0 2 1",
"output": "20"
},
{
"input": "100 50 100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 0 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "10"
},
{
"input": "100 50 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 0 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "490"
},
{
"input": "100 77 50\n50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 0 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0",
"output": "10"
},
{
"input": "100 1 1\n0 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0",
"output": "980"
},
{
"input": "100 1 100\n0 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "10"
},
{
"input": "100 10 99\n0 0 0 0 0 0 0 0 0 0 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 98",
"output": "890"
},
{
"input": "7 4 5\n1 0 6 0 5 6 0",
"output": "10"
},
{
"input": "7 4 5\n1 6 5 0 0 6 0",
"output": "10"
},
{
"input": "100 42 59\n50 50 50 50 50 50 50 50 50 50 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 60 60 60 60 60 60 60 60 0 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 0",
"output": "90"
},
{
"input": "2 1 100\n0 1",
"output": "10"
},
{
"input": "2 2 100\n1 0",
"output": "10"
},
{
"input": "10 1 88\n0 95 0 0 0 0 0 94 0 85",
"output": "90"
},
{
"input": "10 2 14\n2 0 1 26 77 39 41 100 13 32",
"output": "10"
},
{
"input": "10 3 11\n0 0 0 0 0 62 0 52 1 35",
"output": "60"
},
{
"input": "20 12 44\n27 40 58 69 53 38 31 39 75 95 8 0 28 81 77 90 38 61 21 88",
"output": "10"
},
{
"input": "30 29 10\n59 79 34 12 100 6 1 58 18 73 54 11 37 46 89 90 80 85 73 45 64 5 31 0 89 19 0 74 0 82",
"output": "70"
},
{
"input": "40 22 1\n7 95 44 53 0 0 19 93 0 68 65 0 24 91 10 58 17 0 71 0 100 0 94 90 79 73 0 73 4 61 54 81 7 13 21 84 5 41 0 1",
"output": "180"
},
{
"input": "40 22 99\n60 0 100 0 0 100 100 0 0 0 0 100 100 0 0 100 100 0 100 100 100 0 100 100 100 0 100 100 0 0 100 100 100 0 0 100 0 100 0 0",
"output": "210"
},
{
"input": "50 10 82\n56 54 0 0 0 0 88 93 0 0 83 93 0 0 91 89 0 30 62 52 24 84 80 8 38 13 92 78 16 87 23 30 71 55 16 63 15 99 4 93 24 6 3 35 4 42 73 27 86 37",
"output": "80"
},
{
"input": "63 49 22\n18 3 97 52 75 2 12 24 58 75 80 97 22 10 79 51 30 60 68 99 75 2 35 3 97 88 9 7 18 5 0 0 0 91 0 91 56 36 76 0 0 0 52 27 35 0 51 72 0 96 57 0 0 0 0 92 55 28 0 30 0 78 77",
"output": "190"
},
{
"input": "74 38 51\n53 36 55 42 64 5 87 9 0 16 86 78 9 22 19 1 25 72 1 0 0 0 79 0 0 0 77 58 70 0 0 100 64 0 99 59 0 0 0 0 65 74 0 96 0 58 89 93 61 88 0 0 82 89 0 0 49 24 7 77 89 87 94 61 100 31 93 70 39 49 39 14 20 84",
"output": "190"
},
{
"input": "89 22 11\n36 0 68 89 0 85 72 0 38 56 0 44 0 94 0 28 71 0 0 18 0 0 0 89 0 0 0 75 0 0 0 32 66 0 0 0 0 0 0 48 63 0 64 58 0 23 48 0 0 52 93 61 57 0 18 0 0 34 62 17 0 41 0 0 53 59 44 0 0 51 40 0 0 100 100 54 0 88 0 5 45 56 57 67 24 16 88 86 15",
"output": "580"
},
{
"input": "97 44 100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 51 19",
"output": "520"
},
{
"input": "100 1 1\n0 0 0 0 10 54 84 6 17 94 65 82 34 0 61 46 42 0 2 16 56 0 100 0 82 0 0 0 89 78 96 56 0 0 0 0 0 0 0 0 77 70 0 96 67 0 0 32 44 1 72 50 14 11 24 61 100 64 19 5 67 69 44 82 93 22 67 93 22 61 53 64 79 41 84 48 43 97 7 24 8 49 23 16 72 52 97 29 69 47 29 49 64 91 4 73 17 18 51 67",
"output": "490"
},
{
"input": "100 1 50\n0 0 0 60 0 0 54 0 80 0 0 0 97 0 68 97 84 0 0 93 0 0 0 0 68 0 0 62 0 0 55 68 65 87 0 69 0 0 0 0 0 52 61 100 0 71 0 82 88 78 0 81 0 95 0 57 0 67 0 0 0 55 86 0 60 72 0 0 73 0 83 0 0 60 64 0 56 0 0 77 84 0 58 63 84 0 0 67 0 16 3 88 0 98 31 52 40 35 85 23",
"output": "890"
},
{
"input": "100 1 100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 91 70 14",
"output": "970"
},
{
"input": "100 1 29\n0 0 0 0 64 0 89 97 0 0 0 59 0 67 62 0 59 0 0 80 0 0 0 0 0 97 0 57 0 64 32 0 44 0 0 48 0 47 38 0 42 0 0 0 0 0 0 46 74 0 86 33 33 0 44 0 79 0 0 0 0 91 59 0 59 65 55 0 0 58 33 95 0 97 76 0 81 0 41 0 38 81 80 0 85 0 31 0 0 92 0 0 45 96 0 85 91 87 0 10",
"output": "990"
},
{
"input": "100 50 20\n3 0 32 0 48 32 64 0 54 26 0 0 0 0 0 28 0 0 54 0 0 45 49 0 38 74 0 0 39 42 62 48 75 96 89 42 0 44 0 0 30 21 76 0 50 0 79 0 0 0 0 99 0 84 62 0 0 0 0 53 80 0 28 0 0 53 0 0 38 0 62 0 0 62 0 0 88 0 44 32 0 81 35 45 49 0 69 73 38 27 72 0 96 72 69 0 0 22 76 10",
"output": "490"
},
{
"input": "100 50 20\n49 0 56 0 87 25 40 0 50 0 0 97 0 0 36 29 0 0 0 0 0 73 29 71 44 0 0 0 91 92 69 0 0 60 81 49 48 38 0 87 0 82 0 32 0 82 46 39 0 0 29 0 0 29 0 79 47 0 0 0 0 0 49 0 24 33 70 0 63 45 97 90 0 0 29 53 55 0 84 0 0 100 26 0 88 0 0 0 0 81 70 0 30 80 0 75 59 98 0 2",
"output": "500"
},
{
"input": "100 2 2\n0 0 43 90 47 5 2 97 52 69 21 48 64 10 34 97 97 74 8 19 68 56 55 24 47 38 43 73 72 72 60 60 51 36 33 44 100 45 13 54 72 52 0 15 3 6 50 8 88 4 78 26 40 27 30 63 67 83 61 91 33 97 54 20 92 27 89 35 10 7 84 50 11 95 74 88 24 44 74 100 18 56 34 91 41 34 51 51 11 91 89 54 19 100 83 89 10 17 76 20",
"output": "50"
},
{
"input": "100 100 34\n5 73 0 0 44 0 0 0 79 55 0 0 0 0 0 0 0 0 83 67 75 0 0 0 0 59 0 74 0 0 47 98 0 0 72 41 0 55 87 0 0 78 84 0 0 39 0 79 72 95 0 0 0 0 0 85 53 84 0 0 0 0 37 75 0 66 0 0 0 0 61 0 70 0 37 60 42 78 92 52 0 0 0 55 77 57 0 63 37 0 0 0 96 70 0 94 97 0 0 0",
"output": "990"
},
{
"input": "100 100 100\n43 79 21 87 84 14 28 69 92 16 3 71 79 37 48 37 72 58 12 72 62 49 37 17 60 54 41 99 15 72 40 89 76 1 99 87 14 56 63 48 69 37 96 64 7 14 1 73 85 33 98 70 97 71 96 28 49 71 56 2 67 22 100 2 98 100 62 77 92 76 98 98 47 26 22 47 50 56 9 16 72 47 5 62 29 78 81 1 0 63 32 65 87 3 40 53 8 80 93 0",
"output": "10"
},
{
"input": "100 38 1\n3 59 12 81 33 95 0 41 36 17 63 76 42 77 85 56 3 96 55 41 24 87 18 9 0 37 0 61 69 0 0 0 67 0 0 0 0 0 0 18 0 0 47 56 74 0 0 80 0 42 0 1 60 59 62 9 19 87 92 48 58 30 98 51 99 10 42 94 51 53 50 89 24 5 52 82 50 39 98 8 95 4 57 21 10 0 44 32 19 14 64 34 79 76 17 3 15 22 71 51",
"output": "140"
},
{
"input": "100 72 1\n56 98 8 27 9 23 16 76 56 1 34 43 96 73 75 49 62 20 18 23 51 55 30 84 4 20 89 40 75 16 69 35 1 0 16 0 80 0 41 17 0 0 76 23 0 92 0 34 0 91 82 54 0 0 0 63 85 59 98 24 29 0 8 77 26 0 34 95 39 0 0 0 74 0 0 0 0 12 0 92 0 0 55 95 66 30 0 0 29 98 0 0 0 47 0 0 80 0 0 4",
"output": "390"
},
{
"input": "100 66 1\n38 50 64 91 37 44 74 21 14 41 80 90 26 51 78 85 80 86 44 14 49 75 93 48 78 89 23 72 35 22 14 48 100 71 62 22 7 95 80 66 32 20 17 47 79 30 41 52 15 62 67 71 1 6 0 9 0 0 0 11 0 0 24 0 31 0 77 0 51 0 0 0 0 0 0 77 0 36 44 19 90 45 6 25 100 87 93 30 4 97 36 88 33 50 26 71 97 71 51 68",
"output": "130"
},
{
"input": "100 55 1\n0 33 45 83 56 96 58 24 45 30 38 60 39 69 21 87 59 21 72 73 27 46 61 61 11 97 77 5 39 3 3 35 76 37 53 84 24 75 9 48 31 90 100 84 74 81 83 83 42 23 29 94 18 1 0 53 52 99 86 37 94 54 28 75 28 80 17 14 98 68 76 20 32 23 42 31 57 79 60 14 18 27 1 98 32 3 96 25 15 38 2 6 3 28 59 54 63 2 43 59",
"output": "10"
},
{
"input": "100 55 1\n24 52 41 6 55 11 58 25 63 12 70 39 23 28 72 17 96 85 7 84 21 13 34 37 97 43 36 32 15 30 58 5 14 71 40 70 9 92 44 73 31 58 96 90 19 35 29 91 25 36 48 95 61 78 0 1 99 61 81 88 42 53 61 57 42 55 74 45 41 92 99 30 20 25 89 50 37 4 17 24 6 65 15 44 40 2 38 43 7 90 38 59 75 87 96 28 12 67 24 32",
"output": "10"
},
{
"input": "100 21 1\n62 5 97 80 81 28 83 0 26 0 0 0 0 23 0 0 90 0 0 0 0 0 0 0 0 54 71 8 0 0 42 0 73 0 17 0 1 31 71 78 58 72 84 39 54 59 13 29 16 41 71 35 88 55 70 50 33 100 100 60 52 90 7 66 44 55 51 42 90 17 86 44 46 8 52 74 8 22 2 92 34 37 58 98 70 74 19 91 74 25 4 38 71 68 50 68 63 14 60 98",
"output": "160"
},
{
"input": "5 2 20\n27 0 32 21 19",
"output": "30"
},
{
"input": "6 4 10\n10 0 0 0 0 10",
"output": "20"
},
{
"input": "8 7 100\n1 0 0 0 0 0 0 1",
"output": "10"
},
{
"input": "5 3 20\n1 21 0 0 1",
"output": "20"
},
{
"input": "4 3 1\n0 0 0 1",
"output": "10"
},
{
"input": "5 2 3\n4 0 5 6 1",
"output": "30"
},
{
"input": "5 3 87\n88 89 0 1 90",
"output": "10"
},
{
"input": "5 3 20\n15 30 0 15 35",
"output": "10"
},
{
"input": "6 3 50\n0 0 0 1 2 0",
"output": "10"
},
{
"input": "6 4 9\n100 9 10 0 0 9",
"output": "20"
},
{
"input": "5 4 20\n0 20 0 0 20",
"output": "10"
},
{
"input": "6 3 3\n1 5 0 2 2 0",
"output": "10"
},
{
"input": "5 4 100\n0 1 0 0 1",
"output": "10"
}
] | 1,598,440,066 | 2,147,483,647 | Python 3 | OK | TESTS | 58 | 109 | 307,200 | n,m,k=map(int,input().split())
l=list(map(int,input().split()))
x=[]
y=[]
for i in range(n):
if l[i]==0:
continue
elif l[i]<=k:
x.append(i+1)
for i in range(len(x)):
if x[i]>m:
y.append(x[i]-m)
elif x[i]<m:
y.append(m-x[i])
print(min(y)*10) | Title: Buying A House
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.
The girl lives in house *m* of a village. There are *n* houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house *n*. The village is also well-structured: house *i* and house *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.
You will be given *n* integers *a*1,<=*a*2,<=...,<=*a**n* that denote the availability and the prices of the houses. If house *i* is occupied, and therefore cannot be bought, then *a**i* equals 0. Otherwise, house *i* can be bought, and *a**i* represents the money required to buy it, in dollars.
As Zane has only *k* dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love.
Input Specification:
The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — denoting the availability and the prices of the houses.
It is guaranteed that *a**m*<==<=0 and that it is possible to purchase some house with no more than *k* dollars.
Output Specification:
Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy.
Demo Input:
['5 1 20\n0 27 32 21 19\n', '7 3 50\n62 0 0 0 99 33 22\n', '10 5 100\n1 0 1 0 0 0 0 0 1 1\n']
Demo Output:
['40', '30', '20']
Note:
In the first sample, with *k* = 20 dollars, Zane can buy only house 5. The distance from house *m* = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.
In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house *m* = 3 and house 6 are only 30 meters away, while house *m* = 3 and house 7 are 40 meters away. | ```python
n,m,k=map(int,input().split())
l=list(map(int,input().split()))
x=[]
y=[]
for i in range(n):
if l[i]==0:
continue
elif l[i]<=k:
x.append(i+1)
for i in range(len(x)):
if x[i]>m:
y.append(x[i]-m)
elif x[i]<m:
y.append(m-x[i])
print(min(y)*10)
``` | 3 |
|
686 | A | Free Ice Cream | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids. | The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place. | Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress. | [
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] | [
"22 1\n",
"3 2\n"
] | Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | 500 | [
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
"input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46",
"output": "46 0"
},
{
"input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100",
"output": "700 0"
},
{
"input": "1 0\n- 526403222",
"output": "0 1"
},
{
"input": "1 897986543\n- 371188251",
"output": "526798292 0"
},
{
"input": "1 0\n+ 1",
"output": "1 0"
},
{
"input": "1 0\n- 1",
"output": "0 1"
},
{
"input": "1 10\n+ 10",
"output": "20 0"
},
{
"input": "1 3\n- 5",
"output": "3 1"
},
{
"input": "1 0\n- 5",
"output": "0 1"
},
{
"input": "1 0\n+ 5",
"output": "5 0"
}
] | 1,631,545,277 | 2,147,483,647 | PyPy 3 | OK | TESTS | 34 | 139 | 21,708,800 | a,b=[int(v) for v in input().split()]
sum=0
for i in range(a):
x,t=[v for v in input().split()]
t=int(t)
if x=="+": b+=t
else:
if b>=t: b-=t
else: sum+=1
print(b,sum) | Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | ```python
a,b=[int(v) for v in input().split()]
sum=0
for i in range(a):
x,t=[v for v in input().split()]
t=int(t)
if x=="+": b+=t
else:
if b>=t: b-=t
else: sum+=1
print(b,sum)
``` | 3 |
|
779 | B | Weird Rounding | PROGRAMMING | 1,100 | [
"brute force",
"greedy"
] | null | null | Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10*k*.
In the given number of *n* Polycarp wants to remove the least number of digits to get a number that is divisible by 10*k*. For example, if *k*<==<=3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103<==<=1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number *n*, so that the result is divisible by 10*k*. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists. | The only line of the input contains two integer numbers *n* and *k* (0<=≤<=*n*<=≤<=2<=000<=000<=000, 1<=≤<=*k*<=≤<=9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros. | Print *w* — the required minimal number of digits to erase. After removing the appropriate *w* digits from the number *n*, the result should have a value that is divisible by 10*k*. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0). | [
"30020 3\n",
"100 9\n",
"10203049 2\n"
] | [
"1\n",
"2\n",
"3\n"
] | In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number. | 1,000 | [
{
"input": "30020 3",
"output": "1"
},
{
"input": "100 9",
"output": "2"
},
{
"input": "10203049 2",
"output": "3"
},
{
"input": "0 1",
"output": "0"
},
{
"input": "0 9",
"output": "0"
},
{
"input": "100 2",
"output": "0"
},
{
"input": "102030404 2",
"output": "2"
},
{
"input": "1000999999 3",
"output": "6"
},
{
"input": "12000000 4",
"output": "0"
},
{
"input": "1090090090 5",
"output": "2"
},
{
"input": "10 1",
"output": "0"
},
{
"input": "10 2",
"output": "1"
},
{
"input": "10 9",
"output": "1"
},
{
"input": "100 1",
"output": "0"
},
{
"input": "100 3",
"output": "2"
},
{
"input": "101010110 3",
"output": "3"
},
{
"input": "101010110 1",
"output": "0"
},
{
"input": "101010110 2",
"output": "2"
},
{
"input": "101010110 4",
"output": "4"
},
{
"input": "101010110 5",
"output": "8"
},
{
"input": "101010110 9",
"output": "8"
},
{
"input": "1234567890 1",
"output": "0"
},
{
"input": "1234567890 2",
"output": "9"
},
{
"input": "1234567890 9",
"output": "9"
},
{
"input": "2000000000 1",
"output": "0"
},
{
"input": "2000000000 2",
"output": "0"
},
{
"input": "2000000000 3",
"output": "0"
},
{
"input": "2000000000 9",
"output": "0"
},
{
"input": "1010101010 1",
"output": "0"
},
{
"input": "1010101010 2",
"output": "1"
},
{
"input": "1010101010 3",
"output": "2"
},
{
"input": "1010101010 4",
"output": "3"
},
{
"input": "1010101010 5",
"output": "4"
},
{
"input": "1010101010 6",
"output": "9"
},
{
"input": "1010101010 7",
"output": "9"
},
{
"input": "1010101010 8",
"output": "9"
},
{
"input": "1010101010 9",
"output": "9"
},
{
"input": "10001000 1",
"output": "0"
},
{
"input": "10001000 2",
"output": "0"
},
{
"input": "10001000 3",
"output": "0"
},
{
"input": "10001000 4",
"output": "1"
},
{
"input": "10001000 5",
"output": "1"
},
{
"input": "10001000 6",
"output": "1"
},
{
"input": "10001000 7",
"output": "7"
},
{
"input": "10001000 8",
"output": "7"
},
{
"input": "10001000 9",
"output": "7"
},
{
"input": "1000000001 1",
"output": "1"
},
{
"input": "1000000001 2",
"output": "1"
},
{
"input": "1000000001 3",
"output": "1"
},
{
"input": "1000000001 6",
"output": "1"
},
{
"input": "1000000001 7",
"output": "1"
},
{
"input": "1000000001 8",
"output": "1"
},
{
"input": "1000000001 9",
"output": "9"
},
{
"input": "1000 1",
"output": "0"
},
{
"input": "100001100 3",
"output": "2"
},
{
"input": "7057 6",
"output": "3"
},
{
"input": "30000000 5",
"output": "0"
},
{
"input": "470 1",
"output": "0"
},
{
"input": "500500000 4",
"output": "0"
},
{
"input": "2103 8",
"output": "3"
},
{
"input": "600000000 2",
"output": "0"
},
{
"input": "708404442 1",
"output": "4"
},
{
"input": "5000140 6",
"output": "6"
},
{
"input": "1100047 3",
"output": "2"
},
{
"input": "309500 5",
"output": "5"
},
{
"input": "70053160 4",
"output": "7"
},
{
"input": "44000 1",
"output": "0"
},
{
"input": "400370000 3",
"output": "0"
},
{
"input": "5800 6",
"output": "3"
},
{
"input": "20700050 1",
"output": "0"
},
{
"input": "650 1",
"output": "0"
},
{
"input": "320005070 6",
"output": "8"
},
{
"input": "370000 4",
"output": "0"
},
{
"input": "1011 2",
"output": "3"
},
{
"input": "1000111 5",
"output": "6"
},
{
"input": "1001111 5",
"output": "6"
},
{
"input": "99990 3",
"output": "4"
},
{
"input": "10100200 6",
"output": "7"
},
{
"input": "200 3",
"output": "2"
},
{
"input": "103055 3",
"output": "5"
},
{
"input": "1030555 3",
"output": "6"
},
{
"input": "100111 4",
"output": "5"
},
{
"input": "101 2",
"output": "2"
},
{
"input": "1001 3",
"output": "3"
},
{
"input": "100000 6",
"output": "5"
},
{
"input": "1100000 6",
"output": "6"
},
{
"input": "123450 2",
"output": "5"
},
{
"input": "1003 3",
"output": "3"
},
{
"input": "1111100 4",
"output": "6"
},
{
"input": "532415007 8",
"output": "8"
},
{
"input": "801 2",
"output": "2"
},
{
"input": "1230 2",
"output": "3"
},
{
"input": "9900 3",
"output": "3"
},
{
"input": "14540444 2",
"output": "7"
},
{
"input": "11111100 4",
"output": "7"
},
{
"input": "11001 3",
"output": "4"
},
{
"input": "1011110 3",
"output": "6"
},
{
"input": "15450112 2",
"output": "7"
},
{
"input": "2220 3",
"output": "3"
},
{
"input": "90099 3",
"output": "4"
},
{
"input": "10005 4",
"output": "4"
},
{
"input": "1010 3",
"output": "3"
},
{
"input": "444444400 3",
"output": "8"
},
{
"input": "10020 4",
"output": "4"
},
{
"input": "10303 3",
"output": "4"
},
{
"input": "123000 4",
"output": "5"
},
{
"input": "12300 3",
"output": "4"
},
{
"input": "101 1",
"output": "1"
},
{
"input": "500001 8",
"output": "5"
},
{
"input": "121002 3",
"output": "5"
},
{
"input": "10011 3",
"output": "4"
},
{
"input": "505050 4",
"output": "5"
},
{
"input": "1421011 2",
"output": "6"
},
{
"input": "1202022 3",
"output": "6"
},
{
"input": "1000023 7",
"output": "6"
},
{
"input": "110 2",
"output": "2"
},
{
"input": "111000 4",
"output": "5"
},
{
"input": "10340 3",
"output": "4"
},
{
"input": "101 9",
"output": "2"
},
{
"input": "2001 3",
"output": "3"
},
{
"input": "122320 2",
"output": "5"
},
{
"input": "22200 3",
"output": "4"
},
{
"input": "11110 2",
"output": "4"
},
{
"input": "11010 3",
"output": "4"
},
{
"input": "1000002333 6",
"output": "9"
},
{
"input": "101010 4",
"output": "5"
},
{
"input": "210 9",
"output": "2"
},
{
"input": "500555 3",
"output": "5"
},
{
"input": "1110111 3",
"output": "6"
},
{
"input": "1100000000 9",
"output": "9"
},
{
"input": "11000 4",
"output": "4"
},
{
"input": "100 4",
"output": "2"
},
{
"input": "234560 3",
"output": "5"
},
{
"input": "10230 3",
"output": "4"
},
{
"input": "10030234 5",
"output": "7"
},
{
"input": "1200 3",
"output": "3"
},
{
"input": "123400 3",
"output": "5"
},
{
"input": "1034543 4",
"output": "6"
},
{
"input": "10100 4",
"output": "4"
},
{
"input": "10 5",
"output": "1"
},
{
"input": "4501022 3",
"output": "6"
},
{
"input": "12340 2",
"output": "4"
},
{
"input": "30020 4",
"output": "4"
},
{
"input": "1111100 6",
"output": "6"
},
{
"input": "10101 5",
"output": "4"
},
{
"input": "32132100 3",
"output": "7"
},
{
"input": "1000023 6",
"output": "6"
},
{
"input": "12300 4",
"output": "4"
},
{
"input": "78400 3",
"output": "4"
},
{
"input": "10203049 5",
"output": "7"
},
{
"input": "404044 3",
"output": "5"
},
{
"input": "1024 2",
"output": "3"
},
{
"input": "505 2",
"output": "2"
},
{
"input": "20 2",
"output": "1"
},
{
"input": "1111100 3",
"output": "6"
},
{
"input": "1000 9",
"output": "3"
},
{
"input": "3333300 3",
"output": "6"
},
{
"input": "1100 3",
"output": "3"
},
{
"input": "963000 4",
"output": "5"
},
{
"input": "100457 5",
"output": "5"
},
{
"input": "10049 3",
"output": "4"
}
] | 1,488,216,568 | 1,768 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 4,608,000 | a,b=map(int,input().split())
b=10**b
k=a//b
a=a-(k*b)
a=str(a)
v=0
if b>int(a):
v=len(a)-1
else:
for i in range(len(a)):
if int(a[i])>1:
v=v+1
print(v)
| Title: Weird Rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10*k*.
In the given number of *n* Polycarp wants to remove the least number of digits to get a number that is divisible by 10*k*. For example, if *k*<==<=3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103<==<=1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number *n*, so that the result is divisible by 10*k*. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input Specification:
The only line of the input contains two integer numbers *n* and *k* (0<=≤<=*n*<=≤<=2<=000<=000<=000, 1<=≤<=*k*<=≤<=9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output Specification:
Print *w* — the required minimal number of digits to erase. After removing the appropriate *w* digits from the number *n*, the result should have a value that is divisible by 10*k*. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Demo Input:
['30020 3\n', '100 9\n', '10203049 2\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number. | ```python
a,b=map(int,input().split())
b=10**b
k=a//b
a=a-(k*b)
a=str(a)
v=0
if b>int(a):
v=len(a)-1
else:
for i in range(len(a)):
if int(a[i])>1:
v=v+1
print(v)
``` | 0 |
|
381 | A | Sereja and Dima | PROGRAMMING | 800 | [
"greedy",
"implementation",
"two pointers"
] | null | null | Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. | On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. | [
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] | [
"12 5\n",
"16 12\n"
] | In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | 500 | [
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24",
"output": "644 500"
},
{
"input": "1\n3",
"output": "3 0"
},
{
"input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646",
"output": "6848 6568"
},
{
"input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727",
"output": "9562 9561"
},
{
"input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12",
"output": "315 315"
},
{
"input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304",
"output": "3238 2222"
},
{
"input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325",
"output": "5246 4864"
},
{
"input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71",
"output": "8147 7807"
},
{
"input": "1\n1",
"output": "1 0"
}
] | 1,673,722,238 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 46 | 0 | n = int(input())
l = input().split()
for i in range(len(l)):
l[i] = int(l[i])
sereja_score = 0
dima_score = 0
for i in range(len(l)):
if i % 2 == 0:
l2 = [l[0], l[-1]]
l1 = max(l2)
l.remove((l1))
sereja_score += l1
elif i % 2 == 1:
l4 = [l[0], l[-1]]
l3 = max(l4)
l.remove(l3)
dima_score += l3
print(sereja_score, dima_score)
# def score_game(n, l):
# for i in range(len(l)):
# l[i] = int(l[i])
#
# sereja_score = 0
# dima_score = 0
#
# for i in range(len(l)):
# if (i == 0) or (i % 2 == 0):
# l1 = max(l)
# l.remove((l1))
# sereja_score += l1
# else:
# l3 = max(l)
# l.remove(l3)
# dima_score += l3
#
# print(sereja_score, dima_score)
#
# n = int(input())
# l = input().split()
# score_game(n, l)
| Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | ```python
n = int(input())
l = input().split()
for i in range(len(l)):
l[i] = int(l[i])
sereja_score = 0
dima_score = 0
for i in range(len(l)):
if i % 2 == 0:
l2 = [l[0], l[-1]]
l1 = max(l2)
l.remove((l1))
sereja_score += l1
elif i % 2 == 1:
l4 = [l[0], l[-1]]
l3 = max(l4)
l.remove(l3)
dima_score += l3
print(sereja_score, dima_score)
# def score_game(n, l):
# for i in range(len(l)):
# l[i] = int(l[i])
#
# sereja_score = 0
# dima_score = 0
#
# for i in range(len(l)):
# if (i == 0) or (i % 2 == 0):
# l1 = max(l)
# l.remove((l1))
# sereja_score += l1
# else:
# l3 = max(l)
# l.remove(l3)
# dima_score += l3
#
# print(sereja_score, dima_score)
#
# n = int(input())
# l = input().split()
# score_game(n, l)
``` | 3 |
|
725 | A | Jumping Ball | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper. | Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position. | [
"4\n<<><\n",
"5\n>>>>>\n",
"4\n>><<\n"
] | [
"2",
"5",
"0"
] | In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field. | 500 | [
{
"input": "4\n<<><",
"output": "2"
},
{
"input": "5\n>>>>>",
"output": "5"
},
{
"input": "4\n>><<",
"output": "0"
},
{
"input": "3\n<<>",
"output": "3"
},
{
"input": "3\n<<<",
"output": "3"
},
{
"input": "3\n><<",
"output": "0"
},
{
"input": "1\n<",
"output": "1"
},
{
"input": "2\n<>",
"output": "2"
},
{
"input": "3\n<>>",
"output": "3"
},
{
"input": "3\n><>",
"output": "1"
},
{
"input": "2\n><",
"output": "0"
},
{
"input": "2\n>>",
"output": "2"
},
{
"input": "2\n<<",
"output": "2"
},
{
"input": "1\n>",
"output": "1"
},
{
"input": "3\n>><",
"output": "0"
},
{
"input": "3\n>>>",
"output": "3"
},
{
"input": "3\n<><",
"output": "1"
},
{
"input": "10\n<<<><<<>>>",
"output": "6"
},
{
"input": "20\n><><<><<<>>>>>>>>>>>",
"output": "11"
},
{
"input": "20\n<<<<<<<<<<><<<<>>>>>",
"output": "15"
},
{
"input": "50\n<<<<<<<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "50"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<>><<>><<<<<>><>><<<>><><<>>><<>>><<<<><><><<><<<<><>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "49"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<>>>><<>><>><>><<><><><><>>>><><<<>>>><<<>>>>>>><><",
"output": "50"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<",
"output": "100"
},
{
"input": "100\n>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "100"
},
{
"input": "12\n<<>><<>><<>>",
"output": "4"
},
{
"input": "6\n<<><>>",
"output": "4"
},
{
"input": "6\n><>>>>",
"output": "4"
},
{
"input": "8\n>>>><<<>",
"output": "1"
},
{
"input": "4\n<><>",
"output": "2"
},
{
"input": "4\n><><",
"output": "0"
},
{
"input": "7\n<<>>><>",
"output": "3"
},
{
"input": "10\n><><>>>>>>",
"output": "6"
},
{
"input": "5\n<><>>",
"output": "3"
},
{
"input": "12\n<><<<<>>>>>>",
"output": "7"
},
{
"input": "6\n<>><<>",
"output": "2"
},
{
"input": "6\n>>><>>",
"output": "2"
},
{
"input": "10\n><><>>>><>",
"output": "1"
},
{
"input": "5\n><>>>",
"output": "3"
},
{
"input": "5\n<<><>",
"output": "3"
},
{
"input": "5\n<><<<",
"output": "1"
},
{
"input": "4\n<><<",
"output": "1"
},
{
"input": "8\n<<>><<>>",
"output": "4"
},
{
"input": "7\n<<><>>>",
"output": "5"
},
{
"input": "5\n><<>>",
"output": "2"
},
{
"input": "10\n<<<<<>>>>>",
"output": "10"
},
{
"input": "6\n><<<<<",
"output": "0"
},
{
"input": "8\n<<><><>>",
"output": "4"
},
{
"input": "10\n<<<<><<<><",
"output": "4"
},
{
"input": "12\n<<<>>>><<>>>",
"output": "6"
},
{
"input": "4\n><>>",
"output": "2"
},
{
"input": "11\n<<><<>><<>>",
"output": "4"
}
] | 1,479,389,056 | 2,147,483,647 | Python 3 | OK | TESTS | 78 | 77 | 614,400 | if __name__ == '__main__':
n = int(input())
line = str(input())
num = 0
for it in line:
if it == '<':
num += 1
else:
break
for it in line[::-1]:
if it == '>':
num += 1
else:
break
print(num)
| Title: Jumping Ball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper.
Output Specification:
Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position.
Demo Input:
['4\n<<><\n', '5\n>>>>>\n', '4\n>><<\n']
Demo Output:
['2', '5', '0']
Note:
In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field. | ```python
if __name__ == '__main__':
n = int(input())
line = str(input())
num = 0
for it in line:
if it == '<':
num += 1
else:
break
for it in line[::-1]:
if it == '>':
num += 1
else:
break
print(num)
``` | 3 |
|
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,513,140,278 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | s = input()
b = ''
for i in s:
b = i + b
if s == b
print('YES')
else:
print('NO') | Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
s = input()
b = ''
for i in s:
b = i + b
if s == b
print('YES')
else:
print('NO')
``` | -1 |
231 | A | Team | PROGRAMMING | 800 | [
"brute force",
"greedy"
] | null | null | One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. | Print a single integer — the number of problems the friends will implement on the contest. | [
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] | [
"2\n",
"1\n"
] | In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | 500 | [
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,699,743,521 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 92 | 0 | # Input
n = int(input())
# Count the number of problems each friend is sure about
count_sure = 0
for _ in range(n):
p, v, t = map(int, input().split())
# Check if at least two friends are sure about the solution
if p + v + t >= 2:
count_sure += 1
# Output
print(count_sure)
| Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | ```python
# Input
n = int(input())
# Count the number of problems each friend is sure about
count_sure = 0
for _ in range(n):
p, v, t = map(int, input().split())
# Check if at least two friends are sure about the solution
if p + v + t >= 2:
count_sure += 1
# Output
print(count_sure)
``` | 3 |
|
519 | B | A and B and Compilation Errors | PROGRAMMING | 1,100 | [
"data structures",
"implementation",
"sortings"
] | null | null | A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected? | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one. | Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively. | [
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] | [
"8\n123\n",
"1\n3\n"
] | In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | 1,000 | [
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 656422756 9301599 25720377\n277246428 388684428 661369649 460626451 656422756 802090732 9301599 784303821 376287098\n376287098 802090732 388684428 9301599 656422756 784303821 460626451 277246428",
"output": "25720377\n661369649"
},
{
"input": "3\n796067435 964699482 819602309\n964699482 796067435\n964699482",
"output": "819602309\n796067435"
},
{
"input": "3\n374054998 726316780 902899520\n902899520 726316780\n726316780",
"output": "374054998\n902899520"
},
{
"input": "3\n168638990 939116221 323703261\n168638990 323703261\n168638990",
"output": "939116221\n323703261"
},
{
"input": "3\n77 77 77\n77 77\n77",
"output": "77\n77"
},
{
"input": "3\n84 30 9\n9 84\n9",
"output": "30\n84"
},
{
"input": "6\n5 4 3 3 5 5\n3 5 5 4 3\n3 5 4 3",
"output": "5\n5"
},
{
"input": "4\n1 5 7 8\n1 5 7\n1 5",
"output": "8\n7"
}
] | 1,694,479,021 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 140 | 23,449,600 | from sys import stdin
input = stdin.readline
def main(data) -> None:
x = sum(data[0])
y = sum(data[1])
z = sum(data[2])
print(x-y)
print(y-z)
if __name__ == "__main__":
input()
main([list(map(int, input().split())) for _ in range(3)])
| Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | ```python
from sys import stdin
input = stdin.readline
def main(data) -> None:
x = sum(data[0])
y = sum(data[1])
z = sum(data[2])
print(x-y)
print(y-z)
if __name__ == "__main__":
input()
main([list(map(int, input().split())) for _ in range(3)])
``` | 3 |
|
294 | A | Shaass and Oskols | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots. | The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment. | On the *i*-th line of the output print the number of birds on the *i*-th wire. | [
"5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n",
"3\n2 4 1\n1\n2 2\n"
] | [
"0\n12\n5\n0\n16\n",
"3\n0\n3\n"
] | none | 500 | [
{
"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6",
"output": "0\n12\n5\n0\n16"
},
{
"input": "3\n2 4 1\n1\n2 2",
"output": "3\n0\n3"
},
{
"input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43",
"output": "0\n66\n57\n7\n0"
},
{
"input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51",
"output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52"
},
{
"input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1",
"output": "0\n0"
},
{
"input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5",
"output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21"
},
{
"input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0",
"output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95"
},
{
"input": "1\n100\n1\n1 100",
"output": "0"
},
{
"input": "1\n100\n1\n1 1",
"output": "0"
},
{
"input": "1\n50\n1\n1 25",
"output": "0"
},
{
"input": "2\n50 0\n1\n1 1",
"output": "0\n49"
},
{
"input": "1\n10\n0",
"output": "10"
}
] | 1,686,425,194 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | input()
s=[*map(int,input().split())]
input()
for l in open(0):
a,b=map(int,l.split())
if a-2 >=0:
s[a-2]+=b-1
if a < 5:
s[a] +=s[a-1]-b
s[a-1]=0
print(*s,sep="\n") | Title: Shaass and Oskols
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
Input Specification:
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
Output Specification:
On the *i*-th line of the output print the number of birds on the *i*-th wire.
Demo Input:
['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n']
Demo Output:
['0\n12\n5\n0\n16\n', '3\n0\n3\n']
Note:
none | ```python
input()
s=[*map(int,input().split())]
input()
for l in open(0):
a,b=map(int,l.split())
if a-2 >=0:
s[a-2]+=b-1
if a < 5:
s[a] +=s[a-1]-b
s[a-1]=0
print(*s,sep="\n")
``` | 0 |
|
501 | A | Contest | PROGRAMMING | 900 | [
"implementation"
] | null | null | Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth. | The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round). | Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points. | [
"500 1000 20 30\n",
"1000 1000 1 1\n",
"1500 1000 176 177\n"
] | [
"Vasya\n",
"Tie\n",
"Misha\n"
] | none | 500 | [
{
"input": "500 1000 20 30",
"output": "Vasya"
},
{
"input": "1000 1000 1 1",
"output": "Tie"
},
{
"input": "1500 1000 176 177",
"output": "Misha"
},
{
"input": "1500 1000 74 177",
"output": "Misha"
},
{
"input": "750 2500 175 178",
"output": "Vasya"
},
{
"input": "750 1000 54 103",
"output": "Tie"
},
{
"input": "2000 1250 176 130",
"output": "Tie"
},
{
"input": "1250 1750 145 179",
"output": "Tie"
},
{
"input": "2000 2000 176 179",
"output": "Tie"
},
{
"input": "1500 1500 148 148",
"output": "Tie"
},
{
"input": "2750 1750 134 147",
"output": "Misha"
},
{
"input": "3250 250 175 173",
"output": "Misha"
},
{
"input": "500 500 170 176",
"output": "Misha"
},
{
"input": "250 1000 179 178",
"output": "Vasya"
},
{
"input": "3250 1000 160 138",
"output": "Misha"
},
{
"input": "3000 2000 162 118",
"output": "Tie"
},
{
"input": "1500 1250 180 160",
"output": "Tie"
},
{
"input": "1250 2500 100 176",
"output": "Tie"
},
{
"input": "3500 3500 177 178",
"output": "Tie"
},
{
"input": "3000 3250 16 34",
"output": "Tie"
},
{
"input": "1750 3000 137 49",
"output": "Vasya"
},
{
"input": "500 1500 179 71",
"output": "Vasya"
},
{
"input": "1250 2000 101 180",
"output": "Misha"
},
{
"input": "250 750 180 176",
"output": "Vasya"
},
{
"input": "2250 2250 163 145",
"output": "Vasya"
},
{
"input": "3000 3000 176 78",
"output": "Vasya"
},
{
"input": "250 3500 8 178",
"output": "Vasya"
},
{
"input": "1750 1250 179 180",
"output": "Misha"
},
{
"input": "2750 1750 13 164",
"output": "Misha"
},
{
"input": "1750 2250 178 53",
"output": "Vasya"
},
{
"input": "2500 2750 73 179",
"output": "Misha"
},
{
"input": "1000 3500 178 175",
"output": "Vasya"
},
{
"input": "1000 500 7 162",
"output": "Misha"
},
{
"input": "1000 250 175 48",
"output": "Misha"
},
{
"input": "1750 500 166 177",
"output": "Misha"
},
{
"input": "250 250 0 0",
"output": "Tie"
},
{
"input": "250 3500 0 0",
"output": "Vasya"
},
{
"input": "250 3500 0 180",
"output": "Vasya"
},
{
"input": "3500 3500 180 180",
"output": "Tie"
},
{
"input": "3500 250 0 180",
"output": "Misha"
}
] | 1,611,317,269 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 62 | 409,600 | import sys, os
from collections import defaultdict
input = sys.stdin.readline
read = lambda: list(map(int, input().strip().split()))
def f(p, t):
return(max((3*p)//10, 9 - (p//250)*t))
def main():
a,b,c,d = read()
x, y = f(a, c), f(b, d)
if x > y:print("Misha")
elif x < y:print("Vasya")
else:print("Tie")
if __name__ == "__main__":
main()
| Title: Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
Input Specification:
The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).
Output Specification:
Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points.
Demo Input:
['500 1000 20 30\n', '1000 1000 1 1\n', '1500 1000 176 177\n']
Demo Output:
['Vasya\n', 'Tie\n', 'Misha\n']
Note:
none | ```python
import sys, os
from collections import defaultdict
input = sys.stdin.readline
read = lambda: list(map(int, input().strip().split()))
def f(p, t):
return(max((3*p)//10, 9 - (p//250)*t))
def main():
a,b,c,d = read()
x, y = f(a, c), f(b, d)
if x > y:print("Misha")
elif x < y:print("Vasya")
else:print("Tie")
if __name__ == "__main__":
main()
``` | 0 |
|
401 | A | Vanya and Cards | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Vanya loves playing. He even has a special set of cards to play with. Each card has a single integer. The number on the card can be positive, negative and can even be equal to zero. The only limit is, the number on each card doesn't exceed *x* in the absolute value.
Natasha doesn't like when Vanya spends a long time playing, so she hid all of his cards. Vanya became sad and started looking for the cards but he only found *n* of them. Vanya loves the balance, so he wants the sum of all numbers on found cards equal to zero. On the other hand, he got very tired of looking for cards. Help the boy and say what is the minimum number of cards does he need to find to make the sum equal to zero?
You can assume that initially Vanya had infinitely many cards with each integer number from <=-<=*x* to *x*. | The first line contains two integers: *n* (1<=≤<=*n*<=≤<=1000) — the number of found cards and *x* (1<=≤<=*x*<=≤<=1000) — the maximum absolute value of the number on a card. The second line contains *n* space-separated integers — the numbers on found cards. It is guaranteed that the numbers do not exceed *x* in their absolute value. | Print a single number — the answer to the problem. | [
"3 2\n-1 1 2\n",
"2 3\n-2 -2\n"
] | [
"1\n",
"2\n"
] | In the first sample, Vanya needs to find a single card with number -2.
In the second sample, Vanya needs to find two cards with number 2. He can't find a single card with the required number as the numbers on the lost cards do not exceed 3 in their absolute value. | 500 | [
{
"input": "3 2\n-1 1 2",
"output": "1"
},
{
"input": "2 3\n-2 -2",
"output": "2"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
},
{
"input": "2 2\n-1 -1",
"output": "1"
},
{
"input": "15 5\n-2 -1 2 -4 -3 4 -4 -2 -2 2 -2 -1 1 -4 -2",
"output": "4"
},
{
"input": "15 16\n-15 -5 -15 -14 -8 15 -15 -12 -5 -3 5 -7 3 8 -15",
"output": "6"
},
{
"input": "1 4\n-3",
"output": "1"
},
{
"input": "10 7\n6 4 6 6 -3 4 -1 2 3 3",
"output": "5"
},
{
"input": "2 1\n1 -1",
"output": "0"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "8 13\n-11 -1 -11 12 -2 -2 -10 -11",
"output": "3"
},
{
"input": "16 11\n3 -7 7 -9 -2 -3 -4 -2 -6 8 10 7 1 4 6 7",
"output": "2"
},
{
"input": "67 15\n-2 -2 6 -4 -7 4 3 13 -9 -4 11 -7 -6 -11 1 11 -1 11 14 10 -8 7 5 11 -13 1 -1 7 -14 9 -11 -11 13 -4 12 -11 -8 -5 -11 6 10 -2 6 9 9 6 -11 -2 7 -10 -1 9 -8 -5 1 -7 -2 3 -1 -13 -6 -9 -8 10 13 -3 9",
"output": "1"
},
{
"input": "123 222\n44 -190 -188 -185 -55 17 190 176 157 176 -24 -113 -54 -61 -53 53 -77 68 -12 -114 -217 163 -122 37 -37 20 -108 17 -140 -210 218 19 -89 54 18 197 111 -150 -36 -131 -172 36 67 16 -202 72 169 -137 -34 -122 137 -72 196 -17 -104 180 -102 96 -69 -184 21 -15 217 -61 175 -221 62 173 -93 -106 122 -135 58 7 -110 -108 156 -141 -102 -50 29 -204 -46 -76 101 -33 -190 99 52 -197 175 -71 161 -140 155 10 189 -217 -97 -170 183 -88 83 -149 157 -208 154 -3 77 90 74 165 198 -181 -166 -4 -200 -89 -200 131 100 -61 -149",
"output": "8"
},
{
"input": "130 142\n58 -50 43 -126 84 -92 -108 -92 57 127 12 -135 -49 89 141 -112 -31 47 75 -19 80 81 -5 17 10 4 -26 68 -102 -10 7 -62 -135 -123 -16 55 -72 -97 -34 21 21 137 130 97 40 -18 110 -52 73 52 85 103 -134 -107 88 30 66 97 126 82 13 125 127 -87 81 22 45 102 13 95 4 10 -35 39 -43 -112 -5 14 -46 19 61 -44 -116 137 -116 -80 -39 92 -75 29 -65 -15 5 -108 -114 -129 -5 52 -21 118 -41 35 -62 -125 130 -95 -11 -75 19 108 108 127 141 2 -130 54 96 -81 -102 140 -58 -102 132 50 -126 82 6 45 -114 -42",
"output": "5"
},
{
"input": "7 12\n2 5 -1 -4 -7 4 3",
"output": "1"
},
{
"input": "57 53\n-49 7 -41 7 38 -51 -23 8 45 1 -24 26 37 28 -31 -40 38 25 -32 -47 -3 20 -40 -32 -44 -36 5 33 -16 -5 28 10 -22 3 -10 -51 -32 -51 27 -50 -22 -12 41 3 15 24 30 -12 -34 -15 -29 38 -10 -35 -9 6 -51",
"output": "8"
},
{
"input": "93 273\n-268 -170 -163 19 -69 18 -244 35 -34 125 -224 -48 179 -247 127 -150 271 -49 -102 201 84 -151 -70 -46 -16 216 240 127 3 218 -209 223 -227 -201 228 -8 203 46 -100 -207 126 255 40 -58 -217 93 172 -97 23 183 102 -92 -157 -117 173 47 144 -235 -227 -62 -128 13 -151 158 110 -116 68 -2 -148 -206 -52 79 -152 -223 74 -149 -69 232 38 -70 -256 -213 -236 132 -189 -200 199 -57 -108 -53 269 -101 -134",
"output": "8"
},
{
"input": "1 1000\n997",
"output": "1"
},
{
"input": "4 3\n2 -1 -2 -1",
"output": "1"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 2\n1 -1",
"output": "0"
},
{
"input": "2 2\n-1 1",
"output": "0"
},
{
"input": "2 3\n-1 1",
"output": "0"
},
{
"input": "2 2\n-2 2",
"output": "0"
},
{
"input": "2 2\n2 2",
"output": "2"
},
{
"input": "4 2\n-1 -1 -1 -1",
"output": "2"
},
{
"input": "4 1\n-1 -1 -1 1",
"output": "2"
},
{
"input": "3 2\n2 2 2",
"output": "3"
},
{
"input": "10 300\n300 300 300 300 300 300 300 300 300 300",
"output": "10"
}
] | 1,615,903,400 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 0 | n,x=[int(x) for x in input().split()]
ai=[int(x) for x in input().split()]
cnt=0
for i in ai:
num=-1*i
if(num not in ai):
cnt+=1
print(cnt) | Title: Vanya and Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya loves playing. He even has a special set of cards to play with. Each card has a single integer. The number on the card can be positive, negative and can even be equal to zero. The only limit is, the number on each card doesn't exceed *x* in the absolute value.
Natasha doesn't like when Vanya spends a long time playing, so she hid all of his cards. Vanya became sad and started looking for the cards but he only found *n* of them. Vanya loves the balance, so he wants the sum of all numbers on found cards equal to zero. On the other hand, he got very tired of looking for cards. Help the boy and say what is the minimum number of cards does he need to find to make the sum equal to zero?
You can assume that initially Vanya had infinitely many cards with each integer number from <=-<=*x* to *x*.
Input Specification:
The first line contains two integers: *n* (1<=≤<=*n*<=≤<=1000) — the number of found cards and *x* (1<=≤<=*x*<=≤<=1000) — the maximum absolute value of the number on a card. The second line contains *n* space-separated integers — the numbers on found cards. It is guaranteed that the numbers do not exceed *x* in their absolute value.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['3 2\n-1 1 2\n', '2 3\n-2 -2\n']
Demo Output:
['1\n', '2\n']
Note:
In the first sample, Vanya needs to find a single card with number -2.
In the second sample, Vanya needs to find two cards with number 2. He can't find a single card with the required number as the numbers on the lost cards do not exceed 3 in their absolute value. | ```python
n,x=[int(x) for x in input().split()]
ai=[int(x) for x in input().split()]
cnt=0
for i in ai:
num=-1*i
if(num not in ai):
cnt+=1
print(cnt)
``` | 0 |
|
948 | A | Protect Sheep | PROGRAMMING | 900 | [
"brute force",
"dfs and similar",
"graphs",
"implementation"
] | null | null | Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his sheep are protected.
The pasture is a rectangle consisting of *R*<=×<=*C* cells. Each cell is either empty, contains a sheep, a wolf or a dog. Sheep and dogs always stay in place, but wolves can roam freely around the pasture, by repeatedly moving to the left, right, up or down to a neighboring cell. When a wolf enters a cell with a sheep, it consumes it. However, no wolf can enter a cell with a dog.
Initially there are no dogs. Place dogs onto the pasture in such a way that no wolf can reach any sheep, or determine that it is impossible. Note that since you have many dogs, you do not need to minimize their number. | First line contains two integers *R* (1<=≤<=*R*<=≤<=500) and *C* (1<=≤<=*C*<=≤<=500), denoting the number of rows and the numbers of columns respectively.
Each of the following *R* lines is a string consisting of exactly *C* characters, representing one row of the pasture. Here, 'S' means a sheep, 'W' a wolf and '.' an empty cell. | If it is impossible to protect all sheep, output a single line with the word "No".
Otherwise, output a line with the word "Yes". Then print *R* lines, representing the pasture after placing dogs. Again, 'S' means a sheep, 'W' a wolf, 'D' is a dog and '.' an empty space. You are not allowed to move, remove or add a sheep or a wolf.
If there are multiple solutions, you may print any of them. You don't have to minimize the number of dogs. | [
"6 6\n..S...\n..S.W.\n.S....\n..W...\n...W..\n......\n",
"1 2\nSW\n",
"5 5\n.S...\n...S.\nS....\n...S.\n.S...\n"
] | [
"Yes\n..SD..\n..SDW.\n.SD...\n.DW...\nDD.W..\n......\n",
"No\n",
"Yes\n.S...\n...S.\nS.D..\n...S.\n.S...\n"
] | In the first example, we can split the pasture into two halves, one containing wolves and one containing sheep. Note that the sheep at (2,1) is safe, as wolves cannot move diagonally.
In the second example, there are no empty spots to put dogs that would guard the lone sheep.
In the third example, there are no wolves, so the task is very easy. We put a dog in the center to observe the peacefulness of the meadow, but the solution would be correct even without him. | 500 | [
{
"input": "1 2\nSW",
"output": "No"
},
{
"input": "10 10\n....W.W.W.\n.........S\n.S.S...S..\nW.......SS\n.W..W.....\n.W...W....\nS..S...S.S\n....W...S.\n..S..S.S.S\nSS.......S",
"output": "Yes\nDDDDWDWDWD\nDDDDDDDDDS\nDSDSDDDSDD\nWDDDDDDDSS\nDWDDWDDDDD\nDWDDDWDDDD\nSDDSDDDSDS\nDDDDWDDDSD\nDDSDDSDSDS\nSSDDDDDDDS"
},
{
"input": "10 10\n....W.W.W.\n...W.....S\n.S.S...S..\nW......WSS\n.W..W.....\n.W...W....\nS..S...S.S\n...WWW..S.\n..S..S.S.S\nSS.......S",
"output": "No"
},
{
"input": "1 50\nW...S..............W.....S..S...............S...W.",
"output": "Yes\nWDDDSDDDDDDDDDDDDDDWDDDDDSDDSDDDDDDDDDDDDDDDSDDDWD"
},
{
"input": "2 4\n...S\n...W",
"output": "No"
},
{
"input": "4 2\n..\n..\n..\nSW",
"output": "No"
},
{
"input": "4 2\n..\n..\n..\nWS",
"output": "No"
},
{
"input": "2 4\n...W\n...S",
"output": "No"
},
{
"input": "50 1\nS\n.\n.\n.\n.\n.\n.\nS\n.\n.\n.\n.\n.\n.\n.\n.\nS\n.\nW\n.\nS\n.\n.\n.\n.\nS\n.\n.\n.\n.\n.\n.\n.\nW\n.\n.\n.\nW\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.",
"output": "Yes\nS\nD\nD\nD\nD\nD\nD\nS\nD\nD\nD\nD\nD\nD\nD\nD\nS\nD\nW\nD\nS\nD\nD\nD\nD\nS\nD\nD\nD\nD\nD\nD\nD\nW\nD\nD\nD\nW\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD\nD"
},
{
"input": "4 4\nW..S\nW..S\nW..S\nW..S",
"output": "Yes\nWDDS\nWDDS\nWDDS\nWDDS"
},
{
"input": "4 4\nSSSS\n....\n....\nWWWW",
"output": "Yes\nSSSS\nDDDD\nDDDD\nWWWW"
},
{
"input": "4 4\nWWWW\n....\n....\nSSSS",
"output": "Yes\nWWWW\nDDDD\nDDDD\nSSSS"
},
{
"input": "4 4\nS..W\nS..W\nS..W\nS..W",
"output": "Yes\nSDDW\nSDDW\nSDDW\nSDDW"
},
{
"input": "1 1\n.",
"output": "Yes\nD"
},
{
"input": "1 1\nW",
"output": "Yes\nW"
},
{
"input": "1 1\nS",
"output": "Yes\nS"
},
{
"input": "4 2\n..\n..\n.W\n.S",
"output": "No"
},
{
"input": "4 2\n..\n..\n.S\n.W",
"output": "No"
},
{
"input": "4 2\n..\n..\nW.\nS.",
"output": "No"
},
{
"input": "4 2\n..\n..\nS.\nW.",
"output": "No"
},
{
"input": "2 4\n....\n..SW",
"output": "No"
},
{
"input": "2 4\n....\n..WS",
"output": "No"
},
{
"input": "1 2\nS.",
"output": "Yes\nSD"
}
] | 1,520,698,194 | 2,094 | Python 3 | OK | TESTS | 98 | 217 | 6,144,000 | a,b = map(int, input().split())
anslist = [" "*(b+2)]
for i in range (a):
anslist.append(" "+ input() + " ")
anslist.append(" "*(b+2))
flag = 0
for k in range (1,a+1) :
p = anslist[k]
q = anslist[k+1]
r = anslist[k-1]
for j in range (1, b+1):
if (p[j] == "S" and p[j-1] == "W") or (p[j] == "S" and p[j+1] == "W") or (p[j] == "S" and q[j] == "W") or (p[j] == "S" and r[j] == "W"):
flag = 1
break
if flag == 1 :
break
if flag == 0 :
print("Yes")
for i in range (1, a+1):
anslist[i] = anslist[i].replace('.', 'D')
print (anslist[i][1:b+1])
else :
print ("No") | Title: Protect Sheep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his sheep are protected.
The pasture is a rectangle consisting of *R*<=×<=*C* cells. Each cell is either empty, contains a sheep, a wolf or a dog. Sheep and dogs always stay in place, but wolves can roam freely around the pasture, by repeatedly moving to the left, right, up or down to a neighboring cell. When a wolf enters a cell with a sheep, it consumes it. However, no wolf can enter a cell with a dog.
Initially there are no dogs. Place dogs onto the pasture in such a way that no wolf can reach any sheep, or determine that it is impossible. Note that since you have many dogs, you do not need to minimize their number.
Input Specification:
First line contains two integers *R* (1<=≤<=*R*<=≤<=500) and *C* (1<=≤<=*C*<=≤<=500), denoting the number of rows and the numbers of columns respectively.
Each of the following *R* lines is a string consisting of exactly *C* characters, representing one row of the pasture. Here, 'S' means a sheep, 'W' a wolf and '.' an empty cell.
Output Specification:
If it is impossible to protect all sheep, output a single line with the word "No".
Otherwise, output a line with the word "Yes". Then print *R* lines, representing the pasture after placing dogs. Again, 'S' means a sheep, 'W' a wolf, 'D' is a dog and '.' an empty space. You are not allowed to move, remove or add a sheep or a wolf.
If there are multiple solutions, you may print any of them. You don't have to minimize the number of dogs.
Demo Input:
['6 6\n..S...\n..S.W.\n.S....\n..W...\n...W..\n......\n', '1 2\nSW\n', '5 5\n.S...\n...S.\nS....\n...S.\n.S...\n']
Demo Output:
['Yes\n..SD..\n..SDW.\n.SD...\n.DW...\nDD.W..\n......\n', 'No\n', 'Yes\n.S...\n...S.\nS.D..\n...S.\n.S...\n']
Note:
In the first example, we can split the pasture into two halves, one containing wolves and one containing sheep. Note that the sheep at (2,1) is safe, as wolves cannot move diagonally.
In the second example, there are no empty spots to put dogs that would guard the lone sheep.
In the third example, there are no wolves, so the task is very easy. We put a dog in the center to observe the peacefulness of the meadow, but the solution would be correct even without him. | ```python
a,b = map(int, input().split())
anslist = [" "*(b+2)]
for i in range (a):
anslist.append(" "+ input() + " ")
anslist.append(" "*(b+2))
flag = 0
for k in range (1,a+1) :
p = anslist[k]
q = anslist[k+1]
r = anslist[k-1]
for j in range (1, b+1):
if (p[j] == "S" and p[j-1] == "W") or (p[j] == "S" and p[j+1] == "W") or (p[j] == "S" and q[j] == "W") or (p[j] == "S" and r[j] == "W"):
flag = 1
break
if flag == 1 :
break
if flag == 0 :
print("Yes")
for i in range (1, a+1):
anslist[i] = anslist[i].replace('.', 'D')
print (anslist[i][1:b+1])
else :
print ("No")
``` | 3 |
|
387 | B | George and Round | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"two pointers"
] | null | null | George decided to prepare a Codesecrof round, so he has prepared *m* problems for the round. Let's number the problems with integers 1 through *m*. George estimates the *i*-th problem's complexity by integer *b**i*.
To make the round good, he needs to put at least *n* problems there. Besides, he needs to have at least one problem with complexity exactly *a*1, at least one with complexity exactly *a*2, ..., and at least one with complexity exactly *a**n*. Of course, the round can also have problems with other complexities.
George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity *c* to any positive integer complexity *d* (*c*<=≥<=*d*), by changing limits on the input data.
However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the *m* he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*<=≤<=106) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b*1<=≤<=*b*2...<=≤<=*b**m*<=≤<=106) — the complexities of the problems prepared by George. | Print a single integer — the answer to the problem. | [
"3 5\n1 2 3\n1 2 2 3 3\n",
"3 5\n1 2 3\n1 1 1 1 1\n",
"3 1\n2 3 4\n1\n"
] | [
"0\n",
"2\n",
"3\n"
] | In the first sample the set of the prepared problems meets the requirements for a good round.
In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round.
In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4. | 1,000 | [
{
"input": "3 5\n1 2 3\n1 2 2 3 3",
"output": "0"
},
{
"input": "3 5\n1 2 3\n1 1 1 1 1",
"output": "2"
},
{
"input": "3 1\n2 3 4\n1",
"output": "3"
},
{
"input": "29 100\n20 32 41 67 72 155 331 382 399 412 465 470 484 511 515 529 616 637 679 715 733 763 826 843 862 903 925 979 989\n15 15 15 17 18 19 19 20 21 21 22 24 25 26 26 27 28 31 32 32 37 38 38 39 39 40 41 42 43 43 45 45 46 47 49 49 50 50 50 51 52 53 53 55 56 57 59 59 59 60 60 62 62 63 63 64 64 64 66 67 69 69 70 70 72 72 73 74 75 76 77 78 80 80 81 81 83 83 83 84 86 86 86 86 87 88 89 91 91 91 92 93 94 94 96 97 97 97 98 98",
"output": "24"
}
] | 1,667,389,570 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 31 | 0 | n,m = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
a = [i for i in a if i not in b]
print(len(a))
| Title: George and Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George decided to prepare a Codesecrof round, so he has prepared *m* problems for the round. Let's number the problems with integers 1 through *m*. George estimates the *i*-th problem's complexity by integer *b**i*.
To make the round good, he needs to put at least *n* problems there. Besides, he needs to have at least one problem with complexity exactly *a*1, at least one with complexity exactly *a*2, ..., and at least one with complexity exactly *a**n*. Of course, the round can also have problems with other complexities.
George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity *c* to any positive integer complexity *d* (*c*<=≥<=*d*), by changing limits on the input data.
However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the *m* he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*<=≤<=106) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b*1<=≤<=*b*2...<=≤<=*b**m*<=≤<=106) — the complexities of the problems prepared by George.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3 5\n1 2 3\n1 2 2 3 3\n', '3 5\n1 2 3\n1 1 1 1 1\n', '3 1\n2 3 4\n1\n']
Demo Output:
['0\n', '2\n', '3\n']
Note:
In the first sample the set of the prepared problems meets the requirements for a good round.
In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round.
In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4. | ```python
n,m = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
a = [i for i in a if i not in b]
print(len(a))
``` | 0 |
|
129 | A | Cookies | PROGRAMMING | 900 | [
"implementation"
] | null | null | Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even? | The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag. | Print in the only line the only number — the sought number of ways. If there are no such ways print 0. | [
"1\n1\n",
"10\n1 2 2 3 4 4 4 2 2 2\n",
"11\n2 2 2 2 2 2 2 2 2 2 99\n"
] | [
"1\n",
"8\n",
"1\n"
] | In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies. | 500 | [
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 2 2 3 4 4 4 2 2 2",
"output": "8"
},
{
"input": "11\n2 2 2 2 2 2 2 2 2 2 99",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "7\n7 7 7 7 7 7 7",
"output": "7"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "4"
},
{
"input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2",
"output": "50"
},
{
"input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99",
"output": "49"
},
{
"input": "82\n43 44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15",
"output": "50"
},
{
"input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38",
"output": "7"
},
{
"input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87",
"output": "39"
},
{
"input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53",
"output": "37"
},
{
"input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32",
"output": "51"
},
{
"input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80",
"output": "21"
},
{
"input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81",
"output": "37"
},
{
"input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59",
"output": "28"
},
{
"input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94",
"output": "47"
},
{
"input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28",
"output": "15"
},
{
"input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42",
"output": "11"
},
{
"input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35",
"output": "33"
},
{
"input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22",
"output": "37"
},
{
"input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14",
"output": "15"
},
{
"input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77",
"output": "35"
},
{
"input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75",
"output": "42"
},
{
"input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26",
"output": "23"
},
{
"input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78",
"output": "15"
},
{
"input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78",
"output": "13"
},
{
"input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51",
"output": "17"
},
{
"input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59",
"output": "45"
},
{
"input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43",
"output": "53"
},
{
"input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8",
"output": "40"
},
{
"input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54",
"output": "53"
},
{
"input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77",
"output": "55"
},
{
"input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66",
"output": "52"
},
{
"input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89",
"output": "51"
},
{
"input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2",
"output": "53"
},
{
"input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23",
"output": "48"
},
{
"input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25",
"output": "48"
},
{
"input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71",
"output": "58"
},
{
"input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45",
"output": "53"
},
{
"input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17",
"output": "54"
},
{
"input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52",
"output": "49"
},
{
"input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31",
"output": "44"
},
{
"input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74",
"output": "46"
},
{
"input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75",
"output": "53"
},
{
"input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37",
"output": "43"
},
{
"input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82",
"output": "51"
},
{
"input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5",
"output": "47"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "5\n1 1 3 2 2",
"output": "3"
}
] | 1,693,131,949 | 2,147,483,647 | Python 3 | OK | TESTS | 52 | 92 | 0 | n=int(input())
lis=list(map(int,input().strip().split()))
s=sum(lis)
if n==1:
print(1)
else:
c=0
nl=[]
if s%2==0:
for i in lis:
if i not in nl:
if i%2==0:
c=c+lis.count(i)
nl.append(i)
else:
for i in lis:
if i not in nl:
if i%2!=0:
c=c+lis.count(i)
nl.append(i)
print(c)
| Title: Cookies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Output Specification:
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
Demo Input:
['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n']
Demo Output:
['1\n', '8\n', '1\n']
Note:
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies. | ```python
n=int(input())
lis=list(map(int,input().strip().split()))
s=sum(lis)
if n==1:
print(1)
else:
c=0
nl=[]
if s%2==0:
for i in lis:
if i not in nl:
if i%2==0:
c=c+lis.count(i)
nl.append(i)
else:
for i in lis:
if i not in nl:
if i%2!=0:
c=c+lis.count(i)
nl.append(i)
print(c)
``` | 3 |
|
937 | A | Olympiad | PROGRAMMING | 800 | [
"implementation",
"sortings"
] | null | null | The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score. | Print a single integer — the desired number of ways. | [
"4\n1 3 3 2\n",
"3\n1 1 1\n",
"4\n42 0 0 42\n"
] | [
"3\n",
"1\n",
"1\n"
] | There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything. | 500 | [
{
"input": "4\n1 3 3 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "4\n42 0 0 42",
"output": "1"
},
{
"input": "10\n1 0 1 0 1 0 0 0 0 1",
"output": "1"
},
{
"input": "10\n572 471 540 163 50 30 561 510 43 200",
"output": "10"
},
{
"input": "100\n122 575 426 445 172 81 247 429 97 202 175 325 382 384 417 356 132 502 328 537 57 339 518 211 479 306 140 168 268 16 140 263 593 249 391 310 555 468 231 180 157 18 334 328 276 155 21 280 322 545 111 267 467 274 291 304 235 34 365 180 21 95 501 552 325 331 302 353 296 22 289 399 7 466 32 302 568 333 75 192 284 10 94 128 154 512 9 480 243 521 551 492 420 197 207 125 367 117 438 600",
"output": "94"
},
{
"input": "100\n600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600",
"output": "1"
},
{
"input": "78\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12",
"output": "13"
},
{
"input": "34\n220 387 408 343 184 447 197 307 337 414 251 319 426 322 347 242 208 412 188 185 241 235 216 259 331 372 322 284 444 384 214 297 389 391",
"output": "33"
},
{
"input": "100\n1 2 1 0 3 0 2 0 0 1 2 0 1 3 0 3 3 1 3 0 0 2 1 2 2 1 3 3 3 3 3 2 0 0 2 1 2 3 2 3 0 1 1 3 3 2 0 3 1 0 2 2 2 1 2 3 2 1 0 3 0 2 0 3 0 2 1 0 3 1 0 2 2 1 3 1 3 0 2 3 3 1 1 3 1 3 0 3 2 0 2 3 3 0 2 0 2 0 1 3",
"output": "3"
},
{
"input": "100\n572 471 540 163 50 30 561 510 43 200 213 387 500 424 113 487 357 333 294 337 435 202 447 494 485 465 161 344 470 559 104 356 393 207 224 213 511 514 60 386 149 216 392 229 429 173 165 401 395 150 127 579 344 390 529 296 225 425 318 79 465 447 177 110 367 212 459 270 41 500 277 567 125 436 178 9 214 342 203 112 144 24 79 155 495 556 40 549 463 281 241 316 2 246 1 396 510 293 332 55",
"output": "93"
},
{
"input": "99\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12 2 3 9 3 7 13 7 13 0 11 8 12 2 5 9 4 0 6 6 2 13",
"output": "13"
},
{
"input": "99\n1 0 1 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1",
"output": "1"
},
{
"input": "99\n21 74 25 44 71 80 46 28 96 1 74 24 81 83 16 55 31 1 27 36 56 38 17 10 78 5 39 67 67 15 39 62 92 48 90 9 54 67 30 79 56 17 33 27 75 54 20 79 21 44 10 66 66 73 90 3 34 33 64 79 20 94 0 51 24 30 1 52 95 21 88 98 6 65 31 1 67 32 74 91 83 9 93 27 53 11 8 79 42 20 50 91 19 96 6 24 66 16 37",
"output": "61"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n0 600",
"output": "1"
},
{
"input": "4\n1 1 1 2",
"output": "2"
},
{
"input": "4\n0 0 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "2\n0 5",
"output": "1"
},
{
"input": "5\n1 0 0 1 2",
"output": "2"
}
] | 1,617,091,599 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 62 | 0 | import sys
input = sys.stdin.readline
n = int(input())
a = [int(i) for i in input().split()]
res = [int(i) for i in a if i > 0]
print(len(set(res)))
| Title: Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
Output Specification:
Print a single integer — the desired number of ways.
Demo Input:
['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n']
Demo Output:
['3\n', '1\n', '1\n']
Note:
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything. | ```python
import sys
input = sys.stdin.readline
n = int(input())
a = [int(i) for i in input().split()]
res = [int(i) for i in a if i > 0]
print(len(set(res)))
``` | 3 |
|
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,611,482,840 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 154 | 0 | m,n=[int(x) for x in input().split()]
print((m*n)//2) | Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
m,n=[int(x) for x in input().split()]
print((m*n)//2)
``` | 3.9615 |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,602,951,171 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 218 | 0 | def func (a, b):
if (b == 0):
return a
return func(b, a%b)
y, w = map(int, input().split())
a = 7 - max(y, w)
b = 6
c = func(a,b)
a /= c
b /= c
print("%d/%d"%(a, b))
| Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
def func (a, b):
if (b == 0):
return a
return func(b, a%b)
y, w = map(int, input().split())
a = 7 - max(y, w)
b = 6
c = func(a,b)
a /= c
b /= c
print("%d/%d"%(a, b))
``` | 3.891 |
638 | C | Road Improvement | PROGRAMMING | 1,800 | [
"*special",
"dfs and similar",
"graphs",
"greedy",
"trees"
] | null | null | In Berland there are *n* cities and *n*<=-<=1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simultaneously for one day. Both brigades repair one road for the whole day and cannot take part in repairing other roads on that day. But the repair brigade can do nothing on that day.
Determine the minimum number of days needed to repair all the roads. The brigades cannot change the cities where they initially are. | The first line of the input contains a positive integer *n* (2<=≤<=*n*<=≤<=200<=000) — the number of cities in Berland.
Each of the next *n*<=-<=1 lines contains two numbers *u**i*, *v**i*, meaning that the *i*-th road connects city *u**i* and city *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*). | First print number *k* — the minimum number of days needed to repair all the roads in Berland.
In next *k* lines print the description of the roads that should be repaired on each of the *k* days. On the *i*-th line print first number *d**i* — the number of roads that should be repaired on the *i*-th day, and then *d**i* space-separated integers — the numbers of the roads that should be repaired on the *i*-th day. The roads are numbered according to the order in the input, starting from one.
If there are multiple variants, you can print any of them. | [
"4\n1 2\n3 4\n3 2\n",
"6\n3 4\n5 4\n3 2\n1 3\n4 6\n"
] | [
"2\n2 2 1\n1 3\n",
"3\n1 1 \n2 2 3 \n2 4 5 \n"
] | In the first sample you can repair all the roads in two days, for example, if you repair roads 1 and 2 on the first day and road 3 — on the second day. | 1,500 | [
{
"input": "4\n1 2\n3 4\n3 2",
"output": "2\n2 1 2 \n1 3 "
},
{
"input": "6\n3 4\n5 4\n3 2\n1 3\n4 6",
"output": "3\n1 1 \n2 2 3 \n2 4 5 "
},
{
"input": "8\n1 3\n1 6\n3 4\n6 2\n5 6\n6 7\n7 8",
"output": "4\n3 2 3 7 \n2 1 4 \n1 5 \n1 6 "
},
{
"input": "5\n1 2\n1 3\n1 4\n1 5",
"output": "4\n1 1 \n1 2 \n1 3 \n1 4 "
},
{
"input": "2\n1 2",
"output": "1\n1 1 "
},
{
"input": "2\n2 1",
"output": "1\n1 1 "
},
{
"input": "3\n1 2\n3 2",
"output": "2\n1 1 \n1 2 "
},
{
"input": "3\n1 3\n2 3",
"output": "2\n1 1 \n1 2 "
},
{
"input": "4\n1 4\n1 2\n4 3",
"output": "2\n1 1 \n2 2 3 "
},
{
"input": "4\n1 2\n1 3\n1 4",
"output": "3\n1 1 \n1 2 \n1 3 "
},
{
"input": "6\n1 2\n1 3\n1 4\n3 5\n4 6",
"output": "3\n3 1 4 5 \n1 2 \n1 3 "
},
{
"input": "6\n1 2\n1 3\n1 4\n3 5\n3 6",
"output": "3\n2 1 4 \n1 2 \n2 3 5 "
},
{
"input": "8\n1 2\n2 3\n3 4\n1 5\n5 6\n6 7\n1 8",
"output": "3\n3 1 3 5 \n3 2 4 6 \n1 7 "
},
{
"input": "10\n4 1\n9 5\n6 8\n4 9\n3 10\n2 8\n9 3\n10 7\n8 7",
"output": "3\n4 1 3 7 8 \n3 2 5 6 \n2 4 9 "
},
{
"input": "10\n2 4\n6 10\n10 3\n7 4\n7 9\n8 2\n3 1\n4 5\n2 6",
"output": "3\n4 1 2 5 7 \n3 3 4 6 \n2 8 9 "
},
{
"input": "2\n2 1",
"output": "1\n1 1 "
},
{
"input": "3\n3 2\n1 2",
"output": "2\n1 1 \n1 2 "
},
{
"input": "3\n3 1\n1 2",
"output": "2\n1 1 \n1 2 "
},
{
"input": "3\n2 1\n3 2",
"output": "2\n1 1 \n1 2 "
},
{
"input": "3\n1 2\n1 3",
"output": "2\n1 1 \n1 2 "
},
{
"input": "4\n1 2\n4 2\n3 4",
"output": "2\n2 1 3 \n1 2 "
},
{
"input": "4\n3 1\n1 4\n2 1",
"output": "3\n1 1 \n1 2 \n1 3 "
},
{
"input": "4\n1 3\n4 3\n2 4",
"output": "2\n2 1 3 \n1 2 "
},
{
"input": "4\n3 1\n1 2\n1 4",
"output": "3\n1 1 \n1 2 \n1 3 "
},
{
"input": "4\n3 4\n1 4\n3 2",
"output": "2\n1 1 \n2 2 3 "
},
{
"input": "4\n1 2\n1 3\n4 1",
"output": "3\n1 1 \n1 2 \n1 3 "
},
{
"input": "10\n2 8\n5 10\n3 4\n1 6\n3 9\n1 7\n4 8\n10 8\n1 8",
"output": "4\n4 1 2 3 4 \n3 5 6 7 \n1 8 \n1 9 "
},
{
"input": "12\n1 2\n2 3\n2 4\n2 5\n5 7\n5 6\n5 8\n5 12\n12 10\n12 11\n12 9",
"output": "5\n2 4 9 \n3 1 5 10 \n3 2 6 11 \n2 3 7 \n1 8 "
}
] | 1,458,484,187 | 8,987 | Python 3 | WRONG_ANSWER | PRETESTS | 3 | 62 | 4,812,800 | def f(v, u, color): # v - �����
global maxs
s = 0
if s == color:
s += 1
for i in range(len(arr2[v])):
if arr2[v][i][0] != u:
f(arr2[v][i][0], v, s)
colors[s].append(arr2[v][i][1])
s += 1
if s == color:
s += 1
maxs = max(s, maxs)
n = int(input())
arr2 = [[] for i in range(n)]
maxs = 0
for i in range(n - 1):
v, u = list(map(int, input().split()))
arr2[v - 1].append((u - 1, i + 1))
arr2[u - 1].append((v - 1, i + 1))
colors = [[] for i in range(n - 1)]
for i in range(len(arr2[0])):
f(arr2[0][i][0], 0, i)
colors[i].append(arr2[0][i][1])
print(maxs)
for i in range(n - 1):
if len(colors[i]) != 0:
print(len(colors[i]), *colors[i]) | Title: Road Improvement
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland there are *n* cities and *n*<=-<=1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simultaneously for one day. Both brigades repair one road for the whole day and cannot take part in repairing other roads on that day. But the repair brigade can do nothing on that day.
Determine the minimum number of days needed to repair all the roads. The brigades cannot change the cities where they initially are.
Input Specification:
The first line of the input contains a positive integer *n* (2<=≤<=*n*<=≤<=200<=000) — the number of cities in Berland.
Each of the next *n*<=-<=1 lines contains two numbers *u**i*, *v**i*, meaning that the *i*-th road connects city *u**i* and city *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*).
Output Specification:
First print number *k* — the minimum number of days needed to repair all the roads in Berland.
In next *k* lines print the description of the roads that should be repaired on each of the *k* days. On the *i*-th line print first number *d**i* — the number of roads that should be repaired on the *i*-th day, and then *d**i* space-separated integers — the numbers of the roads that should be repaired on the *i*-th day. The roads are numbered according to the order in the input, starting from one.
If there are multiple variants, you can print any of them.
Demo Input:
['4\n1 2\n3 4\n3 2\n', '6\n3 4\n5 4\n3 2\n1 3\n4 6\n']
Demo Output:
['2\n2 2 1\n1 3\n', '3\n1 1 \n2 2 3 \n2 4 5 \n']
Note:
In the first sample you can repair all the roads in two days, for example, if you repair roads 1 and 2 on the first day and road 3 — on the second day. | ```python
def f(v, u, color): # v - �����
global maxs
s = 0
if s == color:
s += 1
for i in range(len(arr2[v])):
if arr2[v][i][0] != u:
f(arr2[v][i][0], v, s)
colors[s].append(arr2[v][i][1])
s += 1
if s == color:
s += 1
maxs = max(s, maxs)
n = int(input())
arr2 = [[] for i in range(n)]
maxs = 0
for i in range(n - 1):
v, u = list(map(int, input().split()))
arr2[v - 1].append((u - 1, i + 1))
arr2[u - 1].append((v - 1, i + 1))
colors = [[] for i in range(n - 1)]
for i in range(len(arr2[0])):
f(arr2[0][i][0], 0, i)
colors[i].append(arr2[0][i][1])
print(maxs)
for i in range(n - 1):
if len(colors[i]) != 0:
print(len(colors[i]), *colors[i])
``` | 0 |
|
876 | A | Trip For Meal | PROGRAMMING | 900 | [
"math"
] | null | null | Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance. | First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits.
Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses.
Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses. | Output one number — minimum distance in meters Winnie must go through to have a meal *n* times. | [
"3\n2\n3\n1\n",
"1\n2\n3\n5\n"
] | [
"3\n",
"0\n"
] | In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all. | 500 | [
{
"input": "3\n2\n3\n1",
"output": "3"
},
{
"input": "1\n2\n3\n5",
"output": "0"
},
{
"input": "10\n1\n8\n3",
"output": "9"
},
{
"input": "7\n10\n5\n6",
"output": "30"
},
{
"input": "9\n9\n7\n5",
"output": "42"
},
{
"input": "9\n37\n85\n76",
"output": "296"
},
{
"input": "76\n46\n77\n11",
"output": "860"
},
{
"input": "80\n42\n1\n37",
"output": "79"
},
{
"input": "8\n80\n55\n1",
"output": "61"
},
{
"input": "10\n13\n72\n17",
"output": "117"
},
{
"input": "9\n24\n1\n63",
"output": "8"
},
{
"input": "65\n5\n8\n7",
"output": "320"
},
{
"input": "56\n8\n9\n3",
"output": "170"
},
{
"input": "59\n8\n1\n2",
"output": "58"
},
{
"input": "75\n50\n50\n5",
"output": "415"
},
{
"input": "75\n54\n76\n66",
"output": "3996"
},
{
"input": "73\n71\n69\n66",
"output": "4755"
},
{
"input": "83\n58\n88\n16",
"output": "1354"
},
{
"input": "74\n31\n11\n79",
"output": "803"
},
{
"input": "62\n27\n16\n72",
"output": "976"
},
{
"input": "72\n95\n27\n9",
"output": "657"
},
{
"input": "1\n2\n2\n1",
"output": "0"
},
{
"input": "1\n1\n1\n1",
"output": "0"
},
{
"input": "1\n1\n1\n99",
"output": "0"
},
{
"input": "100\n100\n100\n100",
"output": "9900"
},
{
"input": "2\n1\n1\n3",
"output": "1"
},
{
"input": "1\n3\n2\n1",
"output": "0"
},
{
"input": "1\n5\n6\n1",
"output": "0"
},
{
"input": "1\n2\n6\n1",
"output": "0"
},
{
"input": "1\n30\n20\n1",
"output": "0"
},
{
"input": "1\n15\n13\n11",
"output": "0"
},
{
"input": "1\n100\n50\n1",
"output": "0"
},
{
"input": "1\n2\n3\n1",
"output": "0"
},
{
"input": "5\n5\n5\n5",
"output": "20"
},
{
"input": "1\n100\n50\n5",
"output": "0"
},
{
"input": "1\n5\n7\n3",
"output": "0"
},
{
"input": "1\n3\n4\n2",
"output": "0"
},
{
"input": "3\n3\n3\n3",
"output": "6"
},
{
"input": "1\n99\n98\n1",
"output": "0"
},
{
"input": "5\n3\n3\n1",
"output": "6"
},
{
"input": "2\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1\n7\n8",
"output": "1"
},
{
"input": "2\n2\n2\n1",
"output": "2"
},
{
"input": "2\n2\n3\n1",
"output": "2"
},
{
"input": "1\n5\n6\n2",
"output": "0"
},
{
"input": "100\n1\n1\n100",
"output": "99"
},
{
"input": "2\n3\n2\n1",
"output": "2"
},
{
"input": "1\n99\n98\n97",
"output": "0"
}
] | 1,508,154,194 | 2,293 | Python 3 | WRONG_ANSWER | PRETESTS | 2 | 61 | 5,529,600 | ip = lambda: int(input())
n = ip()
a = ip()
b = ip()
c = ip()
if (n==1):
print("0")
else:
if (a == max(a,b,c)):
print(b+c)
elif(b==max(a,b,c)):
print(a+c)
else: print(a+b) | Title: Trip For Meal
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance.
Input Specification:
First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits.
Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses.
Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses.
Output Specification:
Output one number — minimum distance in meters Winnie must go through to have a meal *n* times.
Demo Input:
['3\n2\n3\n1\n', '1\n2\n3\n5\n']
Demo Output:
['3\n', '0\n']
Note:
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all. | ```python
ip = lambda: int(input())
n = ip()
a = ip()
b = ip()
c = ip()
if (n==1):
print("0")
else:
if (a == max(a,b,c)):
print(b+c)
elif(b==max(a,b,c)):
print(a+c)
else: print(a+b)
``` | 0 |
|
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,606,894,110 | 2,147,483,647 | PyPy 3 | OK | TESTS | 81 | 310 | 0 | X, Y, Z = 0, 0, 0
for _ in range(int(input())):
x, y, z = [int(j) for j in input().split()]
X += x
Y += y
Z += z
if X == 0 and Y == 0 and Z == 0:
print("YES")
else:
print("NO")
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
X, Y, Z = 0, 0, 0
for _ in range(int(input())):
x, y, z = [int(j) for j in input().split()]
X += x
Y += y
Z += z
if X == 0 and Y == 0 and Z == 0:
print("YES")
else:
print("NO")
``` | 3.9225 |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,669,996,961 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 | m,n,a=int(input()).split()
print(((m+a+-1)/a)*((n+a+-1)/a)) | Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
m,n,a=int(input()).split()
print(((m+a+-1)/a)*((n+a+-1)/a))
``` | -1 |
299 | B | Ksusha the Squirrel | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | Ksusha the Squirrel is standing at the beginning of a straight road, divided into *n* sectors. The sectors are numbered 1 to *n*, from left to right. Initially, Ksusha stands in sector 1.
Ksusha wants to walk to the end of the road, that is, get to sector *n*. Unfortunately, there are some rocks on the road. We know that Ksusha hates rocks, so she doesn't want to stand in sectors that have rocks.
Ksusha the squirrel keeps fit. She can jump from sector *i* to any of the sectors *i*<=+<=1,<=*i*<=+<=2,<=...,<=*i*<=+<=*k*.
Help Ksusha! Given the road description, say if she can reach the end of the road (note, she cannot stand on a rock)? | The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=3·105,<=1<=≤<=*k*<=≤<=3·105). The next line contains *n* characters — the description of the road: the *i*-th character equals ".", if the *i*-th sector contains no rocks. Otherwise, it equals "#".
It is guaranteed that the first and the last characters equal ".". | Print "YES" (without the quotes) if Ksusha can reach the end of the road, otherwise print "NO" (without the quotes). | [
"2 1\n..\n",
"5 2\n.#.#.\n",
"7 3\n.#.###.\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | none | 1,000 | [
{
"input": "2 1\n..",
"output": "YES"
},
{
"input": "5 2\n.#.#.",
"output": "YES"
},
{
"input": "7 3\n.#.###.",
"output": "NO"
},
{
"input": "2 200\n..",
"output": "YES"
},
{
"input": "2 1\n..",
"output": "YES"
},
{
"input": "2 2\n..",
"output": "YES"
},
{
"input": "2 100000\n..",
"output": "YES"
},
{
"input": "3 1\n.#.",
"output": "NO"
},
{
"input": "3 2\n.#.",
"output": "YES"
},
{
"input": "3 10000\n.#.",
"output": "YES"
}
] | 1,682,842,132 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 154 | 0 | n, k = map(int, input().split())
road = input()
def can_reach_end(n, k, road):
current = 0
last_no_rock_pos = -1
while current < n:
if road[current] == '#':
current = last_no_rock_pos + k + 1
if current >= n:
return 'NO'
else:
if current == n - 1:
return 'YES'
elif current - last_no_rock_pos >= k:
last_no_rock_pos = current
current += 1
return 'NO'
print(can_reach_end(n, k, road)) | Title: Ksusha the Squirrel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ksusha the Squirrel is standing at the beginning of a straight road, divided into *n* sectors. The sectors are numbered 1 to *n*, from left to right. Initially, Ksusha stands in sector 1.
Ksusha wants to walk to the end of the road, that is, get to sector *n*. Unfortunately, there are some rocks on the road. We know that Ksusha hates rocks, so she doesn't want to stand in sectors that have rocks.
Ksusha the squirrel keeps fit. She can jump from sector *i* to any of the sectors *i*<=+<=1,<=*i*<=+<=2,<=...,<=*i*<=+<=*k*.
Help Ksusha! Given the road description, say if she can reach the end of the road (note, she cannot stand on a rock)?
Input Specification:
The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=3·105,<=1<=≤<=*k*<=≤<=3·105). The next line contains *n* characters — the description of the road: the *i*-th character equals ".", if the *i*-th sector contains no rocks. Otherwise, it equals "#".
It is guaranteed that the first and the last characters equal ".".
Output Specification:
Print "YES" (without the quotes) if Ksusha can reach the end of the road, otherwise print "NO" (without the quotes).
Demo Input:
['2 1\n..\n', '5 2\n.#.#.\n', '7 3\n.#.###.\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none | ```python
n, k = map(int, input().split())
road = input()
def can_reach_end(n, k, road):
current = 0
last_no_rock_pos = -1
while current < n:
if road[current] == '#':
current = last_no_rock_pos + k + 1
if current >= n:
return 'NO'
else:
if current == n - 1:
return 'YES'
elif current - last_no_rock_pos >= k:
last_no_rock_pos = current
current += 1
return 'NO'
print(can_reach_end(n, k, road))
``` | 0 |
|
437 | C | The Child and Toy | PROGRAMMING | 1,400 | [
"graphs",
"greedy",
"sortings"
] | null | null | On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy.
The toy consists of *n* parts and *m* ropes. Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let's define an energy value of part *i* as *v**i*. The child spend *v**f*1<=+<=*v**f*2<=+<=...<=+<=*v**f**k* energy for removing part *i* where *f*1,<=*f*2,<=...,<=*f**k* are the parts that are directly connected to the *i*-th and haven't been removed.
Help the child to find out, what is the minimum total energy he should spend to remove all *n* parts. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000; 0<=≤<=*m*<=≤<=2000). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (0<=≤<=*v**i*<=≤<=105). Then followed *m* lines, each line contains two integers *x**i* and *y**i*, representing a rope from part *x**i* to part *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*; *x**i*<=≠<=*y**i*).
Consider all the parts are numbered from 1 to *n*. | Output the minimum total energy the child should spend to remove all *n* parts of the toy. | [
"4 3\n10 20 30 40\n1 4\n1 2\n2 3\n",
"4 4\n100 100 100 100\n1 2\n2 3\n2 4\n3 4\n",
"7 10\n40 10 20 10 20 80 40\n1 5\n4 7\n4 5\n5 2\n5 7\n6 4\n1 6\n1 3\n4 3\n1 4\n"
] | [
"40\n",
"400\n",
"160\n"
] | One of the optimal sequence of actions in the first sample is:
- First, remove part 3, cost of the action is 20. - Then, remove part 2, cost of the action is 10. - Next, remove part 4, cost of the action is 10. - At last, remove part 1, cost of the action is 0.
So the total energy the child paid is 20 + 10 + 10 + 0 = 40, which is the minimum.
In the second sample, the child will spend 400 no matter in what order he will remove the parts. | 1,500 | [
{
"input": "4 3\n10 20 30 40\n1 4\n1 2\n2 3",
"output": "40"
},
{
"input": "4 4\n100 100 100 100\n1 2\n2 3\n2 4\n3 4",
"output": "400"
},
{
"input": "7 10\n40 10 20 10 20 80 40\n1 5\n4 7\n4 5\n5 2\n5 7\n6 4\n1 6\n1 3\n4 3\n1 4",
"output": "160"
},
{
"input": "1 0\n23333",
"output": "0"
},
{
"input": "5 4\n1 2 2 2 2\n1 2\n1 3\n1 4\n1 5",
"output": "4"
},
{
"input": "10 30\n3 6 17 15 13 15 6 12 9 1\n3 8\n1 10\n4 7\n1 7\n3 7\n2 9\n8 10\n3 1\n3 4\n8 6\n10 3\n3 9\n2 3\n10 4\n2 10\n5 8\n9 5\n6 1\n2 1\n7 2\n7 6\n7 10\n4 8\n5 6\n3 6\n4 1\n8 9\n7 9\n4 2\n5 10",
"output": "188"
},
{
"input": "3 3\n1 1 1\n1 2\n2 3\n3 1",
"output": "3"
}
] | 1,626,176,686 | 2,147,483,647 | PyPy 3 | OK | TESTS | 29 | 140 | 23,654,400 | n, m =map(int, input().split())
v = list(map(int, input().split()))
s = 0
for _ in range(m):
x, y = map(int, input().split())
s += min(v[x-1], v[y-1])
print(s) | Title: The Child and Toy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy.
The toy consists of *n* parts and *m* ropes. Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let's define an energy value of part *i* as *v**i*. The child spend *v**f*1<=+<=*v**f*2<=+<=...<=+<=*v**f**k* energy for removing part *i* where *f*1,<=*f*2,<=...,<=*f**k* are the parts that are directly connected to the *i*-th and haven't been removed.
Help the child to find out, what is the minimum total energy he should spend to remove all *n* parts.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000; 0<=≤<=*m*<=≤<=2000). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (0<=≤<=*v**i*<=≤<=105). Then followed *m* lines, each line contains two integers *x**i* and *y**i*, representing a rope from part *x**i* to part *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*; *x**i*<=≠<=*y**i*).
Consider all the parts are numbered from 1 to *n*.
Output Specification:
Output the minimum total energy the child should spend to remove all *n* parts of the toy.
Demo Input:
['4 3\n10 20 30 40\n1 4\n1 2\n2 3\n', '4 4\n100 100 100 100\n1 2\n2 3\n2 4\n3 4\n', '7 10\n40 10 20 10 20 80 40\n1 5\n4 7\n4 5\n5 2\n5 7\n6 4\n1 6\n1 3\n4 3\n1 4\n']
Demo Output:
['40\n', '400\n', '160\n']
Note:
One of the optimal sequence of actions in the first sample is:
- First, remove part 3, cost of the action is 20. - Then, remove part 2, cost of the action is 10. - Next, remove part 4, cost of the action is 10. - At last, remove part 1, cost of the action is 0.
So the total energy the child paid is 20 + 10 + 10 + 0 = 40, which is the minimum.
In the second sample, the child will spend 400 no matter in what order he will remove the parts. | ```python
n, m =map(int, input().split())
v = list(map(int, input().split()))
s = 0
for _ in range(m):
x, y = map(int, input().split())
s += min(v[x-1], v[y-1])
print(s)
``` | 3 |
|
550 | A | Two Substrings | PROGRAMMING | 1,500 | [
"brute force",
"dp",
"greedy",
"implementation",
"strings"
] | null | null | You are given string *s*. Your task is to determine if the given string *s* contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). | The only line of input contains a string *s* of length between 1 and 105 consisting of uppercase Latin letters. | Print "YES" (without the quotes), if string *s* contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. | [
"ABA\n",
"BACFAB\n",
"AXBYBXA\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | 1,000 | [
{
"input": "ABA",
"output": "NO"
},
{
"input": "BACFAB",
"output": "YES"
},
{
"input": "AXBYBXA",
"output": "NO"
},
{
"input": "ABABAB",
"output": "YES"
},
{
"input": "BBBBBBBBBB",
"output": "NO"
},
{
"input": "ABBA",
"output": "YES"
},
{
"input": "ABAXXXAB",
"output": "YES"
},
{
"input": "TESTABAXXABTEST",
"output": "YES"
},
{
"input": "A",
"output": "NO"
},
{
"input": "B",
"output": "NO"
},
{
"input": "X",
"output": "NO"
},
{
"input": "BA",
"output": "NO"
},
{
"input": "AB",
"output": "NO"
},
{
"input": "AA",
"output": "NO"
},
{
"input": "BB",
"output": "NO"
},
{
"input": "BAB",
"output": "NO"
},
{
"input": "AAB",
"output": "NO"
},
{
"input": "BAA",
"output": "NO"
},
{
"input": "ABB",
"output": "NO"
},
{
"input": "BBA",
"output": "NO"
},
{
"input": "AAA",
"output": "NO"
},
{
"input": "BBB",
"output": "NO"
},
{
"input": "AXBXBXA",
"output": "NO"
},
{
"input": "SKDSKDJABSDBADKFJDK",
"output": "YES"
},
{
"input": "ABAXXBBXXAA",
"output": "NO"
},
{
"input": "ABAB",
"output": "NO"
},
{
"input": "BABA",
"output": "NO"
},
{
"input": "AAAB",
"output": "NO"
},
{
"input": "AAAA",
"output": "NO"
},
{
"input": "AABA",
"output": "NO"
},
{
"input": "ABAA",
"output": "NO"
},
{
"input": "BAAA",
"output": "NO"
},
{
"input": "AABB",
"output": "NO"
},
{
"input": "BAAB",
"output": "YES"
},
{
"input": "BBAA",
"output": "NO"
},
{
"input": "BBBA",
"output": "NO"
},
{
"input": "BBAB",
"output": "NO"
},
{
"input": "BABB",
"output": "NO"
},
{
"input": "ABBB",
"output": "NO"
},
{
"input": "BBBB",
"output": "NO"
},
{
"input": "BABAB",
"output": "YES"
},
{
"input": "ABABA",
"output": "YES"
},
{
"input": "AAABAAACBBBC",
"output": "NO"
},
{
"input": "AABBBACBBBBBBAACBCCACBBAABBBBBCAACABAACABCACCBCBCCCBCBCABCBBCCBCBBAACBACAABACBBCACCBCCACCABBCBABBAAC",
"output": "YES"
},
{
"input": "CBBABDDBBADAC",
"output": "YES"
},
{
"input": "ABYFAB",
"output": "NO"
},
{
"input": "BABXXAB",
"output": "YES"
},
{
"input": "ABAXAB",
"output": "YES"
},
{
"input": "ABABXAB",
"output": "YES"
},
{
"input": "ABXABXABXABXABXBAXBAXBAXBA",
"output": "YES"
},
{
"input": "QQQQQQQQQABABQQQQQQQQ",
"output": "NO"
}
] | 1,683,879,504 | 2,147,483,647 | PyPy 3 | OK | TESTS | 85 | 93 | 614,400 | def contains_AB_and_BA(s):
if "AB" in s and "BA" in s[s.index("AB")+2:]:
return True
if "BA" in s and "AB" in s[s.index("BA")+2:]:
return True
return False
s = input()
if contains_AB_and_BA(s):
print("Yes")
else:
print("No") | Title: Two Substrings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given string *s*. Your task is to determine if the given string *s* contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).
Input Specification:
The only line of input contains a string *s* of length between 1 and 105 consisting of uppercase Latin letters.
Output Specification:
Print "YES" (without the quotes), if string *s* contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.
Demo Input:
['ABA\n', 'BACFAB\n', 'AXBYBXA\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | ```python
def contains_AB_and_BA(s):
if "AB" in s and "BA" in s[s.index("AB")+2:]:
return True
if "BA" in s and "AB" in s[s.index("BA")+2:]:
return True
return False
s = input()
if contains_AB_and_BA(s):
print("Yes")
else:
print("No")
``` | 3 |
|
16 | A | Flag | PROGRAMMING | 800 | [
"implementation"
] | A. Flag | 2 | 64 | According to a new ISO standard, a flag of every country should have a chequered field *n*<=×<=*m*, each square should be of one of 10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard. | The first line of the input contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), *n* — the amount of rows, *m* — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following *n* lines contain *m* characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square. | Output YES, if the flag meets the new ISO standard, and NO otherwise. | [
"3 3\n000\n111\n222\n",
"3 3\n000\n000\n111\n",
"3 3\n000\n111\n002\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 0 | [
{
"input": "3 3\n000\n111\n222",
"output": "YES"
},
{
"input": "3 3\n000\n000\n111",
"output": "NO"
},
{
"input": "3 3\n000\n111\n002",
"output": "NO"
},
{
"input": "10 10\n2222222222\n5555555555\n0000000000\n4444444444\n1111111111\n3333333393\n3333333333\n5555555555\n0000000000\n8888888888",
"output": "NO"
},
{
"input": "10 13\n4442444444444\n8888888888888\n6666666666666\n0000000000000\n3333333333333\n4444444444444\n7777777777777\n8388888888888\n1111111111111\n5555555555555",
"output": "NO"
},
{
"input": "10 8\n33333333\n44444444\n11111115\n81888888\n44444444\n11111111\n66666666\n33330333\n33333333\n33333333",
"output": "NO"
},
{
"input": "5 5\n88888\n44444\n66666\n55555\n88888",
"output": "YES"
},
{
"input": "20 19\n1111111111111111111\n5555555555555555555\n0000000000000000000\n3333333333333333333\n1111111111111111111\n2222222222222222222\n4444444444444444444\n5555555555555555555\n0000000000000000000\n4444444444444444444\n0000000000000000000\n5555555555555555555\n7777777777777777777\n9999999999999999999\n2222222222222222222\n4444444444444444444\n1111111111111111111\n6666666666666666666\n7777777777777777777\n2222222222222222222",
"output": "YES"
},
{
"input": "1 100\n8888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888",
"output": "YES"
},
{
"input": "100 1\n5\n7\n9\n4\n7\n2\n5\n1\n6\n7\n2\n7\n6\n8\n7\n4\n0\n2\n9\n8\n9\n1\n6\n4\n3\n4\n7\n1\n9\n3\n0\n8\n3\n1\n7\n5\n3\n9\n5\n1\n3\n5\n8\n1\n9\n3\n9\n0\n6\n0\n7\n6\n5\n2\n8\n3\n7\n6\n5\n1\n8\n3\n6\n9\n6\n0\n5\n8\n5\n2\n9\n1\n0\n1\n8\n3\n2\n1\n0\n3\n9\n0\n5\n1\n0\n4\n9\n3\n0\n4\n8\n4\n8\n6\n3\n0\n4\n6\n8\n4",
"output": "YES"
},
{
"input": "1 1\n2",
"output": "YES"
},
{
"input": "1 100\n1111111111111111111111111111111111111111111111111111111111111111111111111111181111111111111111111111",
"output": "NO"
},
{
"input": "100 1\n3\n6\n4\n3\n0\n2\n8\n7\n3\n2\n1\n7\n1\n3\n2\n3\n6\n9\n0\n8\n5\n9\n7\n9\n2\n1\n4\n5\n1\n9\n2\n5\n1\n4\n6\n4\n9\n1\n0\n2\n1\n4\n7\n1\n4\n8\n0\n9\n2\n1\n6\n2\n8\n6\n9\n5\n8\n6\n4\n5\n9\n2\n7\n4\n1\n5\n8\n0\n9\n5\n4\n6\n5\n0\n6\n3\n6\n9\n7\n2\n0\n9\n7\n3\n2\n4\n9\n4\n7\n1\n2\n3\n1\n7\n9\n1\n9\n0\n4\n0",
"output": "YES"
}
] | 1,631,912,913 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | a,b=map(int, input().split(' ')
m= [[int(input()) for x in range(a)] for y in range(b)]
flag= True
for i in range(a):
if m[i]!=m[i+1]:
flag= True
else:
flag= False
break
if flag==True:
print("Yes")
else:
print("No") | Title: Flag
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
According to a new ISO standard, a flag of every country should have a chequered field *n*<=×<=*m*, each square should be of one of 10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard.
Input Specification:
The first line of the input contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), *n* — the amount of rows, *m* — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following *n* lines contain *m* characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.
Output Specification:
Output YES, if the flag meets the new ISO standard, and NO otherwise.
Demo Input:
['3 3\n000\n111\n222\n', '3 3\n000\n000\n111\n', '3 3\n000\n111\n002\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
a,b=map(int, input().split(' ')
m= [[int(input()) for x in range(a)] for y in range(b)]
flag= True
for i in range(a):
if m[i]!=m[i+1]:
flag= True
else:
flag= False
break
if flag==True:
print("Yes")
else:
print("No")
``` | -1 |
725 | A | Jumping Ball | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper. | Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position. | [
"4\n<<><\n",
"5\n>>>>>\n",
"4\n>><<\n"
] | [
"2",
"5",
"0"
] | In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field. | 500 | [
{
"input": "4\n<<><",
"output": "2"
},
{
"input": "5\n>>>>>",
"output": "5"
},
{
"input": "4\n>><<",
"output": "0"
},
{
"input": "3\n<<>",
"output": "3"
},
{
"input": "3\n<<<",
"output": "3"
},
{
"input": "3\n><<",
"output": "0"
},
{
"input": "1\n<",
"output": "1"
},
{
"input": "2\n<>",
"output": "2"
},
{
"input": "3\n<>>",
"output": "3"
},
{
"input": "3\n><>",
"output": "1"
},
{
"input": "2\n><",
"output": "0"
},
{
"input": "2\n>>",
"output": "2"
},
{
"input": "2\n<<",
"output": "2"
},
{
"input": "1\n>",
"output": "1"
},
{
"input": "3\n>><",
"output": "0"
},
{
"input": "3\n>>>",
"output": "3"
},
{
"input": "3\n<><",
"output": "1"
},
{
"input": "10\n<<<><<<>>>",
"output": "6"
},
{
"input": "20\n><><<><<<>>>>>>>>>>>",
"output": "11"
},
{
"input": "20\n<<<<<<<<<<><<<<>>>>>",
"output": "15"
},
{
"input": "50\n<<<<<<<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "50"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<>><<>><<<<<>><>><<<>><><<>>><<>>><<<<><><><<><<<<><>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "49"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<>>>><<>><>><>><<><><><><>>>><><<<>>>><<<>>>>>>><><",
"output": "50"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<",
"output": "100"
},
{
"input": "100\n>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "100"
},
{
"input": "12\n<<>><<>><<>>",
"output": "4"
},
{
"input": "6\n<<><>>",
"output": "4"
},
{
"input": "6\n><>>>>",
"output": "4"
},
{
"input": "8\n>>>><<<>",
"output": "1"
},
{
"input": "4\n<><>",
"output": "2"
},
{
"input": "4\n><><",
"output": "0"
},
{
"input": "7\n<<>>><>",
"output": "3"
},
{
"input": "10\n><><>>>>>>",
"output": "6"
},
{
"input": "5\n<><>>",
"output": "3"
},
{
"input": "12\n<><<<<>>>>>>",
"output": "7"
},
{
"input": "6\n<>><<>",
"output": "2"
},
{
"input": "6\n>>><>>",
"output": "2"
},
{
"input": "10\n><><>>>><>",
"output": "1"
},
{
"input": "5\n><>>>",
"output": "3"
},
{
"input": "5\n<<><>",
"output": "3"
},
{
"input": "5\n<><<<",
"output": "1"
},
{
"input": "4\n<><<",
"output": "1"
},
{
"input": "8\n<<>><<>>",
"output": "4"
},
{
"input": "7\n<<><>>>",
"output": "5"
},
{
"input": "5\n><<>>",
"output": "2"
},
{
"input": "10\n<<<<<>>>>>",
"output": "10"
},
{
"input": "6\n><<<<<",
"output": "0"
},
{
"input": "8\n<<><><>>",
"output": "4"
},
{
"input": "10\n<<<<><<<><",
"output": "4"
},
{
"input": "12\n<<<>>>><<>>>",
"output": "6"
},
{
"input": "4\n><>>",
"output": "2"
},
{
"input": "11\n<<><<>><<>>",
"output": "4"
}
] | 1,663,609,076 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 2,000 | 716,800 | n=int(input())
s=input()
c=0
for x in range(n):
pos=x
co=0
while True:
co+=1
if s[pos]==">":
pos+=1
else:
pos-=1
if pos<0 or pos>=n:
c+=1
break
if co>=n:
break
print(c) | Title: Jumping Ball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper.
Output Specification:
Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position.
Demo Input:
['4\n<<><\n', '5\n>>>>>\n', '4\n>><<\n']
Demo Output:
['2', '5', '0']
Note:
In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field. | ```python
n=int(input())
s=input()
c=0
for x in range(n):
pos=x
co=0
while True:
co+=1
if s[pos]==">":
pos+=1
else:
pos-=1
if pos<0 or pos>=n:
c+=1
break
if co>=n:
break
print(c)
``` | 0 |
|
940 | A | Points on the line | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"sortings"
] | null | null | We've got no test cases. A big olympiad is coming up. But the problemsetters' number one priority should be adding another problem to the round.
The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset {1,<=3,<=2,<=1} is 2.
Diameter of multiset consisting of one point is 0.
You are given *n* points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed *d*? | The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=100,<=0<=≤<=*d*<=≤<=100) — the amount of points and the maximum allowed diameter respectively.
The second line contains *n* space separated integers (1<=≤<=*x**i*<=≤<=100) — the coordinates of the points. | Output a single integer — the minimum number of points you have to remove. | [
"3 1\n2 1 4\n",
"3 0\n7 7 7\n",
"6 3\n1 3 4 6 9 10\n"
] | [
"1\n",
"0\n",
"3\n"
] | In the first test case the optimal strategy is to remove the point with coordinate 4. The remaining points will have coordinates 1 and 2, so the diameter will be equal to 2 - 1 = 1.
In the second test case the diameter is equal to 0, so its is unnecessary to remove any points.
In the third test case the optimal strategy is to remove points with coordinates 1, 9 and 10. The remaining points will have coordinates 3, 4 and 6, so the diameter will be equal to 6 - 3 = 3. | 500 | [
{
"input": "3 1\n2 1 4",
"output": "1"
},
{
"input": "3 0\n7 7 7",
"output": "0"
},
{
"input": "6 3\n1 3 4 6 9 10",
"output": "3"
},
{
"input": "11 5\n10 11 12 13 14 15 16 17 18 19 20",
"output": "5"
},
{
"input": "1 100\n1",
"output": "0"
},
{
"input": "100 10\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "84"
},
{
"input": "100 70\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "27"
},
{
"input": "1 10\n25",
"output": "0"
},
{
"input": "70 80\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70",
"output": "0"
},
{
"input": "3 1\n25 26 27",
"output": "1"
},
{
"input": "100 5\n51 56 52 60 52 53 52 60 56 54 55 50 53 51 57 53 52 54 54 52 51 55 50 56 60 51 58 50 60 59 50 54 60 55 55 57 54 59 59 55 55 52 56 57 59 54 53 57 52 50 50 55 59 54 54 56 51 58 52 51 56 56 58 56 54 54 57 52 51 58 56 57 54 59 58 53 50 52 50 60 57 51 54 59 54 54 52 55 53 55 51 53 52 54 51 56 55 53 58 56",
"output": "34"
},
{
"input": "100 11\n44 89 57 64 94 96 73 96 55 52 91 73 73 93 51 62 63 85 43 75 60 78 98 55 80 84 65 75 61 88 62 71 53 57 94 85 60 96 66 96 61 72 97 64 51 44 63 82 67 86 60 57 74 85 57 79 61 94 86 78 84 56 60 75 91 91 92 62 89 85 79 57 76 97 65 56 46 78 51 69 50 52 85 80 76 71 81 51 90 71 77 60 63 62 84 59 79 84 69 81",
"output": "70"
},
{
"input": "100 0\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "96"
},
{
"input": "100 100\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "0"
},
{
"input": "76 32\n50 53 69 58 55 39 40 42 40 55 58 73 55 72 75 44 45 55 46 60 60 42 41 64 77 39 68 51 61 49 38 41 56 57 64 43 78 36 39 63 40 66 52 76 39 68 39 73 40 68 54 60 35 67 69 52 58 52 38 63 69 38 69 60 73 64 65 41 59 55 37 57 40 34 35 35",
"output": "13"
},
{
"input": "100 1\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "93"
},
{
"input": "100 5\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "89"
},
{
"input": "98 64\n2 29 36 55 58 15 25 33 7 16 61 1 4 24 63 26 36 16 16 3 57 39 56 7 11 24 20 12 22 10 56 5 11 39 61 52 27 54 21 6 61 36 40 52 54 5 15 52 58 23 45 39 65 16 27 40 13 64 47 24 51 29 9 18 49 49 8 47 2 64 7 63 49 10 20 26 34 3 45 66 8 46 16 32 16 38 3 6 15 17 35 48 36 5 57 29 61 15",
"output": "1"
},
{
"input": "100 56\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "43"
},
{
"input": "100 0\n14 13 14 13 14 13 13 13 13 14 13 13 14 14 13 14 14 14 14 13 13 13 14 13 13 14 14 14 14 14 14 13 13 13 13 14 13 14 13 14 13 14 14 14 14 13 13 14 14 13 13 13 13 14 13 14 13 14 13 14 13 13 13 14 13 13 14 13 14 14 13 13 13 14 14 14 14 13 13 14 14 14 14 14 14 14 13 14 13 13 13 14 14 13 13 13 13 13 14 14",
"output": "50"
},
{
"input": "100 0\n14 17 18 22 19 18 19 21 19 19 22 22 19 21 24 23 24 19 25 24 24 21 20 13 26 18 17 15 25 13 17 20 20 21 13 22 27 15 18 27 19 15 16 25 18 17 18 22 19 17 18 24 14 16 18 16 22 16 17 27 18 17 18 24 22 13 14 20 23 19 16 21 19 13 14 14 25 15 27 24 26 22 16 20 16 14 21 27 15 23 23 24 27 14 24 17 19 24 15 27",
"output": "89"
},
{
"input": "100 100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0"
},
{
"input": "1 100\n22",
"output": "0"
},
{
"input": "1 0\n22",
"output": "0"
},
{
"input": "1 99\n99",
"output": "0"
},
{
"input": "1 5\n6",
"output": "0"
},
{
"input": "3 1\n10 20 30",
"output": "2"
},
{
"input": "3 0\n1 2 3",
"output": "2"
},
{
"input": "3 2\n1 50 99",
"output": "2"
},
{
"input": "7 4\n1 3 4 9 10 11 12",
"output": "3"
},
{
"input": "2 5\n67 23",
"output": "1"
},
{
"input": "4 2\n1 4 7 9",
"output": "2"
},
{
"input": "2 0\n1 2",
"output": "1"
},
{
"input": "8 1\n3 3 3 5 5 5 5 5",
"output": "3"
},
{
"input": "5 1\n3 5 5 5 6",
"output": "1"
}
] | 1,554,867,283 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 109 | 0 | import sys
def minp():
return sys.stdin.readline().strip()
r = 0
n, d = map(int,minp().split())
a = sorted(map(int,minp().split()))
l = 0
r = 0
b = []
while l < n:
while (r < n-1) and (a[r+1]-a[l] <= d):
r += 1
b.append(r-l+1)
l += 1
print(n-max(b)) | Title: Points on the line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We've got no test cases. A big olympiad is coming up. But the problemsetters' number one priority should be adding another problem to the round.
The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset {1,<=3,<=2,<=1} is 2.
Diameter of multiset consisting of one point is 0.
You are given *n* points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed *d*?
Input Specification:
The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=100,<=0<=≤<=*d*<=≤<=100) — the amount of points and the maximum allowed diameter respectively.
The second line contains *n* space separated integers (1<=≤<=*x**i*<=≤<=100) — the coordinates of the points.
Output Specification:
Output a single integer — the minimum number of points you have to remove.
Demo Input:
['3 1\n2 1 4\n', '3 0\n7 7 7\n', '6 3\n1 3 4 6 9 10\n']
Demo Output:
['1\n', '0\n', '3\n']
Note:
In the first test case the optimal strategy is to remove the point with coordinate 4. The remaining points will have coordinates 1 and 2, so the diameter will be equal to 2 - 1 = 1.
In the second test case the diameter is equal to 0, so its is unnecessary to remove any points.
In the third test case the optimal strategy is to remove points with coordinates 1, 9 and 10. The remaining points will have coordinates 3, 4 and 6, so the diameter will be equal to 6 - 3 = 3. | ```python
import sys
def minp():
return sys.stdin.readline().strip()
r = 0
n, d = map(int,minp().split())
a = sorted(map(int,minp().split()))
l = 0
r = 0
b = []
while l < n:
while (r < n-1) and (a[r+1]-a[l] <= d):
r += 1
b.append(r-l+1)
l += 1
print(n-max(b))
``` | 3 |
|
226 | B | Naughty Stone Piles | PROGRAMMING | 1,900 | [
"greedy"
] | null | null | There are *n* piles of stones of sizes *a*1,<=*a*2,<=...,<=*a**n* lying on the table in front of you.
During one move you can take one pile and add it to the other. As you add pile *i* to pile *j*, the size of pile *j* increases by the current size of pile *i*, and pile *i* stops existing. The cost of the adding operation equals the size of the added pile.
Your task is to determine the minimum cost at which you can gather all stones in one pile.
To add some challenge, the stone piles built up conspiracy and decided that each pile will let you add to it not more than *k* times (after that it can only be added to another pile).
Moreover, the piles decided to puzzle you completely and told you *q* variants (not necessarily distinct) of what *k* might equal.
Your task is to find the minimum cost for each of *q* variants. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of stone piles. The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial sizes of the stone piles.
The third line contains integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries. The last line contains *q* space-separated integers *k*1,<=*k*2,<=...,<=*k**q* (1<=≤<=*k**i*<=≤<=105) — the values of number *k* for distinct queries. Note that numbers *k**i* can repeat. | Print *q* whitespace-separated integers — the answers to the queries in the order, in which the queries are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"5\n2 3 4 1 1\n2\n2 3\n"
] | [
"9 8 "
] | In the first sample one way to get the optimal answer goes like this: we add in turns the 4-th and the 5-th piles to the 2-nd one; then we add the 1-st pile to the 3-rd one; we add the 2-nd pile to the 3-rd one. The first two operations cost 1 each; the third one costs 2, the fourth one costs 5 (the size of the 2-nd pile after the first two operations is not 3, it already is 5).
In the second sample you can add the 2-nd pile to the 3-rd one (the operations costs 3); then the 1-st one to the 3-th one (the cost is 2); then the 5-th one to the 4-th one (the costs is 1); and at last, the 4-th one to the 3-rd one (the cost is 2). | 1,000 | [
{
"input": "5\n2 3 4 1 1\n2\n2 3",
"output": "9 8 "
},
{
"input": "2\n2 9\n5\n4 10 7 3 4",
"output": "2 2 2 2 2 "
},
{
"input": "1\n7\n4\n6 2 3 3",
"output": "0 0 0 0 "
},
{
"input": "2\n7 10\n2\n2 4",
"output": "7 7 "
},
{
"input": "1\n10\n5\n5 3 7 7 1",
"output": "0 0 0 0 0 "
},
{
"input": "1\n2\n5\n7 3 9 8 1",
"output": "0 0 0 0 0 "
},
{
"input": "4\n8 10 4 4\n3\n7 8 1",
"output": "16 16 28 "
},
{
"input": "2\n7 9\n1\n9",
"output": "7 "
},
{
"input": "3\n4 5 4\n2\n10 2",
"output": "8 8 "
},
{
"input": "3\n1 6 8\n1\n6",
"output": "7 "
},
{
"input": "2\n9 3\n1\n6",
"output": "3 "
},
{
"input": "5\n9 5 7 3 3\n1\n3",
"output": "21 "
},
{
"input": "2\n7 4\n1\n7",
"output": "4 "
},
{
"input": "4\n7 4 1 7\n3\n6 8 3",
"output": "12 12 12 "
},
{
"input": "3\n3 7 3\n1\n5",
"output": "6 "
},
{
"input": "1\n3\n1\n2",
"output": "0 "
},
{
"input": "1\n1\n3\n2 1 10",
"output": "0 0 0 "
}
] | 1,650,741,529 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | n = int(input())
a = list(map(int, input().split()))
a.sort()
for i in range(1, n):
a[i] += a[i - 1];
q = int(input())
queries = list(map(int, input().split()))
for qq in range(q):
k = queries[qq]
if k == 1:
print(a[n - 2])
continue
d = k
s = 0
i = n - 2
while i >= 0:
s += a[i]
i -= d
d *= k
print(s)
| Title: Naughty Stone Piles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* piles of stones of sizes *a*1,<=*a*2,<=...,<=*a**n* lying on the table in front of you.
During one move you can take one pile and add it to the other. As you add pile *i* to pile *j*, the size of pile *j* increases by the current size of pile *i*, and pile *i* stops existing. The cost of the adding operation equals the size of the added pile.
Your task is to determine the minimum cost at which you can gather all stones in one pile.
To add some challenge, the stone piles built up conspiracy and decided that each pile will let you add to it not more than *k* times (after that it can only be added to another pile).
Moreover, the piles decided to puzzle you completely and told you *q* variants (not necessarily distinct) of what *k* might equal.
Your task is to find the minimum cost for each of *q* variants.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of stone piles. The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the initial sizes of the stone piles.
The third line contains integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries. The last line contains *q* space-separated integers *k*1,<=*k*2,<=...,<=*k**q* (1<=≤<=*k**i*<=≤<=105) — the values of number *k* for distinct queries. Note that numbers *k**i* can repeat.
Output Specification:
Print *q* whitespace-separated integers — the answers to the queries in the order, in which the queries are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['5\n2 3 4 1 1\n2\n2 3\n']
Demo Output:
['9 8 ']
Note:
In the first sample one way to get the optimal answer goes like this: we add in turns the 4-th and the 5-th piles to the 2-nd one; then we add the 1-st pile to the 3-rd one; we add the 2-nd pile to the 3-rd one. The first two operations cost 1 each; the third one costs 2, the fourth one costs 5 (the size of the 2-nd pile after the first two operations is not 3, it already is 5).
In the second sample you can add the 2-nd pile to the 3-rd one (the operations costs 3); then the 1-st one to the 3-th one (the cost is 2); then the 5-th one to the 4-th one (the costs is 1); and at last, the 4-th one to the 3-rd one (the cost is 2). | ```python
n = int(input())
a = list(map(int, input().split()))
a.sort()
for i in range(1, n):
a[i] += a[i - 1];
q = int(input())
queries = list(map(int, input().split()))
for qq in range(q):
k = queries[qq]
if k == 1:
print(a[n - 2])
continue
d = k
s = 0
i = n - 2
while i >= 0:
s += a[i]
i -= d
d *= k
print(s)
``` | 0 |
|
144 | A | Arrival of the General | PROGRAMMING | 800 | [
"implementation"
] | null | null | A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all *n* squad soldiers to line up on the parade ground.
By the military charter the soldiers should stand in the order of non-increasing of their height. But as there's virtually no time to do that, the soldiers lined up in the arbitrary order. However, the general is rather short-sighted and he thinks that the soldiers lined up correctly if the first soldier in the line has the maximum height and the last soldier has the minimum height. Please note that the way other solders are positioned does not matter, including the case when there are several soldiers whose height is maximum or minimum. Only the heights of the first and the last soldier are important.
For example, the general considers the sequence of heights (4, 3, 4, 2, 1, 1) correct and the sequence (4, 3, 1, 2, 2) wrong.
Within one second the colonel can swap any two neighboring soldiers. Help him count the minimum time needed to form a line-up which the general will consider correct. | The first input line contains the only integer *n* (2<=≤<=*n*<=≤<=100) which represents the number of soldiers in the line. The second line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) the values of the soldiers' heights in the order of soldiers' heights' increasing in the order from the beginning of the line to its end. The numbers are space-separated. Numbers *a*1,<=*a*2,<=...,<=*a**n* are not necessarily different. | Print the only integer — the minimum number of seconds the colonel will need to form a line-up the general will like. | [
"4\n33 44 11 22\n",
"7\n10 10 58 31 63 40 76\n"
] | [
"2\n",
"10\n"
] | In the first sample the colonel will need to swap the first and second soldier and then the third and fourth soldier. That will take 2 seconds. The resulting position of the soldiers is (44, 33, 22, 11).
In the second sample the colonel may swap the soldiers in the following sequence:
1. (10, 10, 58, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 76, 40) 1. (10, 58, 10, 31, 76, 63, 40) 1. (10, 58, 31, 10, 76, 63, 40) 1. (10, 58, 31, 76, 10, 63, 40) 1. (10, 58, 31, 76, 63, 10, 40) 1. (10, 58, 76, 31, 63, 10, 40) 1. (10, 76, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 40, 10) | 500 | [
{
"input": "4\n33 44 11 22",
"output": "2"
},
{
"input": "7\n10 10 58 31 63 40 76",
"output": "10"
},
{
"input": "2\n88 89",
"output": "1"
},
{
"input": "5\n100 95 100 100 88",
"output": "0"
},
{
"input": "7\n48 48 48 48 45 45 45",
"output": "0"
},
{
"input": "10\n68 47 67 29 63 71 71 65 54 56",
"output": "10"
},
{
"input": "15\n77 68 96 60 92 75 61 60 66 79 80 65 60 95 92",
"output": "4"
},
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "20\n30 30 30 14 30 14 30 30 30 14 30 14 14 30 14 14 30 14 14 14",
"output": "0"
},
{
"input": "35\n37 41 46 39 47 39 44 47 44 42 44 43 47 39 46 39 38 42 39 37 40 44 41 42 41 42 39 42 36 36 42 36 42 42 42",
"output": "7"
},
{
"input": "40\n99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 98 99 99 99 99 99 99 99 99 100 99 99 99 99 99 99",
"output": "47"
},
{
"input": "50\n48 52 44 54 53 56 62 49 39 41 53 39 40 64 53 50 62 48 40 52 51 48 40 52 61 62 62 61 48 64 55 57 56 40 48 58 41 60 60 56 64 50 64 45 48 45 46 63 59 57",
"output": "50"
},
{
"input": "57\n7 24 17 19 6 19 10 11 12 22 14 5 5 11 13 10 24 19 24 24 24 11 21 20 4 14 24 24 18 13 24 3 20 3 3 3 3 9 3 9 22 22 16 3 3 3 15 11 3 3 8 17 10 13 3 14 13",
"output": "3"
},
{
"input": "65\n58 50 35 44 35 37 36 58 38 36 58 56 56 49 48 56 58 43 40 44 52 44 58 58 57 50 43 35 55 39 38 49 53 56 50 42 41 56 34 57 49 38 34 51 56 38 58 40 53 46 48 34 38 43 49 49 58 56 41 43 44 34 38 48 36",
"output": "3"
},
{
"input": "69\n70 48 49 48 49 71 48 53 55 69 48 53 54 58 53 63 48 48 69 67 72 75 71 75 74 74 57 63 65 60 48 48 65 48 48 51 50 49 62 53 76 68 76 56 76 76 64 76 76 57 61 76 73 51 59 76 65 50 69 50 76 67 76 63 62 74 74 58 73",
"output": "73"
},
{
"input": "75\n70 65 64 71 71 64 71 64 68 71 65 64 65 68 71 66 66 69 68 63 69 65 71 69 68 68 71 67 71 65 65 65 71 71 65 69 63 66 62 67 64 63 62 64 67 65 62 69 62 64 69 62 67 64 67 70 64 63 64 64 69 62 62 64 70 62 62 68 67 69 62 64 66 70 68",
"output": "7"
},
{
"input": "84\n92 95 84 85 94 80 90 86 80 92 95 84 86 83 86 83 93 91 95 92 84 88 82 84 84 84 80 94 93 80 94 80 95 83 85 80 95 95 80 84 86 92 83 81 90 87 81 89 92 93 80 87 90 85 93 85 93 94 93 89 94 83 93 91 80 83 90 94 95 80 95 92 85 84 93 94 94 82 91 95 95 89 85 94",
"output": "15"
},
{
"input": "90\n86 87 72 77 82 71 75 78 61 67 79 90 64 94 94 74 85 87 73 76 71 71 60 69 77 73 76 80 82 57 62 57 57 83 76 72 75 87 72 94 77 85 59 82 86 69 62 80 95 73 83 94 79 85 91 68 85 74 93 95 68 75 89 93 83 78 95 78 83 77 81 85 66 92 63 65 75 78 67 91 77 74 59 86 77 76 90 67 70 64",
"output": "104"
},
{
"input": "91\n94 98 96 94 95 98 98 95 98 94 94 98 95 95 99 97 97 94 95 98 94 98 96 98 96 98 97 95 94 94 94 97 94 96 98 98 98 94 96 95 94 95 97 97 97 98 94 98 96 95 98 96 96 98 94 97 96 98 97 95 97 98 94 95 94 94 97 94 96 97 97 93 94 95 95 94 96 98 97 96 94 98 98 96 96 96 96 96 94 96 97",
"output": "33"
},
{
"input": "92\n44 28 32 29 41 41 36 39 40 39 41 35 41 28 35 27 41 34 28 38 43 43 41 38 27 26 28 36 30 29 39 32 35 35 32 30 39 30 37 27 41 41 28 30 43 31 35 33 36 28 44 40 41 35 31 42 37 38 37 34 39 40 27 40 33 33 44 43 34 33 34 34 35 38 38 37 30 39 35 41 45 42 41 32 33 33 31 30 43 41 43 43",
"output": "145"
},
{
"input": "93\n46 32 52 36 39 30 57 63 63 30 32 44 27 59 46 38 40 45 44 62 35 36 51 48 39 58 36 51 51 51 48 58 59 36 29 35 31 49 64 60 34 38 42 56 33 42 52 31 63 34 45 51 35 45 33 53 33 62 31 38 66 29 51 54 28 61 32 45 57 41 36 34 47 36 31 28 67 48 52 46 32 40 64 58 27 53 43 57 34 66 43 39 26",
"output": "76"
},
{
"input": "94\n56 55 54 31 32 42 46 29 24 54 40 40 20 45 35 56 32 33 51 39 26 56 21 56 51 27 29 39 56 52 54 43 43 55 48 51 44 49 52 49 23 19 19 28 20 26 45 33 35 51 42 36 25 25 38 23 21 35 54 50 41 20 37 28 42 20 22 43 37 34 55 21 24 38 19 41 45 34 19 33 44 54 38 31 23 53 35 32 47 40 39 31 20 34",
"output": "15"
},
{
"input": "95\n57 71 70 77 64 64 76 81 81 58 63 75 81 77 71 71 71 60 70 70 69 67 62 64 78 64 69 62 76 76 57 70 68 77 70 68 73 77 79 73 60 57 69 60 74 65 58 75 75 74 73 73 65 75 72 57 81 62 62 70 67 58 76 57 79 81 68 64 58 77 70 59 79 64 80 58 71 59 81 71 80 64 78 80 78 65 70 68 78 80 57 63 64 76 81",
"output": "11"
},
{
"input": "96\n96 95 95 95 96 97 95 97 96 95 98 96 97 95 98 96 98 96 98 96 98 95 96 95 95 95 97 97 95 95 98 98 95 96 96 95 97 96 98 96 95 97 97 95 97 97 95 94 96 96 97 96 97 97 96 94 94 97 95 95 95 96 95 96 95 97 97 95 97 96 95 94 97 97 97 96 97 95 96 94 94 95 97 94 94 97 97 97 95 97 97 95 94 96 95 95",
"output": "13"
},
{
"input": "97\n14 15 12 12 13 15 12 15 12 12 12 12 12 14 15 15 13 12 15 15 12 12 12 13 14 15 15 13 14 15 14 14 14 14 12 13 12 13 13 12 15 12 13 13 15 12 15 13 12 13 13 13 14 13 12 15 14 13 14 15 13 14 14 13 14 12 15 12 14 12 13 14 15 14 13 15 13 12 15 15 15 13 15 15 13 14 16 16 16 13 15 13 15 14 15 15 15",
"output": "104"
},
{
"input": "98\n37 69 35 70 58 69 36 47 41 63 60 54 49 35 55 50 35 53 52 43 35 41 40 49 38 35 48 70 42 35 35 65 56 54 44 59 59 48 51 49 59 67 35 60 69 35 58 50 35 44 48 69 41 58 44 45 35 47 70 61 49 47 37 39 35 51 44 70 72 65 36 41 63 63 48 66 45 50 50 71 37 52 72 67 72 39 72 39 36 64 48 72 69 49 45 72 72 67",
"output": "100"
},
{
"input": "99\n31 31 16 15 19 31 19 22 29 27 12 22 28 30 25 33 26 25 19 22 34 21 17 33 31 22 16 26 22 30 31 17 13 33 13 17 28 25 18 33 27 22 31 22 13 27 20 22 23 15 24 32 29 13 16 20 32 33 14 33 19 27 16 28 25 17 17 28 18 26 32 33 19 23 30 13 14 23 24 28 14 28 22 20 30 14 24 23 17 29 18 28 29 21 28 18 16 24 32",
"output": "107"
},
{
"input": "100\n37 54 39 29 32 49 21 13 34 21 16 42 34 27 16 26 7 34 51 9 11 27 16 40 36 7 48 52 30 42 42 52 51 11 32 26 6 7 28 54 48 51 6 54 42 20 51 48 46 4 4 31 47 6 9 16 8 23 36 50 49 30 47 37 45 24 48 26 8 50 18 24 27 13 39 7 37 34 5 43 42 19 34 23 52 38 35 6 29 11 53 49 21 29 36 22 45 33 51 22",
"output": "50"
},
{
"input": "100\n41 5 20 23 5 47 28 28 64 5 12 14 34 56 22 48 6 52 39 55 62 40 24 54 39 28 56 27 35 51 14 24 38 10 55 20 39 62 52 2 30 55 49 48 15 23 59 44 31 28 66 29 62 59 31 13 28 12 50 45 41 47 29 27 64 66 14 39 62 22 44 63 24 5 63 17 42 7 25 41 49 33 31 51 63 3 37 66 19 9 2 40 16 45 56 21 34 29 39 61",
"output": "59"
},
{
"input": "100\n66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 66 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95 95",
"output": "99"
},
{
"input": "100\n81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 81 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "1"
},
{
"input": "2\n100 99",
"output": "0"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "197"
},
{
"input": "100\n87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87 87",
"output": "0"
},
{
"input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "3\n2 2 2",
"output": "0"
},
{
"input": "3\n1 3 2",
"output": "2"
}
] | 1,693,296,374 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | n = int(input())
heights = list(map(int, input().split()))
max_height = max(heights)
min_height = min(heights)
# Count the number of soldiers to move from the beginning and end
count_beginning = heights.index(max_height)
count_end = n - 1 - heights.index(min_height)
# Calculate the minimum number of seconds needed
min_seconds = n - (count_beginning + count_end)
print(min_seconds) | Title: Arrival of the General
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all *n* squad soldiers to line up on the parade ground.
By the military charter the soldiers should stand in the order of non-increasing of their height. But as there's virtually no time to do that, the soldiers lined up in the arbitrary order. However, the general is rather short-sighted and he thinks that the soldiers lined up correctly if the first soldier in the line has the maximum height and the last soldier has the minimum height. Please note that the way other solders are positioned does not matter, including the case when there are several soldiers whose height is maximum or minimum. Only the heights of the first and the last soldier are important.
For example, the general considers the sequence of heights (4, 3, 4, 2, 1, 1) correct and the sequence (4, 3, 1, 2, 2) wrong.
Within one second the colonel can swap any two neighboring soldiers. Help him count the minimum time needed to form a line-up which the general will consider correct.
Input Specification:
The first input line contains the only integer *n* (2<=≤<=*n*<=≤<=100) which represents the number of soldiers in the line. The second line contains integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) the values of the soldiers' heights in the order of soldiers' heights' increasing in the order from the beginning of the line to its end. The numbers are space-separated. Numbers *a*1,<=*a*2,<=...,<=*a**n* are not necessarily different.
Output Specification:
Print the only integer — the minimum number of seconds the colonel will need to form a line-up the general will like.
Demo Input:
['4\n33 44 11 22\n', '7\n10 10 58 31 63 40 76\n']
Demo Output:
['2\n', '10\n']
Note:
In the first sample the colonel will need to swap the first and second soldier and then the third and fourth soldier. That will take 2 seconds. The resulting position of the soldiers is (44, 33, 22, 11).
In the second sample the colonel may swap the soldiers in the following sequence:
1. (10, 10, 58, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 40, 76) 1. (10, 58, 10, 31, 63, 76, 40) 1. (10, 58, 10, 31, 76, 63, 40) 1. (10, 58, 31, 10, 76, 63, 40) 1. (10, 58, 31, 76, 10, 63, 40) 1. (10, 58, 31, 76, 63, 10, 40) 1. (10, 58, 76, 31, 63, 10, 40) 1. (10, 76, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 10, 40) 1. (76, 10, 58, 31, 63, 40, 10) | ```python
n = int(input())
heights = list(map(int, input().split()))
max_height = max(heights)
min_height = min(heights)
# Count the number of soldiers to move from the beginning and end
count_beginning = heights.index(max_height)
count_end = n - 1 - heights.index(min_height)
# Calculate the minimum number of seconds needed
min_seconds = n - (count_beginning + count_end)
print(min_seconds)
``` | 0 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to *n* computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to *n*. So the *i*-th hacked computer is located at the point *x**i*. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of *F*(*a*) for all *a*, where *a* is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote *A* the set of all integers from 1 to *n*. Noora asks the hacker to find value of the expression . Here *F*(*a*) is calculated as the maximum among the distances between all pairs of computers from the set *a*. Formally, . Since the required sum can be quite large Noora asks to find it modulo 109<=+<=7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=3·105) denoting the number of hacked computers.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) denoting the coordinates of hacked computers. It is guaranteed that all *x**i* are distinct. | Print a single integer — the required sum modulo 109<=+<=7. | [
"2\n4 7\n",
"3\n4 3 1\n"
] | [
"3\n",
"9\n"
] | There are three non-empty subsets in the first sample test:<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/02b2d12556dad85f1c6c6912786eb87d4be2ea17.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/22f6a537962c86b3e28ddb8aaca28a7cdd219a8c.png" style="max-width: 100.0%;max-height: 100.0%;"/> and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7d0f73b3e94e13cb797f39e93d9da74835c5a02d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f368c407c8e85e2b5fedfffaff39d471d765f026.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bb8f2118a3ac352db393b1f067b28e398ce7f816.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/049032074c04b16bc0cc153f95471c40b222072b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc93c7f5b3d122314c9c5a707fae556a8f72a574.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9. | 0 | [
{
"input": "2\n4 7",
"output": "3"
},
{
"input": "3\n4 3 1",
"output": "9"
},
{
"input": "20\n8 11 13 19 21 34 36 44 57 58 61 63 76 78 79 81 85 86 90 95",
"output": "83396599"
},
{
"input": "20\n1 8 9 12 15 17 18 24 30 33 36 41 53 54 59 62 64 66 72 73",
"output": "68059140"
},
{
"input": "20\n2 6 8 9 20 23 27 36 43 49 63 65 70 71 85 87 89 91 94 97",
"output": "92743989"
},
{
"input": "1\n78091781",
"output": "0"
},
{
"input": "2\n1000000000 1",
"output": "999999999"
},
{
"input": "3\n999999998 999999999 999999992",
"output": "21"
},
{
"input": "3\n465343471 465343474 465343473",
"output": "9"
},
{
"input": "10\n10 3 6 2 1 9 8 4 5 7",
"output": "7181"
},
{
"input": "10\n756734546 756734524 756734550 756734529 756734553 756734538 756734541 756734536 756734579 756734537",
"output": "36489"
},
{
"input": "10\n877105545 939360757 849826701 845946140 803128820 926787996 967305000 904694971 921301848 971203310",
"output": "861364152"
},
{
"input": "5\n4 7 13 17 18",
"output": "270"
},
{
"input": "5\n20 17 13 7 2",
"output": "330"
},
{
"input": "5\n3 17 2 5 4",
"output": "237"
},
{
"input": "5\n999999980 999999985 999999986 999999990 999999992",
"output": "210"
},
{
"input": "5\n1000000000 999999988 999999982 999999981 999999980",
"output": "342"
},
{
"input": "5\n999999984 999999997 999999994 999999991 999999982",
"output": "285"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "5\n9 10 7 4 5",
"output": "114"
}
] | 1,495,340,392 | 2,147,483,647 | PyPy 3 | OK | TESTS | 69 | 1,434 | 30,515,200 | ###################################################
import atexit
import io
import sys
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER
@atexit.register
def write():
sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
###################################################
n = int(input())
a = list(map(int,input().split()))
m = 1000000007
s = 0
a.sort()
for i in range(n):
s = (s + a[i]*(pow(2,i,m) - pow(2,n-1-i,m)))%m
print(s) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to *n* computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 1 to *n*. So the *i*-th hacked computer is located at the point *x**i*. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
Leha should calculate a sum of *F*(*a*) for all *a*, where *a* is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote *A* the set of all integers from 1 to *n*. Noora asks the hacker to find value of the expression . Here *F*(*a*) is calculated as the maximum among the distances between all pairs of computers from the set *a*. Formally, . Since the required sum can be quite large Noora asks to find it modulo 109<=+<=7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3·105) denoting the number of hacked computers.
The second line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109) denoting the coordinates of hacked computers. It is guaranteed that all *x**i* are distinct.
Output Specification:
Print a single integer — the required sum modulo 109<=+<=7.
Demo Input:
['2\n4 7\n', '3\n4 3 1\n']
Demo Output:
['3\n', '9\n']
Note:
There are three non-empty subsets in the first sample test:<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/02b2d12556dad85f1c6c6912786eb87d4be2ea17.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/22f6a537962c86b3e28ddb8aaca28a7cdd219a8c.png" style="max-width: 100.0%;max-height: 100.0%;"/> and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7d0f73b3e94e13cb797f39e93d9da74835c5a02d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f368c407c8e85e2b5fedfffaff39d471d765f026.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bb8f2118a3ac352db393b1f067b28e398ce7f816.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/049032074c04b16bc0cc153f95471c40b222072b.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc93c7f5b3d122314c9c5a707fae556a8f72a574.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9. | ```python
###################################################
import atexit
import io
import sys
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER
@atexit.register
def write():
sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
###################################################
n = int(input())
a = list(map(int,input().split()))
m = 1000000007
s = 0
a.sort()
for i in range(n):
s = (s + a[i]*(pow(2,i,m) - pow(2,n-1-i,m)))%m
print(s)
``` | 3 |
|
272 | A | Dima and Friends | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space. | In a single line print the answer to the problem. | [
"1\n1\n",
"1\n2\n",
"2\n3 5\n"
] | [
"3\n",
"2\n",
"3\n"
] | In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers. | 500 | [
{
"input": "1\n1",
"output": "3"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "1\n5",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 2 1 3",
"output": "4"
},
{
"input": "8\n2 2 5 3 4 3 3 2",
"output": "4"
},
{
"input": "7\n4 1 3 2 2 4 5",
"output": "4"
},
{
"input": "3\n3 5 1",
"output": "4"
},
{
"input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2",
"output": "5"
},
{
"input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1",
"output": "4"
},
{
"input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4",
"output": "5"
},
{
"input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3",
"output": "5"
},
{
"input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3",
"output": "4"
},
{
"input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2",
"output": "4"
},
{
"input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3",
"output": "5"
},
{
"input": "6\n4 2 3 1 3 5",
"output": "4"
},
{
"input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4",
"output": "5"
},
{
"input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2",
"output": "5"
},
{
"input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3",
"output": "5"
},
{
"input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1",
"output": "5"
},
{
"input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5",
"output": "5"
},
{
"input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3",
"output": "5"
},
{
"input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2",
"output": "5"
},
{
"input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1",
"output": "5"
},
{
"input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5",
"output": "5"
},
{
"input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3",
"output": "5"
},
{
"input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "4"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "5"
}
] | 1,590,674,671 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 186 | 0 | n = int(input())+1
l = list(map(int,input().split()))
s = sum(l)
count =0
for i in range(1,6):
s1 = s+i
if(s1%n!=1):
count+=1
print(count)
| Title: Dima and Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
Output Specification:
In a single line print the answer to the problem.
Demo Input:
['1\n1\n', '1\n2\n', '2\n3 5\n']
Demo Output:
['3\n', '2\n', '3\n']
Note:
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers. | ```python
n = int(input())+1
l = list(map(int,input().split()))
s = sum(l)
count =0
for i in range(1,6):
s1 = s+i
if(s1%n!=1):
count+=1
print(count)
``` | 3 |
|
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,673,510,861 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | n, m, a = [int(i) for i in input().split()
row = n // a
if n % a != 0:
row += 1
column = m // a
if m % a != 0:
column += 1
print(row * column) | Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n, m, a = [int(i) for i in input().split()
row = n // a
if n % a != 0:
row += 1
column = m // a
if m % a != 0:
column += 1
print(row * column)
``` | -1 |
832 | A | Sasha and Sticks | PROGRAMMING | 800 | [
"games",
"math"
] | null | null | It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him. | The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn. | If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower). | [
"1 1\n",
"10 4\n"
] | [
"YES\n",
"NO\n"
] | In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win. | 500 | [
{
"input": "1 1",
"output": "YES"
},
{
"input": "10 4",
"output": "NO"
},
{
"input": "251656215122324104 164397544865601257",
"output": "YES"
},
{
"input": "963577813436662285 206326039287271924",
"output": "NO"
},
{
"input": "1000000000000000000 1",
"output": "NO"
},
{
"input": "253308697183523656 25332878317796706",
"output": "YES"
},
{
"input": "669038685745448997 501718093668307460",
"output": "YES"
},
{
"input": "116453141993601660 87060381463547965",
"output": "YES"
},
{
"input": "766959657 370931668",
"output": "NO"
},
{
"input": "255787422422806632 146884995820359999",
"output": "YES"
},
{
"input": "502007866464507926 71266379084204128",
"output": "YES"
},
{
"input": "257439908778973480 64157133126869976",
"output": "NO"
},
{
"input": "232709385 91708542",
"output": "NO"
},
{
"input": "252482458300407528 89907711721009125",
"output": "NO"
},
{
"input": "6 2",
"output": "YES"
},
{
"input": "6 3",
"output": "NO"
},
{
"input": "6 4",
"output": "YES"
},
{
"input": "6 5",
"output": "YES"
},
{
"input": "6 6",
"output": "YES"
},
{
"input": "258266151957056904 30153168463725364",
"output": "NO"
},
{
"input": "83504367885565783 52285355047292458",
"output": "YES"
},
{
"input": "545668929424440387 508692735816921376",
"output": "YES"
},
{
"input": "547321411485639939 36665750286082900",
"output": "NO"
},
{
"input": "548973893546839491 183137237979822911",
"output": "NO"
},
{
"input": "544068082 193116851",
"output": "NO"
},
{
"input": "871412474 749817171",
"output": "YES"
},
{
"input": "999999999 1247",
"output": "NO"
},
{
"input": "851941088 712987048",
"output": "YES"
},
{
"input": "559922900 418944886",
"output": "YES"
},
{
"input": "293908937 37520518",
"output": "YES"
},
{
"input": "650075786 130049650",
"output": "NO"
},
{
"input": "1000000000 1000000000",
"output": "YES"
},
{
"input": "548147654663723363 107422751713800746",
"output": "YES"
},
{
"input": "828159210 131819483",
"output": "NO"
},
{
"input": "6242634 4110365",
"output": "YES"
},
{
"input": "458601973 245084155",
"output": "YES"
},
{
"input": "349593257 18089089",
"output": "YES"
},
{
"input": "814768821 312514745",
"output": "NO"
},
{
"input": "697884949 626323363",
"output": "YES"
},
{
"input": "667011589 54866795",
"output": "NO"
},
{
"input": "1000000000000000000 2",
"output": "NO"
},
{
"input": "1000000000000000000 3",
"output": "YES"
},
{
"input": "1000000000000000000 4",
"output": "NO"
},
{
"input": "999999999999999 1",
"output": "YES"
},
{
"input": "17 4",
"output": "NO"
},
{
"input": "2 2",
"output": "YES"
},
{
"input": "1000000000000000 2",
"output": "NO"
},
{
"input": "12 4",
"output": "YES"
},
{
"input": "6 1",
"output": "NO"
},
{
"input": "2 1",
"output": "NO"
},
{
"input": "10000000005 1",
"output": "YES"
},
{
"input": "10000000000000009 2",
"output": "NO"
},
{
"input": "12457895452123 1",
"output": "YES"
},
{
"input": "999999999999999999 9",
"output": "YES"
},
{
"input": "1000000000000 3",
"output": "YES"
},
{
"input": "13099714659575475 6549849616514894",
"output": "NO"
},
{
"input": "100000000000000001 1",
"output": "YES"
},
{
"input": "825175814723458 324",
"output": "YES"
},
{
"input": "20 4",
"output": "YES"
},
{
"input": "100000176877 4",
"output": "YES"
},
{
"input": "100000 3",
"output": "YES"
},
{
"input": "946744073709551614 10",
"output": "YES"
}
] | 1,617,015,570 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 62 | 0 | n,k = map(int, input().split(' '))
if n==k and n%2 == 0:
print("NO")
elif n==k and n%2 !=0:
print("YES")
else:
scount = k
rem = n - k
if rem >= k:
print('NO')
else:
print("YES") | Title: Sasha and Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
Output Specification:
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
Demo Input:
['1 1\n', '10 4\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win. | ```python
n,k = map(int, input().split(' '))
if n==k and n%2 == 0:
print("NO")
elif n==k and n%2 !=0:
print("YES")
else:
scount = k
rem = n - k
if rem >= k:
print('NO')
else:
print("YES")
``` | 0 |
|
842 | A | Kirill And The Game | PROGRAMMING | 1,200 | [
"brute force",
"two pointers"
] | null | null | Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number.
For each two integer numbers *a* and *b* such that *l*<=≤<=*a*<=≤<=*r* and *x*<=≤<=*b*<=≤<=*y* there is a potion with experience *a* and cost *b* in the store (that is, there are (*r*<=-<=*l*<=+<=1)·(*y*<=-<=*x*<=+<=1) potions).
Kirill wants to buy a potion which has efficiency *k*. Will he be able to do this? | First string contains five integer numbers *l*, *r*, *x*, *y*, *k* (1<=≤<=*l*<=≤<=*r*<=≤<=107, 1<=≤<=*x*<=≤<=*y*<=≤<=107, 1<=≤<=*k*<=≤<=107). | Print "YES" without quotes if a potion with efficiency exactly *k* can be bought in the store and "NO" without quotes otherwise.
You can output each of the letters in any register. | [
"1 10 1 10 1\n",
"1 5 6 10 1\n"
] | [
"YES",
"NO"
] | none | 500 | [
{
"input": "1 10 1 10 1",
"output": "YES"
},
{
"input": "1 5 6 10 1",
"output": "NO"
},
{
"input": "1 1 1 1 1",
"output": "YES"
},
{
"input": "1 1 1 1 2",
"output": "NO"
},
{
"input": "1 100000 1 100000 100000",
"output": "YES"
},
{
"input": "1 100000 1 100000 100001",
"output": "NO"
},
{
"input": "25 10000 200 10000 5",
"output": "YES"
},
{
"input": "1 100000 10 100000 50000",
"output": "NO"
},
{
"input": "91939 94921 10197 89487 1",
"output": "NO"
},
{
"input": "30518 58228 74071 77671 1",
"output": "NO"
},
{
"input": "46646 79126 78816 91164 5",
"output": "NO"
},
{
"input": "30070 83417 92074 99337 2",
"output": "NO"
},
{
"input": "13494 17544 96820 99660 6",
"output": "NO"
},
{
"input": "96918 97018 10077 86510 9",
"output": "YES"
},
{
"input": "13046 45594 14823 52475 1",
"output": "YES"
},
{
"input": "29174 40572 95377 97669 4",
"output": "NO"
},
{
"input": "79894 92433 8634 86398 4",
"output": "YES"
},
{
"input": "96022 98362 13380 94100 6",
"output": "YES"
},
{
"input": "79446 95675 93934 96272 3",
"output": "NO"
},
{
"input": "5440 46549 61481 99500 10",
"output": "NO"
},
{
"input": "21569 53580 74739 87749 3",
"output": "NO"
},
{
"input": "72289 78297 79484 98991 7",
"output": "NO"
},
{
"input": "88417 96645 92742 98450 5",
"output": "NO"
},
{
"input": "71841 96625 73295 77648 8",
"output": "NO"
},
{
"input": "87969 99230 78041 94736 4",
"output": "NO"
},
{
"input": "4 4 1 2 3",
"output": "NO"
},
{
"input": "150 150 1 2 100",
"output": "NO"
},
{
"input": "99 100 1 100 50",
"output": "YES"
},
{
"input": "7 7 3 6 2",
"output": "NO"
},
{
"input": "10 10 1 10 1",
"output": "YES"
},
{
"input": "36 36 5 7 6",
"output": "YES"
},
{
"input": "73 96 1 51 51",
"output": "NO"
},
{
"input": "3 3 1 3 2",
"output": "NO"
},
{
"input": "10000000 10000000 1 100000 10000000",
"output": "YES"
},
{
"input": "9222174 9829060 9418763 9955619 9092468",
"output": "NO"
},
{
"input": "70 70 1 2 50",
"output": "NO"
},
{
"input": "100 200 1 20 5",
"output": "YES"
},
{
"input": "1 200000 65536 65536 65537",
"output": "NO"
},
{
"input": "15 15 1 100 1",
"output": "YES"
},
{
"input": "10000000 10000000 1 10000000 100000",
"output": "YES"
},
{
"input": "10 10 2 5 4",
"output": "NO"
},
{
"input": "67 69 7 7 9",
"output": "NO"
},
{
"input": "100000 10000000 1 10000000 100000",
"output": "YES"
},
{
"input": "9 12 1 2 7",
"output": "NO"
},
{
"input": "5426234 6375745 2636512 8492816 4409404",
"output": "NO"
},
{
"input": "6134912 6134912 10000000 10000000 999869",
"output": "NO"
},
{
"input": "3 3 1 100 1",
"output": "YES"
},
{
"input": "10000000 10000000 10 10000000 100000",
"output": "YES"
},
{
"input": "4 4 1 100 2",
"output": "YES"
},
{
"input": "8 13 1 4 7",
"output": "NO"
},
{
"input": "10 10 100000 10000000 10000000",
"output": "NO"
},
{
"input": "5 6 1 4 2",
"output": "YES"
},
{
"input": "1002 1003 1 2 1000",
"output": "NO"
},
{
"input": "4 5 1 2 2",
"output": "YES"
},
{
"input": "5 6 1 5 1",
"output": "YES"
},
{
"input": "15 21 2 4 7",
"output": "YES"
},
{
"input": "4 5 3 7 1",
"output": "YES"
},
{
"input": "15 15 3 4 4",
"output": "NO"
},
{
"input": "3 6 1 2 2",
"output": "YES"
},
{
"input": "2 10 3 6 3",
"output": "YES"
},
{
"input": "1 10000000 1 10000000 100000",
"output": "YES"
},
{
"input": "8 13 1 2 7",
"output": "NO"
},
{
"input": "98112 98112 100000 100000 128850",
"output": "NO"
},
{
"input": "2 2 1 2 1",
"output": "YES"
},
{
"input": "8 8 3 4 2",
"output": "YES"
},
{
"input": "60 60 2 3 25",
"output": "NO"
},
{
"input": "16 17 2 5 5",
"output": "NO"
},
{
"input": "2 4 1 3 1",
"output": "YES"
},
{
"input": "4 5 1 2 3",
"output": "NO"
},
{
"input": "10 10 3 4 3",
"output": "NO"
},
{
"input": "10 10000000 999999 10000000 300",
"output": "NO"
},
{
"input": "100 120 9 11 10",
"output": "YES"
},
{
"input": "8 20 1 3 4",
"output": "YES"
},
{
"input": "10 14 2 3 4",
"output": "YES"
},
{
"input": "2000 2001 1 3 1000",
"output": "YES"
},
{
"input": "12 13 2 3 5",
"output": "NO"
},
{
"input": "7 7 2 3 3",
"output": "NO"
},
{
"input": "5 8 1 10000000 4",
"output": "YES"
},
{
"input": "5 5 1 1 4",
"output": "NO"
},
{
"input": "5 5 1 6 2",
"output": "NO"
},
{
"input": "200 300 4000381 4000382 4000381",
"output": "NO"
},
{
"input": "11 17 2 5 2",
"output": "NO"
},
{
"input": "9999999 10000000 1 10000000 999997",
"output": "NO"
},
{
"input": "7 8 2 3 3",
"output": "NO"
},
{
"input": "7 7 3 3 2",
"output": "NO"
},
{
"input": "15 15 2 3 7",
"output": "NO"
},
{
"input": "65408 65408 859 859 10000000",
"output": "NO"
},
{
"input": "1000000 10000000 1 100000 1",
"output": "NO"
},
{
"input": "6 12 2 3 2",
"output": "YES"
},
{
"input": "7 8 1 3 3",
"output": "NO"
},
{
"input": "4 4 1 2 2",
"output": "YES"
},
{
"input": "2 3 1 2 2",
"output": "YES"
},
{
"input": "11 14 2 3 5",
"output": "NO"
},
{
"input": "7 7 1 10 3",
"output": "NO"
},
{
"input": "49 50 1 2 27",
"output": "NO"
},
{
"input": "1 10000000 1 10000000 123456",
"output": "YES"
},
{
"input": "100000 10000000 100 10000000 100000",
"output": "YES"
},
{
"input": "17 19 2 3 8",
"output": "NO"
},
{
"input": "4 6 3 9 1",
"output": "YES"
},
{
"input": "19 20 6 7 3",
"output": "NO"
},
{
"input": "5000000 10000000 1 4999999 1",
"output": "NO"
}
] | 1,625,201,819 | 2,147,483,647 | Python 3 | MEMORY_LIMIT_EXCEEDED | TESTS | 39 | 1,309 | 268,390,400 | l,r,x,y,k=map(int,input().split())
set1=set(range(l,r+1))
set2=set(x*k for x in range(x,y+1))
res=set1.intersection(set2)
if len(res)==0:
print("NO")
else:
print("YES") | Title: Kirill And The Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number.
For each two integer numbers *a* and *b* such that *l*<=≤<=*a*<=≤<=*r* and *x*<=≤<=*b*<=≤<=*y* there is a potion with experience *a* and cost *b* in the store (that is, there are (*r*<=-<=*l*<=+<=1)·(*y*<=-<=*x*<=+<=1) potions).
Kirill wants to buy a potion which has efficiency *k*. Will he be able to do this?
Input Specification:
First string contains five integer numbers *l*, *r*, *x*, *y*, *k* (1<=≤<=*l*<=≤<=*r*<=≤<=107, 1<=≤<=*x*<=≤<=*y*<=≤<=107, 1<=≤<=*k*<=≤<=107).
Output Specification:
Print "YES" without quotes if a potion with efficiency exactly *k* can be bought in the store and "NO" without quotes otherwise.
You can output each of the letters in any register.
Demo Input:
['1 10 1 10 1\n', '1 5 6 10 1\n']
Demo Output:
['YES', 'NO']
Note:
none | ```python
l,r,x,y,k=map(int,input().split())
set1=set(range(l,r+1))
set2=set(x*k for x in range(x,y+1))
res=set1.intersection(set2)
if len(res)==0:
print("NO")
else:
print("YES")
``` | 0 |
|
554 | A | Kyoya and Photobooks | PROGRAMMING | 900 | [
"brute force",
"math",
"strings"
] | null | null | Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem. | The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters. | Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make. | [
"a\n",
"hi\n"
] | [
"51\n",
"76\n"
] | In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets. | 250 | [
{
"input": "a",
"output": "51"
},
{
"input": "hi",
"output": "76"
},
{
"input": "y",
"output": "51"
},
{
"input": "kgan",
"output": "126"
},
{
"input": "zoabkyuvus",
"output": "276"
},
{
"input": "spyemhyznjieyhhbk",
"output": "451"
},
{
"input": "xulsyfkuizjauadjjopu",
"output": "526"
},
{
"input": "e",
"output": "51"
},
{
"input": "zv",
"output": "76"
},
{
"input": "jgv",
"output": "101"
},
{
"input": "zsfo",
"output": "126"
},
{
"input": "jselr",
"output": "151"
},
{
"input": "dwemig",
"output": "176"
},
{
"input": "mddoxsf",
"output": "201"
},
{
"input": "jgirkrmi",
"output": "226"
},
{
"input": "spkxurcum",
"output": "251"
},
{
"input": "fykkiubdkt",
"output": "276"
},
{
"input": "fznbcxsxygs",
"output": "301"
},
{
"input": "qcrvrdqcbtou",
"output": "326"
},
{
"input": "qktrbjzrqgmlr",
"output": "351"
},
{
"input": "foamodbvptlxxg",
"output": "376"
},
{
"input": "ydzpjhsidipricw",
"output": "401"
},
{
"input": "lpfpndmjfvqejdgf",
"output": "426"
},
{
"input": "ofkvparuvjtggnmab",
"output": "451"
},
{
"input": "xxncfutrtxcwdzwbgs",
"output": "476"
},
{
"input": "zovhffccflkgqncsdte",
"output": "501"
},
{
"input": "cskgsxywlvfeicoueglr",
"output": "526"
},
{
"input": "gggggggggggggggggggg",
"output": "526"
},
{
"input": "qdqdddqddqqddqddqdqd",
"output": "526"
},
{
"input": "takttttaakaaktakttkt",
"output": "526"
},
{
"input": "coccoooogogcgocccmcg",
"output": "526"
},
{
"input": "kskkskkkssksssk",
"output": "401"
},
{
"input": "lllllllllllllll",
"output": "401"
}
] | 1,693,633,398 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 33 | 62 | 0 | str = input()
a = len(str)
print(a * 25 + 26)
| Title: Kyoya and Photobooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input Specification:
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
Output Specification:
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Demo Input:
['a\n', 'hi\n']
Demo Output:
['51\n', '76\n']
Note:
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets. | ```python
str = input()
a = len(str)
print(a * 25 + 26)
``` | 3 |
|
272 | A | Dima and Friends | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space. | In a single line print the answer to the problem. | [
"1\n1\n",
"1\n2\n",
"2\n3 5\n"
] | [
"3\n",
"2\n",
"3\n"
] | In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers. | 500 | [
{
"input": "1\n1",
"output": "3"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "1\n5",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 2 1 3",
"output": "4"
},
{
"input": "8\n2 2 5 3 4 3 3 2",
"output": "4"
},
{
"input": "7\n4 1 3 2 2 4 5",
"output": "4"
},
{
"input": "3\n3 5 1",
"output": "4"
},
{
"input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2",
"output": "5"
},
{
"input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1",
"output": "4"
},
{
"input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4",
"output": "5"
},
{
"input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3",
"output": "5"
},
{
"input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3",
"output": "4"
},
{
"input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2",
"output": "4"
},
{
"input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3",
"output": "5"
},
{
"input": "6\n4 2 3 1 3 5",
"output": "4"
},
{
"input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4",
"output": "5"
},
{
"input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2",
"output": "5"
},
{
"input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3",
"output": "5"
},
{
"input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1",
"output": "5"
},
{
"input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5",
"output": "5"
},
{
"input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3",
"output": "5"
},
{
"input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2",
"output": "5"
},
{
"input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1",
"output": "5"
},
{
"input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5",
"output": "5"
},
{
"input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3",
"output": "5"
},
{
"input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "4"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "5"
}
] | 1,588,886,307 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 216 | 6,656,000 | n = int(input())
a = list(map(int , input().split()))
d = 0
for i in a:
d+=i
for i in range(2,6,+1):
if (d+i)%(n+1)!=1:
print(i)
break | Title: Dima and Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
Output Specification:
In a single line print the answer to the problem.
Demo Input:
['1\n1\n', '1\n2\n', '2\n3 5\n']
Demo Output:
['3\n', '2\n', '3\n']
Note:
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers. | ```python
n = int(input())
a = list(map(int , input().split()))
d = 0
for i in a:
d+=i
for i in range(2,6,+1):
if (d+i)%(n+1)!=1:
print(i)
break
``` | 0 |
|
508 | B | Anton and currency you all know | PROGRAMMING | 1,300 | [
"greedy",
"math",
"strings"
] | null | null | Berland, 2016. The exchange rate of currency you all know against the burle has increased so much that to simplify the calculations, its fractional part was neglected and the exchange rate is now assumed to be an integer.
Reliable sources have informed the financier Anton of some information about the exchange rate of currency you all know against the burle for tomorrow. Now Anton knows that tomorrow the exchange rate will be an even number, which can be obtained from the present rate by swapping exactly two distinct digits in it. Of all the possible values that meet these conditions, the exchange rate for tomorrow will be the maximum possible. It is guaranteed that today the exchange rate is an odd positive integer *n*. Help Anton to determine the exchange rate of currency you all know for tomorrow! | The first line contains an odd positive integer *n* — the exchange rate of currency you all know for today. The length of number *n*'s representation is within range from 2 to 105, inclusive. The representation of *n* doesn't contain any leading zeroes. | If the information about tomorrow's exchange rate is inconsistent, that is, there is no integer that meets the condition, print <=-<=1.
Otherwise, print the exchange rate of currency you all know against the burle for tomorrow. This should be the maximum possible number of those that are even and that are obtained from today's exchange rate by swapping exactly two digits. Exchange rate representation should not contain leading zeroes. | [
"527\n",
"4573\n",
"1357997531\n"
] | [
"572\n",
"3574\n",
"-1\n"
] | none | 1,000 | [
{
"input": "527",
"output": "572"
},
{
"input": "4573",
"output": "3574"
},
{
"input": "1357997531",
"output": "-1"
},
{
"input": "444443",
"output": "444434"
},
{
"input": "22227",
"output": "72222"
},
{
"input": "24683",
"output": "34682"
},
{
"input": "11",
"output": "-1"
},
{
"input": "1435678543",
"output": "1435678534"
},
{
"input": "4250769",
"output": "9250764"
},
{
"input": "4052769",
"output": "9052764"
},
{
"input": "5685341",
"output": "5685314"
},
{
"input": "1111111111111111231",
"output": "1111111111111111132"
},
{
"input": "333333332379",
"output": "333333339372"
},
{
"input": "85",
"output": "58"
},
{
"input": "7700016673",
"output": "7730016670"
},
{
"input": "35451519805848712272404365322858764249299938505103",
"output": "35451519835848712272404365322858764249299938505100"
},
{
"input": "4314752277691991627730686134692292422155753465948025897701703862445837045929984759093775762579123919",
"output": "9314752277691991627730686134692292422155753465948025897701703862445837045929984759093775762579123914"
},
{
"input": "21",
"output": "12"
},
{
"input": "101",
"output": "110"
},
{
"input": "503147",
"output": "573140"
},
{
"input": "333333333333333333333",
"output": "-1"
},
{
"input": "55555555555555555555555",
"output": "-1"
},
{
"input": "99",
"output": "-1"
},
{
"input": "23759",
"output": "93752"
},
{
"input": "235749",
"output": "935742"
},
{
"input": "435729",
"output": "935724"
},
{
"input": "8623",
"output": "8632"
},
{
"input": "109",
"output": "190"
},
{
"input": "20000000000000000000001",
"output": "21000000000000000000000"
},
{
"input": "1001",
"output": "1100"
}
] | 1,650,598,672 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main()
{
long long int n,act;
cin>>n;
act=n;
vector<int> v;
vector<int> q;
while(n>0)
{
v.push_back(n%10);
n=n/10;
}
reverse(v.begin(),v.end());
long long int x=v[v.size()-1],g=v.size()-1;
for(int j=0;j<v.size();j++)
{
if(v[j]%2==0 && v[j]<x)
{
if(j<g)
{
g=j;
}
}
else if(v[j]%2==0)
{
q.push_back(j);
}
}
if(g<v.size()-1)
{
int temp;
temp=v[v.size()-1];
v[v.size()-1]=v[g];
v[g]=temp;
}
else if(q.size()>0)
{
int temp;
temp=v[v.size()-1];
v[v.size()-1]=v[q[0]];
v[q[0]]=temp;
}
long long int s=0;
for(auto h:v)
{
s=10*s+h;
}
if(s==act)
{
cout<<-1;
}
else
{
cout<<s;
}
return 0;
} | Title: Anton and currency you all know
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland, 2016. The exchange rate of currency you all know against the burle has increased so much that to simplify the calculations, its fractional part was neglected and the exchange rate is now assumed to be an integer.
Reliable sources have informed the financier Anton of some information about the exchange rate of currency you all know against the burle for tomorrow. Now Anton knows that tomorrow the exchange rate will be an even number, which can be obtained from the present rate by swapping exactly two distinct digits in it. Of all the possible values that meet these conditions, the exchange rate for tomorrow will be the maximum possible. It is guaranteed that today the exchange rate is an odd positive integer *n*. Help Anton to determine the exchange rate of currency you all know for tomorrow!
Input Specification:
The first line contains an odd positive integer *n* — the exchange rate of currency you all know for today. The length of number *n*'s representation is within range from 2 to 105, inclusive. The representation of *n* doesn't contain any leading zeroes.
Output Specification:
If the information about tomorrow's exchange rate is inconsistent, that is, there is no integer that meets the condition, print <=-<=1.
Otherwise, print the exchange rate of currency you all know against the burle for tomorrow. This should be the maximum possible number of those that are even and that are obtained from today's exchange rate by swapping exactly two digits. Exchange rate representation should not contain leading zeroes.
Demo Input:
['527\n', '4573\n', '1357997531\n']
Demo Output:
['572\n', '3574\n', '-1\n']
Note:
none | ```python
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main()
{
long long int n,act;
cin>>n;
act=n;
vector<int> v;
vector<int> q;
while(n>0)
{
v.push_back(n%10);
n=n/10;
}
reverse(v.begin(),v.end());
long long int x=v[v.size()-1],g=v.size()-1;
for(int j=0;j<v.size();j++)
{
if(v[j]%2==0 && v[j]<x)
{
if(j<g)
{
g=j;
}
}
else if(v[j]%2==0)
{
q.push_back(j);
}
}
if(g<v.size()-1)
{
int temp;
temp=v[v.size()-1];
v[v.size()-1]=v[g];
v[g]=temp;
}
else if(q.size()>0)
{
int temp;
temp=v[v.size()-1];
v[v.size()-1]=v[q[0]];
v[q[0]]=temp;
}
long long int s=0;
for(auto h:v)
{
s=10*s+h;
}
if(s==act)
{
cout<<-1;
}
else
{
cout<<s;
}
return 0;
}
``` | -1 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as follows. Consider an undirected graph *G*. It is required to find a subset of vertices *C* of the maximum size such that any two of them are connected by an edge in graph *G*. Sounds simple, doesn't it? Nobody yet knows an algorithm that finds a solution to this problem in polynomial time of the size of the graph. However, as with many other NP-complete problems, the clique problem is easier if you consider a specific type of a graph.
Consider *n* distinct points on a line. Let the *i*-th point have the coordinate *x**i* and weight *w**i*. Let's form graph *G*, whose vertices are these points and edges connect exactly the pairs of points (*i*,<=*j*), such that the distance between them is not less than the sum of their weights, or more formally: |*x**i*<=-<=*x**j*|<=≥<=*w**i*<=+<=*w**j*.
Find the size of the maximum clique in such graph. | The first line contains the integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of points.
Each of the next *n* lines contains two numbers *x**i*, *w**i* (0<=≤<=*x**i*<=≤<=109,<=1<=≤<=*w**i*<=≤<=109) — the coordinate and the weight of a point. All *x**i* are different. | Print a single number — the number of vertexes in the maximum clique of the given graph. | [
"4\n2 3\n3 1\n6 1\n0 2\n"
] | [
"3\n"
] | If you happen to know how to solve this problem without using the specific properties of the graph formulated in the problem statement, then you are able to get a prize of one million dollars!
The picture for the sample test. | 0 | [
{
"input": "4\n2 3\n3 1\n6 1\n0 2",
"output": "3"
},
{
"input": "1\n42 23",
"output": "1"
},
{
"input": "2\n1 5\n2 6",
"output": "1"
},
{
"input": "2\n1 5\n12 6",
"output": "2"
},
{
"input": "1\n0 1",
"output": "1"
},
{
"input": "1\n1000000000 1000000000",
"output": "1"
},
{
"input": "2\n4 4\n12 5",
"output": "1"
},
{
"input": "2\n4 4\n12 4",
"output": "2"
},
{
"input": "2\n4 4\n12 3",
"output": "2"
},
{
"input": "3\n0 1\n2 1\n4 1",
"output": "3"
},
{
"input": "3\n0 1\n2 2\n4 1",
"output": "2"
},
{
"input": "2\n0 1\n1000000000 1",
"output": "2"
},
{
"input": "2\n0 1000000000\n1000000000 1000000000",
"output": "1"
},
{
"input": "1\n76438 10",
"output": "1"
},
{
"input": "10\n6 15\n4 5\n1 4\n2 4\n0 6\n9 5\n8 14\n5 4\n7 20\n10 20",
"output": "1"
},
{
"input": "10\n0 3\n30 3\n54 3\n6 3\n36 3\n12 3\n42 3\n24 3\n48 3\n18 3",
"output": "10"
},
{
"input": "10\n48 4\n54 4\n12 4\n6 4\n30 4\n36 4\n24 4\n0 4\n42 4\n18 4",
"output": "5"
},
{
"input": "11\n0 4\n54 4\n48 4\n18 4\n24 4\n42 4\n6 4\n36 4\n12 4\n30 4\n60 4",
"output": "6"
},
{
"input": "12\n66 4\n12 4\n60 4\n24 4\n48 4\n0 4\n36 4\n30 4\n6 4\n54 4\n42 4\n18 4",
"output": "6"
},
{
"input": "1\n0 1000000000",
"output": "1"
}
] | 1,566,436,332 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 139 | 409,600 | from collections import defaultdict
def dfs(key, count):
global XW
if XW.get(key):
dfs(XW[key], count+1)
else:
res[0]=max(res[0], count)
n = int(input())
XW = defaultdict()
for i in range(n):
x, w = map(int, input().split())
XW[x] = w
res = [0]
for key, num in XW.items():
dfs(key, 0)
print(res[0])
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as follows. Consider an undirected graph *G*. It is required to find a subset of vertices *C* of the maximum size such that any two of them are connected by an edge in graph *G*. Sounds simple, doesn't it? Nobody yet knows an algorithm that finds a solution to this problem in polynomial time of the size of the graph. However, as with many other NP-complete problems, the clique problem is easier if you consider a specific type of a graph.
Consider *n* distinct points on a line. Let the *i*-th point have the coordinate *x**i* and weight *w**i*. Let's form graph *G*, whose vertices are these points and edges connect exactly the pairs of points (*i*,<=*j*), such that the distance between them is not less than the sum of their weights, or more formally: |*x**i*<=-<=*x**j*|<=≥<=*w**i*<=+<=*w**j*.
Find the size of the maximum clique in such graph.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of points.
Each of the next *n* lines contains two numbers *x**i*, *w**i* (0<=≤<=*x**i*<=≤<=109,<=1<=≤<=*w**i*<=≤<=109) — the coordinate and the weight of a point. All *x**i* are different.
Output Specification:
Print a single number — the number of vertexes in the maximum clique of the given graph.
Demo Input:
['4\n2 3\n3 1\n6 1\n0 2\n']
Demo Output:
['3\n']
Note:
If you happen to know how to solve this problem without using the specific properties of the graph formulated in the problem statement, then you are able to get a prize of one million dollars!
The picture for the sample test. | ```python
from collections import defaultdict
def dfs(key, count):
global XW
if XW.get(key):
dfs(XW[key], count+1)
else:
res[0]=max(res[0], count)
n = int(input())
XW = defaultdict()
for i in range(n):
x, w = map(int, input().split())
XW[x] = w
res = [0]
for key, num in XW.items():
dfs(key, 0)
print(res[0])
``` | 0 |
|
960 | A | Check the string | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes). | The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'. | Print "YES" or "NO", according to the condition. | [
"aaabccc\n",
"bbacc\n",
"aabc\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | 500 | [
{
"input": "aaabccc",
"output": "YES"
},
{
"input": "bbacc",
"output": "NO"
},
{
"input": "aabc",
"output": "YES"
},
{
"input": "aabbcc",
"output": "YES"
},
{
"input": "aaacccbb",
"output": "NO"
},
{
"input": "abc",
"output": "YES"
},
{
"input": "acba",
"output": "NO"
},
{
"input": "bbabbc",
"output": "NO"
},
{
"input": "bbbabacca",
"output": "NO"
},
{
"input": "aabcbcaca",
"output": "NO"
},
{
"input": "aaaaabbbbbb",
"output": "NO"
},
{
"input": "c",
"output": "NO"
},
{
"input": "cc",
"output": "NO"
},
{
"input": "bbb",
"output": "NO"
},
{
"input": "bc",
"output": "NO"
},
{
"input": "ccbcc",
"output": "NO"
},
{
"input": "aaa",
"output": "NO"
},
{
"input": "aaccaa",
"output": "NO"
},
{
"input": "a",
"output": "NO"
},
{
"input": "b",
"output": "NO"
},
{
"input": "abca",
"output": "NO"
},
{
"input": "aabbcccc",
"output": "NO"
},
{
"input": "abac",
"output": "NO"
},
{
"input": "abcc",
"output": "NO"
},
{
"input": "abcb",
"output": "NO"
},
{
"input": "aacc",
"output": "NO"
},
{
"input": "aabbaacccc",
"output": "NO"
},
{
"input": "aabb",
"output": "NO"
},
{
"input": "ac",
"output": "NO"
},
{
"input": "abbacc",
"output": "NO"
},
{
"input": "abacc",
"output": "NO"
},
{
"input": "ababc",
"output": "NO"
},
{
"input": "aa",
"output": "NO"
},
{
"input": "aabaccc",
"output": "NO"
},
{
"input": "bbcc",
"output": "NO"
},
{
"input": "aaabcbc",
"output": "NO"
},
{
"input": "acbbc",
"output": "NO"
},
{
"input": "babc",
"output": "NO"
},
{
"input": "bbbcc",
"output": "NO"
},
{
"input": "bbc",
"output": "NO"
},
{
"input": "abababccc",
"output": "NO"
},
{
"input": "ccbbaa",
"output": "NO"
}
] | 1,608,193,302 | 2,147,483,647 | PyPy 3 | OK | TESTS | 55 | 140 | 0 | s = input(); a = s.count('a'); b = s.count('b'); c = s.count('c')
if a==0 or b==0:
print('No')
exit(0)
if a*'a' + b*'b' + c*'c' == s and (a==c or b==c):
print('Yes')
else:
print('No') | Title: Check the string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).
Input Specification:
The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.
Output Specification:
Print "YES" or "NO", according to the condition.
Demo Input:
['aaabccc\n', 'bbacc\n', 'aabc\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | ```python
s = input(); a = s.count('a'); b = s.count('b'); c = s.count('c')
if a==0 or b==0:
print('No')
exit(0)
if a*'a' + b*'b' + c*'c' == s and (a==c or b==c):
print('Yes')
else:
print('No')
``` | 3 |
|
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,684,676,090 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 46 | 0 | n,m,a=map(int,input().split())
dem=0
if n>a or m>a:
dem+=1
n-=a
m-=a
while n>0 or m>0:
kt1=False
kt2=False
if n>0:
n-=a
kt1=True
dem+=1
if m>0:
m-=a
kt2=True
dem+=1
if kt1 and kt2:
dem+=1
print(dem) | Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n,m,a=map(int,input().split())
dem=0
if n>a or m>a:
dem+=1
n-=a
m-=a
while n>0 or m>0:
kt1=False
kt2=False
if n>0:
n-=a
kt1=True
dem+=1
if m>0:
m-=a
kt2=True
dem+=1
if kt1 and kt2:
dem+=1
print(dem)
``` | 0 |
976 | A | Minimum Binary Number | PROGRAMMING | 800 | [
"implementation"
] | null | null | String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all). | The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct. | Print one string — the minimum correct string that you can obtain from the given one. | [
"4\n1001\n",
"1\n1\n"
] | [
"100\n",
"1\n"
] | In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use. | 0 | [
{
"input": "4\n1001",
"output": "100"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "100\n1110111100001111011111111010110011111111011110000111101101011100110110001011000000101010110101011100",
"output": "1000000000000000000000000000000000000000"
},
{
"input": "100\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "1"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111110111111111111111111111111111111111111111111",
"output": "10"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "8\n10101010",
"output": "10000"
},
{
"input": "2\n10",
"output": "10"
},
{
"input": "3\n111",
"output": "1"
},
{
"input": "5\n11100",
"output": "100"
},
{
"input": "2\n11",
"output": "1"
},
{
"input": "3\n110",
"output": "10"
},
{
"input": "50\n10010010000000000000000000000000000000001000000000",
"output": "10000000000000000000000000000000000000000000000"
}
] | 1,550,566,475 | 2,147,483,647 | Python 3 | OK | TESTS | 14 | 124 | 0 | n=input()
s=input()
print('1'*int(bool(s.count("1")))+"0"*s.count("0")) | Title: Minimum Binary Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all).
Input Specification:
The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct.
Output Specification:
Print one string — the minimum correct string that you can obtain from the given one.
Demo Input:
['4\n1001\n', '1\n1\n']
Demo Output:
['100\n', '1\n']
Note:
In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use. | ```python
n=input()
s=input()
print('1'*int(bool(s.count("1")))+"0"*s.count("0"))
``` | 3 |
|
746 | B | Decoding | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Polycarp is mad about coding, that is why he writes Sveta encoded messages. He calls the median letter in a word the letter which is in the middle of the word. If the word's length is even, the median letter is the left of the two middle letters. In the following examples, the median letter is highlighted: contest, info. If the word consists of single letter, then according to above definition this letter is the median letter.
Polycarp encodes each word in the following way: he writes down the median letter of the word, then deletes it and repeats the process until there are no letters left. For example, he encodes the word volga as logva.
You are given an encoding *s* of some word, your task is to decode it. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=2000) — the length of the encoded word.
The second line contains the string *s* of length *n* consisting of lowercase English letters — the encoding. | Print the word that Polycarp encoded. | [
"5\nlogva\n",
"2\nno\n",
"4\nabba\n"
] | [
"volga\n",
"no\n",
"baba\n"
] | In the first example Polycarp encoded the word volga. At first, he wrote down the letter l from the position 3, after that his word looked like voga. After that Polycarp wrote down the letter o from the position 2, his word became vga. Then Polycarp wrote down the letter g which was at the second position, the word became va. Then he wrote down the letter v, then the letter a. Thus, the encoding looked like logva.
In the second example Polycarp encoded the word no. He wrote down the letter n, the word became o, and he wrote down the letter o. Thus, in this example, the word and its encoding are the same.
In the third example Polycarp encoded the word baba. At first, he wrote down the letter a, which was at the position 2, after that the word looked like bba. Then he wrote down the letter b, which was at the position 2, his word looked like ba. After that he wrote down the letter b, which was at the position 1, the word looked like a, and he wrote down that letter a. Thus, the encoding is abba. | 1,000 | [
{
"input": "5\nlogva",
"output": "volga"
},
{
"input": "2\nno",
"output": "no"
},
{
"input": "4\nabba",
"output": "baba"
},
{
"input": "51\nkfsmpaeviowvkdbuhdagquxxqniselafnfbrgbhmsugcbbnlrvv",
"output": "vlbcumbrfflsnxugdudvovamfkspeiwkbhaqxqieanbghsgbnrv"
},
{
"input": "1\nw",
"output": "w"
},
{
"input": "2\ncb",
"output": "cb"
},
{
"input": "3\nqok",
"output": "oqk"
},
{
"input": "4\naegi",
"output": "gaei"
},
{
"input": "5\noqquy",
"output": "uqoqy"
},
{
"input": "6\nulhpnm",
"output": "nhulpm"
},
{
"input": "7\nijvxljt",
"output": "jxjivlt"
},
{
"input": "8\nwwmiwkeo",
"output": "ewmwwiko"
},
{
"input": "9\ngmwqmpfow",
"output": "opqmgwmfw"
},
{
"input": "10\nhncmexsslh",
"output": "lsechnmxsh"
},
{
"input": "20\nrtcjbjlbtjfmvzdqutuw",
"output": "uudvftlbcrtjjbjmzqtw"
},
{
"input": "21\ngjyiqoebcnpsdegxnsauh",
"output": "usxesnboijgyqecpdgnah"
},
{
"input": "30\nudotcwvcwxajkadxqvxvwgmwmnqrby",
"output": "bqmmwxqdkawvcoudtwcxjaxvvgwnry"
},
{
"input": "31\nipgfrxxcgckksfgexlicjvtnhvrfbmb",
"output": "mfvnvclefkccxfpigrxgksgxijthrbb"
},
{
"input": "50\nwobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy",
"output": "vsolrruoeqehviaqtycivhrbwoevvkhujmhagaholrmsatdjjy"
},
{
"input": "200\nhvayscqiwpcfykibwyudkzuzdkgqqvbnrfeupjefevlvojngmlcjwzijrkzbsaovabkvvwmjgoonyhuiphwmqdoiuueuyqtychbsklflnvghipdgaxhuhiiqlqocpvhldgvnsrtcwxpidrjffwvwcirluyyxzxrglheczeuouklzkvnyubsvgvmdbrylimztotdbmjph",
"output": "pmdoziybmgsunkluuzelrzyurcvfjdpwtsvdhpolihhadignfkbctyeuoqwpuyogmvkaoszriwcmnoleeperbqgdukuwiycwqsahvycipfkbydzzkqvnfujfvvjgljzjkbavbvwjonhihmdiuuqyhsllvhpgxuiqqcvlgnrcxirfwwilyxxghceokzvybvvdrlmttbjh"
},
{
"input": "201\nrpkghhfibtmlkpdiklegblbuyshfirheatjkfoqkfayfbxeeqijwqdwkkrkbdxlhzkhyiifemsghwovorlqedngldskfbhmwrnzmtjuckxoqdszmsdnbuqnlqzswdfhagasmfswanifrjjcuwdsplytvmnfarchgqteedgfpumkssindxndliozojzlpznwedodzwrrus",
"output": "urzoenpzoolndismpgetgcanvypdujriasmaafwzlqbdmsqxcjmnwhfslneloohseiykhxbrkdwiexfakokterfsulglipltihgprkhfbmkdkebbyhihajfqfybeqjqwkkdlzhifmgwvrqdgdkbmrztukodzsnunqsdhgsfwnfjcwsltmfrhqedfuksnxdizjlzwddwrs"
},
{
"input": "500\naopxumqciwxewxvlxzebsztskjvjzwyewjztqrsuvamtvklhqrbodtncqdchjrlpywvmtgnkkwtvpggktewdgvnhydkexwoxkgltaesrtifbwpciqsvrgjtqrdnyqkgqwrryacluaqmgdwxinqieiblolyekcbzahlhxdwqcgieyfgmicvgbbitbzhejkshjunzjteyyfngigjwyqqndtjrdykzrnrpinkwtrlchhxvycrhstpecadszilicrqdeyyidohqvzfnsqfyuemigacysxvtrgxyjcvejkjstsnatfqlkeytxgsksgpcooypsmqgcluzwofaupegxppbupvtumjerohdteuenwcmqaoazohkilgpkjavcrjcslhzkyjcgfzxxzjfufichxcodcawonkxhbqgfimmlycswdzwbnmjwhbwihfoftpcqplncavmbxuwnsabiyvpcrhfgtqyaguoaigknushbqjwqmmyvsxwabrub",
"output": "ubwsymwqhukiogytfrpybswxmanpctohwhjnwdsymigbxnwcoxcffzxfcyzlcrvjplkoaamweedoemtpbpgpaozlgmpocgkgtelfasskecygtxyaieyqnzqoiydriisaethcvhcrwnpnzyrtnqwggfytzuhkeztbgcmfegqdhhzcelliinxdmalarwgqnrtgvqcwftsalkoxkyngwtgptkntvyljcqndbqlvmvsqzwyzvktsexvwxiqupaoxmcwexlzbzsjjwejtruatkhrotcdhrpwmgkwvgkedvhdewxgteribpisrjqdykqrycuqgwiqeboykbalxwciygivbibhjsjnjeynijyqdjdkrriktlhxyrspcdzlcqeydhvfsfumgcsvrxjvjjtntqkyxsspoysqcuwfuexpuvujrhtuncqozhigkacjshkjgzxjuihcdaokhqfmlcwzbmwbiffpqlcvbunaivchgqauagnsbjqmvxarb"
},
{
"input": "501\noilesjbgowlnayckhpoaitijewsyhgavnthycaecwnvzpxgjqfjyxnjcjknvvsmjbjwtcoyfbegmnnheeamvtfjkigqoanhvgdfrjchdqgowrstlmrjmcsuuwvvoeucfyhnxivosrxblfoqwikfxjnnyejdiihpenfcahtjwcnzwvxxseicvdfgqhtvefswznuyohmmljlnxubhevywpmnitnkhecsgccpstxkmdzabsnwxkokdfsogzbpnfvgudvqlstxojzfzugxbfrozveaiofrzksegdelxsdhcjlqwqlgjcqiujptoenxozhkqhcpkarretqzfkwuvbmfdcdoqliyzmlfrfhzrnkbhofuctkpeacqejwvdrlyvepudrlzncbhwrgmxrczphsoymvtuzqjscvbtqpymogupgzctepccravjcrfsadueyraqvwasravkubebojqspdntewnjohvccamvoxdauyakvehjhabpdyzyme",
"output": "mzdbhhvaudomcvonenpqoeuvravayuafcvrcecguoyqbcjztmohzrmrhczrueyrvjqaptuobnzffmylocfbukztraphkzxetjicgqqjhsldekroavofxufjxsqdgfpzofkkwsadktpcsektnpyebxllmouzsetqfvisxwnwtafehijynxkwolxsvxhfuovusmrlswgdcrdvnogkfvaenmefotjjsvkcnyfjxzncaytvgywjtapkynwgjeiolsbolachoiieshanhcewvpgqjxjjnvmbwcybgnhemtjiqahgfjhqortmjcuwvecyniorbfqifjnedipnchjczvxecdghvfwnyhmjnuhvwminhcgcsxmzbnxodsgbnvuvltozzgbrzeifzsgexdclwljquponohqckreqfwvmddqizlrhrkhfckecewdlvpdlnbwgxcpsyvuqsvtpmgpztpcajrsderqwsakbbjsdtwjhcavxaykejapyye"
}
] | 1,653,066,956 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | n1 = int( input() )
n2 = int( input() )
n3 = int( input() )
print( min( n1 , n2 // 2 , n3 // 4 ) * 7 )
| Title: Decoding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is mad about coding, that is why he writes Sveta encoded messages. He calls the median letter in a word the letter which is in the middle of the word. If the word's length is even, the median letter is the left of the two middle letters. In the following examples, the median letter is highlighted: contest, info. If the word consists of single letter, then according to above definition this letter is the median letter.
Polycarp encodes each word in the following way: he writes down the median letter of the word, then deletes it and repeats the process until there are no letters left. For example, he encodes the word volga as logva.
You are given an encoding *s* of some word, your task is to decode it.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=2000) — the length of the encoded word.
The second line contains the string *s* of length *n* consisting of lowercase English letters — the encoding.
Output Specification:
Print the word that Polycarp encoded.
Demo Input:
['5\nlogva\n', '2\nno\n', '4\nabba\n']
Demo Output:
['volga\n', 'no\n', 'baba\n']
Note:
In the first example Polycarp encoded the word volga. At first, he wrote down the letter l from the position 3, after that his word looked like voga. After that Polycarp wrote down the letter o from the position 2, his word became vga. Then Polycarp wrote down the letter g which was at the second position, the word became va. Then he wrote down the letter v, then the letter a. Thus, the encoding looked like logva.
In the second example Polycarp encoded the word no. He wrote down the letter n, the word became o, and he wrote down the letter o. Thus, in this example, the word and its encoding are the same.
In the third example Polycarp encoded the word baba. At first, he wrote down the letter a, which was at the position 2, after that the word looked like bba. Then he wrote down the letter b, which was at the position 2, his word looked like ba. After that he wrote down the letter b, which was at the position 1, the word looked like a, and he wrote down that letter a. Thus, the encoding is abba. | ```python
n1 = int( input() )
n2 = int( input() )
n3 = int( input() )
print( min( n1 , n2 // 2 , n3 // 4 ) * 7 )
``` | -1 |
|
435 | A | Queue on Bus Stop | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has *n* groups of people. The *i*-th group from the beginning has *a**i* people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most *m* people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all *n* groups to the dacha countryside. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The next line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*m*). | Print a single integer — the number of buses that is needed to transport all *n* groups to the dacha countryside. | [
"4 3\n2 3 2 1\n",
"3 4\n1 2 1\n"
] | [
"3\n",
"1\n"
] | none | 500 | [
{
"input": "4 3\n2 3 2 1",
"output": "3"
},
{
"input": "3 4\n1 2 1",
"output": "1"
},
{
"input": "1 5\n4",
"output": "1"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "5"
},
{
"input": "6 4\n1 3 2 3 4 1",
"output": "5"
},
{
"input": "6 8\n6 1 1 1 4 5",
"output": "3"
},
{
"input": "10 10\n1 10 1 10 1 1 7 8 6 7",
"output": "8"
},
{
"input": "100 100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52",
"output": "63"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "10"
},
{
"input": "10 2\n2 2 1 1 1 1 1 2 1 2",
"output": "8"
},
{
"input": "10 3\n1 3 1 1 3 2 2 2 3 3",
"output": "9"
},
{
"input": "10 4\n2 1 1 1 3 4 4 4 1 2",
"output": "6"
},
{
"input": "10 5\n2 2 3 4 4 1 5 3 1 2",
"output": "7"
},
{
"input": "100 3\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2",
"output": "83"
},
{
"input": "100 7\n4 7 4 7 7 4 7 3 5 6 3 5 4 3 7 2 7 2 4 1 6 3 3 7 4 4 5 4 3 6 4 3 2 2 1 4 4 1 7 3 7 7 1 3 1 5 4 1 5 3 5 2 2 1 5 5 1 5 2 7 5 5 1 5 5 4 6 5 1 3 5 6 7 4 1 3 3 4 3 2 7 6 5 7 2 7 1 1 2 2 3 1 3 7 1 3 2 1 1 7",
"output": "71"
},
{
"input": "100 10\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2",
"output": "64"
},
{
"input": "100 15\n3 12 8 3 11 14 12 14 1 11 13 3 5 13 4 14 2 11 7 8 12 9 15 7 15 1 4 11 6 12 1 3 8 13 1 8 14 4 3 14 1 3 1 6 10 15 13 11 12 1 14 13 11 14 11 3 12 7 3 15 14 4 5 6 5 14 7 14 6 2 6 12 6 13 13 1 9 13 15 11 6 3 15 11 9 4 15 8 15 12 1 15 10 10 4 1 15 1 4 1",
"output": "71"
},
{
"input": "100 30\n7 14 22 16 11 13 7 29 20 19 22 6 12 16 1 8 27 21 22 3 15 27 20 12 4 19 1 26 26 22 25 17 29 25 16 29 29 28 16 26 25 14 16 20 5 21 5 15 19 13 17 21 17 19 23 13 1 25 6 30 16 19 12 10 28 8 15 13 14 24 19 30 12 19 22 1 3 14 16 3 20 26 15 19 9 10 19 27 2 16 10 22 15 13 19 3 24 9 8 13",
"output": "71"
},
{
"input": "100 40\n39 19 13 36 11 21 32 12 1 2 39 26 32 39 24 1 4 19 10 4 16 39 32 34 13 24 30 35 3 10 8 18 13 12 39 27 31 40 37 20 17 17 37 5 10 12 22 17 7 1 31 13 11 10 2 6 22 16 2 4 9 27 6 35 22 16 22 30 33 2 26 20 35 19 40 37 19 17 21 28 37 28 40 4 5 4 35 19 26 36 19 12 21 20 21 30 9 16 9 32",
"output": "65"
},
{
"input": "100 50\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1",
"output": "60"
},
{
"input": "100 60\n34 21 39 17 48 46 23 56 46 52 50 39 55 48 54 38 32 38 24 26 44 12 28 9 25 26 10 52 42 60 41 3 16 60 44 29 27 55 19 19 19 57 45 59 29 35 5 14 50 47 57 48 16 7 12 36 58 31 37 58 30 50 19 11 10 41 59 57 49 41 33 9 12 11 53 50 60 51 21 9 44 23 1 16 4 15 17 57 15 17 46 50 18 52 43 24 47 50 19 18",
"output": "74"
},
{
"input": "100 90\n74 65 49 41 3 79 61 83 50 40 13 57 90 14 62 77 36 10 3 5 5 40 50 75 32 26 3 71 79 54 88 50 46 20 42 59 30 36 83 86 60 62 82 68 62 80 18 65 28 28 81 74 62 33 61 35 33 83 90 72 6 6 51 4 22 20 29 10 8 3 84 69 12 17 24 16 12 64 80 74 68 59 1 59 15 59 37 58 79 83 51 56 81 14 37 45 19 31 61 90",
"output": "67"
},
{
"input": "100 99\n69 46 76 47 71 9 66 46 78 17 96 83 56 96 29 3 43 48 79 23 93 61 19 9 29 72 15 84 93 46 71 87 11 43 96 44 54 75 3 66 2 95 46 32 69 52 79 38 57 53 37 60 71 82 28 31 84 58 89 40 62 74 22 50 45 38 99 67 24 28 28 12 69 88 33 10 31 71 46 7 42 81 54 81 96 44 8 1 20 24 28 19 54 35 69 32 71 13 66 15",
"output": "68"
},
{
"input": "90 100\n25 52 88 89 36 17 57 64 66 11 89 61 54 92 48 51 18 42 44 92 6 14 67 100 16 21 17 88 85 73 33 11 94 84 56 72 4 80 90 78 96 5 62 70 54 70 94 80 10 91 100 89 98 87 69 74 88 63 53 79 38 94 89 52 21 82 67 79 100 81 2 40 30 69 34 15 12 33 87 52 95 18 51 30 15 39 30 99 46 84",
"output": "67"
},
{
"input": "5 100\n14 67 15 28 21",
"output": "2"
},
{
"input": "10 100\n2 17 53 94 95 57 36 47 68 48",
"output": "7"
},
{
"input": "1 100\n18",
"output": "1"
},
{
"input": "100 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "30 100\n56 7 99 83 2 65 35 53 99 36 42 57 13 37 68 52 87 11 50 23 86 24 32 39 97 6 64 1 18 86",
"output": "18"
},
{
"input": "60 100\n18 75 43 88 45 43 20 59 59 79 62 39 53 21 28 46 54 53 97 81 18 15 2 95 84 9 36 70 30 76 17 19 83 40 45 32 31 70 23 14 44 35 79 84 97 96 99 60 3 73 64 83 6 12 67 86 70 89 18 61",
"output": "40"
},
{
"input": "1 73\n24",
"output": "1"
},
{
"input": "4 4\n1 4 4 4",
"output": "4"
},
{
"input": "2 6\n1 4",
"output": "1"
},
{
"input": "2 5\n5 5",
"output": "2"
}
] | 1,542,881,368 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 109 | 0 | import math
a = list(map(int,input().split()))
people = list(map(int,input().split()))
total_group = a[0]
seat_bus = a[1]
max_bus = total_group * seat_bus
check = 0
###check if people have same number of seat_bus
##check_same = [i for i in range(total_group) if people[i] == seat_bus]
###remove the same numbers of seat_bus
##print("people before: ",people)
##for i in check_same:
## people.remove(seat_bus)
##
##print("people now: ",people)
##check+=len(check_same)
####print("ini check abis",check)
loop = math.ceil(total_group)
##sums = sum(people)
skip = False
rem = 0
temp = people[0]
prev = 0
cookies = 0
for i in range(0,loop-1):
if temp + people[i+1] <= seat_bus:
skip = False
if people[i] + people[i+1] <= seat_bus and skip == False:
temp = people[i] + people[i+1]
people[i+1] = temp
cookies +=1
else:
cookies = 0
check+=1
last = people[-1]
if last <= seat_bus and skip == False:
check+=1
print(check)
| Title: Queue on Bus Stop
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has *n* groups of people. The *i*-th group from the beginning has *a**i* people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most *m* people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all *n* groups to the dacha countryside.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). The next line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*m*).
Output Specification:
Print a single integer — the number of buses that is needed to transport all *n* groups to the dacha countryside.
Demo Input:
['4 3\n2 3 2 1\n', '3 4\n1 2 1\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
import math
a = list(map(int,input().split()))
people = list(map(int,input().split()))
total_group = a[0]
seat_bus = a[1]
max_bus = total_group * seat_bus
check = 0
###check if people have same number of seat_bus
##check_same = [i for i in range(total_group) if people[i] == seat_bus]
###remove the same numbers of seat_bus
##print("people before: ",people)
##for i in check_same:
## people.remove(seat_bus)
##
##print("people now: ",people)
##check+=len(check_same)
####print("ini check abis",check)
loop = math.ceil(total_group)
##sums = sum(people)
skip = False
rem = 0
temp = people[0]
prev = 0
cookies = 0
for i in range(0,loop-1):
if temp + people[i+1] <= seat_bus:
skip = False
if people[i] + people[i+1] <= seat_bus and skip == False:
temp = people[i] + people[i+1]
people[i+1] = temp
cookies +=1
else:
cookies = 0
check+=1
last = people[-1]
if last <= seat_bus and skip == False:
check+=1
print(check)
``` | 3 |
|
962 | A | Equator | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator. | The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day. | Print the index of the day when Polycarp will celebrate the equator. | [
"4\n1 3 2 1\n",
"6\n2 2 2 2 2 2\n"
] | [
"2\n",
"3\n"
] | In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training. | 0 | [
{
"input": "4\n1 3 2 1",
"output": "2"
},
{
"input": "6\n2 2 2 2 2 2",
"output": "3"
},
{
"input": "1\n10000",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "2\n1 3",
"output": "2"
},
{
"input": "4\n2 1 1 3",
"output": "3"
},
{
"input": "3\n1 1 3",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "2"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "5\n1 2 4 3 5",
"output": "4"
},
{
"input": "5\n2 2 2 4 3",
"output": "4"
},
{
"input": "4\n1 2 3 1",
"output": "3"
},
{
"input": "6\n7 3 10 7 3 11",
"output": "4"
},
{
"input": "2\n3 4",
"output": "2"
},
{
"input": "5\n1 1 1 1 1",
"output": "3"
},
{
"input": "4\n1 3 2 3",
"output": "3"
},
{
"input": "2\n2 3",
"output": "2"
},
{
"input": "3\n32 10 23",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "4"
},
{
"input": "3\n1 2 4",
"output": "3"
},
{
"input": "6\n3 3 3 2 4 4",
"output": "4"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "5"
},
{
"input": "5\n1 3 3 1 1",
"output": "3"
},
{
"input": "4\n1 1 1 2",
"output": "3"
},
{
"input": "4\n1 2 1 3",
"output": "3"
},
{
"input": "3\n2 2 1",
"output": "2"
},
{
"input": "4\n2 3 3 3",
"output": "3"
},
{
"input": "4\n3 2 3 3",
"output": "3"
},
{
"input": "4\n2 1 1 1",
"output": "2"
},
{
"input": "3\n2 1 4",
"output": "3"
},
{
"input": "2\n6 7",
"output": "2"
},
{
"input": "4\n3 3 4 3",
"output": "3"
},
{
"input": "4\n1 1 2 5",
"output": "4"
},
{
"input": "4\n1 8 7 3",
"output": "3"
},
{
"input": "6\n2 2 2 2 2 3",
"output": "4"
},
{
"input": "3\n2 2 5",
"output": "3"
},
{
"input": "4\n1 1 2 1",
"output": "3"
},
{
"input": "5\n1 1 2 2 3",
"output": "4"
},
{
"input": "5\n9 5 3 4 8",
"output": "3"
},
{
"input": "3\n3 3 1",
"output": "2"
},
{
"input": "4\n1 2 2 2",
"output": "3"
},
{
"input": "3\n1 3 5",
"output": "3"
},
{
"input": "4\n1 1 3 6",
"output": "4"
},
{
"input": "6\n1 2 1 1 1 1",
"output": "3"
},
{
"input": "3\n3 1 3",
"output": "2"
},
{
"input": "5\n3 4 5 1 2",
"output": "3"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "5\n3 1 2 5 2",
"output": "4"
},
{
"input": "4\n1 1 1 4",
"output": "4"
},
{
"input": "4\n2 6 1 10",
"output": "4"
},
{
"input": "4\n2 2 3 2",
"output": "3"
},
{
"input": "4\n4 2 2 1",
"output": "2"
},
{
"input": "6\n1 1 1 1 1 4",
"output": "5"
},
{
"input": "3\n3 2 2",
"output": "2"
},
{
"input": "6\n1 3 5 1 7 4",
"output": "5"
},
{
"input": "5\n1 2 4 8 16",
"output": "5"
},
{
"input": "5\n1 2 4 4 4",
"output": "4"
},
{
"input": "6\n4 2 1 2 3 1",
"output": "3"
},
{
"input": "4\n3 2 1 5",
"output": "3"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "3\n2 4 7",
"output": "3"
},
{
"input": "5\n1 1 1 1 3",
"output": "4"
},
{
"input": "3\n3 1 5",
"output": "3"
},
{
"input": "4\n1 2 3 7",
"output": "4"
},
{
"input": "3\n1 4 6",
"output": "3"
},
{
"input": "4\n2 1 2 2",
"output": "3"
},
{
"input": "2\n4 5",
"output": "2"
},
{
"input": "5\n1 2 1 2 1",
"output": "3"
},
{
"input": "3\n2 3 6",
"output": "3"
},
{
"input": "6\n1 1 4 1 1 5",
"output": "4"
},
{
"input": "5\n2 2 2 2 1",
"output": "3"
},
{
"input": "2\n5 6",
"output": "2"
},
{
"input": "4\n2 2 1 4",
"output": "3"
},
{
"input": "5\n2 2 3 4 4",
"output": "4"
},
{
"input": "4\n3 1 1 2",
"output": "2"
},
{
"input": "5\n3 4 1 4 5",
"output": "4"
},
{
"input": "4\n1 3 1 6",
"output": "4"
},
{
"input": "5\n1 1 1 2 2",
"output": "4"
},
{
"input": "4\n1 4 2 4",
"output": "3"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 8",
"output": "9"
},
{
"input": "4\n1 4 5 1",
"output": "3"
},
{
"input": "5\n1 1 1 1 5",
"output": "5"
},
{
"input": "4\n1 3 4 1",
"output": "3"
},
{
"input": "4\n2 2 2 3",
"output": "3"
},
{
"input": "4\n2 3 2 4",
"output": "3"
},
{
"input": "5\n2 2 1 2 2",
"output": "3"
},
{
"input": "3\n4 3 2",
"output": "2"
},
{
"input": "3\n6 5 2",
"output": "2"
},
{
"input": "69\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "35"
},
{
"input": "6\n1 1 1 1 1 2",
"output": "4"
},
{
"input": "5\n1 2 5 4 5",
"output": "4"
},
{
"input": "2\n9 10",
"output": "2"
},
{
"input": "3\n1 1 5",
"output": "3"
},
{
"input": "4\n3 4 3 5",
"output": "3"
},
{
"input": "4\n1 4 3 3",
"output": "3"
},
{
"input": "4\n7 1 3 4",
"output": "2"
},
{
"input": "3\n100 100 1",
"output": "2"
},
{
"input": "4\n5 2 2 2",
"output": "2"
}
] | 1,658,508,898 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 106 | 93 | 20,070,400 | from bisect import bisect_left
n = int(input())
a = list(map(int, input().split()))
for i in range(1, n):
a[i] += a[i-1]
if a[-1] % 2:
mid = (a[-1]+1)//2
else:
mid = a[-1]//2
print(bisect_left(a, mid)+1)
| Title: Equator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day.
Output Specification:
Print the index of the day when Polycarp will celebrate the equator.
Demo Input:
['4\n1 3 2 1\n', '6\n2 2 2 2 2 2\n']
Demo Output:
['2\n', '3\n']
Note:
In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training. | ```python
from bisect import bisect_left
n = int(input())
a = list(map(int, input().split()))
for i in range(1, n):
a[i] += a[i-1]
if a[-1] % 2:
mid = (a[-1]+1)//2
else:
mid = a[-1]//2
print(bisect_left(a, mid)+1)
``` | 3 |
|
915 | A | Garden | PROGRAMMING | 900 | [
"implementation"
] | null | null | Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding. | The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket. | Print one integer number — the minimum number of hours required to water the garden. | [
"3 6\n2 3 5\n",
"6 7\n1 2 3 4 5 6\n"
] | [
"2\n",
"7\n"
] | In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1. | 0 | [
{
"input": "3 6\n2 3 5",
"output": "2"
},
{
"input": "6 7\n1 2 3 4 5 6",
"output": "7"
},
{
"input": "5 97\n1 10 50 97 2",
"output": "1"
},
{
"input": "5 97\n1 10 50 100 2",
"output": "97"
},
{
"input": "100 100\n2 46 24 18 86 90 31 38 84 49 58 28 15 80 14 24 87 56 62 87 41 87 55 71 87 32 41 56 91 32 24 75 43 42 35 30 72 53 31 26 54 61 87 85 36 75 44 31 7 38 77 57 61 54 70 77 45 96 39 57 11 8 91 42 52 15 42 30 92 41 27 26 34 27 3 80 32 86 26 97 63 91 30 75 14 7 19 23 45 11 8 43 44 73 11 56 3 55 63 16",
"output": "50"
},
{
"input": "100 91\n13 13 62 96 74 47 81 46 78 21 20 42 4 73 25 30 76 74 58 28 25 52 42 48 74 40 82 9 25 29 17 22 46 64 57 95 81 39 47 86 40 95 97 35 31 98 45 98 47 78 52 63 58 14 89 97 17 95 28 22 20 36 68 38 95 16 2 26 54 47 42 31 31 81 21 21 65 40 82 53 60 71 75 33 96 98 6 22 95 12 5 48 18 27 58 62 5 96 36 75",
"output": "7"
},
{
"input": "8 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "3 8\n4 3 2",
"output": "2"
},
{
"input": "3 8\n2 4 2",
"output": "2"
},
{
"input": "3 6\n1 3 2",
"output": "2"
},
{
"input": "3 6\n3 2 5",
"output": "2"
},
{
"input": "3 8\n4 2 1",
"output": "2"
},
{
"input": "5 6\n2 3 5 1 2",
"output": "2"
},
{
"input": "2 6\n5 3",
"output": "2"
},
{
"input": "4 12\n6 4 3 1",
"output": "2"
},
{
"input": "3 18\n1 9 6",
"output": "2"
},
{
"input": "3 9\n3 2 1",
"output": "3"
},
{
"input": "3 6\n5 3 2",
"output": "2"
},
{
"input": "2 10\n5 2",
"output": "2"
},
{
"input": "2 18\n6 3",
"output": "3"
},
{
"input": "4 12\n1 2 12 3",
"output": "1"
},
{
"input": "3 7\n3 2 1",
"output": "7"
},
{
"input": "3 6\n3 2 1",
"output": "2"
},
{
"input": "5 10\n5 4 3 2 1",
"output": "2"
},
{
"input": "5 16\n8 4 2 1 7",
"output": "2"
},
{
"input": "6 7\n6 5 4 3 7 1",
"output": "1"
},
{
"input": "2 6\n3 2",
"output": "2"
},
{
"input": "2 4\n4 1",
"output": "1"
},
{
"input": "6 8\n2 4 1 3 5 7",
"output": "2"
},
{
"input": "6 8\n6 5 4 3 2 1",
"output": "2"
},
{
"input": "6 15\n5 2 3 6 4 3",
"output": "3"
},
{
"input": "4 8\n2 4 8 1",
"output": "1"
},
{
"input": "2 5\n5 1",
"output": "1"
},
{
"input": "4 18\n3 1 1 2",
"output": "6"
},
{
"input": "2 1\n2 1",
"output": "1"
},
{
"input": "3 10\n2 10 5",
"output": "1"
},
{
"input": "5 12\n12 4 4 4 3",
"output": "1"
},
{
"input": "3 6\n6 3 2",
"output": "1"
},
{
"input": "2 2\n2 1",
"output": "1"
},
{
"input": "3 18\n1 9 3",
"output": "2"
},
{
"input": "3 8\n7 2 4",
"output": "2"
},
{
"input": "2 100\n99 1",
"output": "100"
},
{
"input": "4 12\n1 3 4 2",
"output": "3"
},
{
"input": "3 6\n2 3 1",
"output": "2"
},
{
"input": "4 6\n3 2 5 12",
"output": "2"
},
{
"input": "4 97\n97 1 50 10",
"output": "1"
},
{
"input": "3 12\n1 12 2",
"output": "1"
},
{
"input": "4 12\n1 4 3 2",
"output": "3"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 19\n7 1 1",
"output": "19"
},
{
"input": "5 12\n12 4 3 4 4",
"output": "1"
},
{
"input": "3 8\n8 4 2",
"output": "1"
},
{
"input": "3 3\n3 2 1",
"output": "1"
},
{
"input": "5 6\n3 2 4 2 2",
"output": "2"
},
{
"input": "2 16\n8 4",
"output": "2"
},
{
"input": "3 6\n10 2 3",
"output": "2"
},
{
"input": "5 3\n2 4 5 3 6",
"output": "1"
},
{
"input": "11 99\n1 2 3 6 5 4 7 8 99 33 66",
"output": "1"
},
{
"input": "3 12\n3 12 2",
"output": "1"
},
{
"input": "5 25\n24 5 15 25 23",
"output": "1"
},
{
"input": "2 4\n8 1",
"output": "4"
},
{
"input": "4 100\n2 50 4 1",
"output": "2"
},
{
"input": "3 28\n7 14 1",
"output": "2"
},
{
"input": "4 8\n2 8 4 1",
"output": "1"
},
{
"input": "4 6\n6 1 2 3",
"output": "1"
},
{
"input": "2 12\n4 3",
"output": "3"
},
{
"input": "4 12\n1 2 4 3",
"output": "3"
},
{
"input": "5 12\n2 3 12 6 4",
"output": "1"
},
{
"input": "4 4\n1 2 2 4",
"output": "1"
},
{
"input": "3 6\n2 3 2",
"output": "2"
},
{
"input": "4 21\n21 20 21 2",
"output": "1"
},
{
"input": "3 8\n3 4 2",
"output": "2"
},
{
"input": "1 25\n25",
"output": "1"
},
{
"input": "99 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99",
"output": "1"
},
{
"input": "98 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98",
"output": "1"
},
{
"input": "79 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79",
"output": "1"
},
{
"input": "4 32\n1 1 1 1",
"output": "32"
},
{
"input": "1 100\n1",
"output": "100"
},
{
"input": "2 100\n7 1",
"output": "100"
},
{
"input": "7 24\n1 3 6 4 5 2 7",
"output": "4"
},
{
"input": "6 87\n1 2 8 4 5 7",
"output": "87"
},
{
"input": "1 88\n1",
"output": "88"
},
{
"input": "1 89\n1",
"output": "89"
}
] | 1,594,291,089 | 2,147,483,647 | PyPy 3 | OK | TESTS | 83 | 140 | 20,172,800 | l1 = [int(x) for x in input().split()]
total=l1[1]
l2 = [int(x) for x in input().split()]
l2.sort(reverse=True)
ans=1
for x in l2:
if total%x==0:
ans=x
break
ans = total//ans
print(ans) | Title: Garden
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
Input Specification:
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
Output Specification:
Print one integer number — the minimum number of hours required to water the garden.
Demo Input:
['3 6\n2 3 5\n', '6 7\n1 2 3 4 5 6\n']
Demo Output:
['2\n', '7\n']
Note:
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1. | ```python
l1 = [int(x) for x in input().split()]
total=l1[1]
l2 = [int(x) for x in input().split()]
l2.sort(reverse=True)
ans=1
for x in l2:
if total%x==0:
ans=x
break
ans = total//ans
print(ans)
``` | 3 |
|
677 | A | Vanya and Fence | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard? | The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person. | Print a single integer — the minimum possible valid width of the road. | [
"3 7\n4 5 14\n",
"6 1\n1 1 1 1 1 1\n",
"6 5\n7 6 8 9 10 5\n"
] | [
"4\n",
"6\n",
"11\n"
] | In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | 500 | [
{
"input": "3 7\n4 5 14",
"output": "4"
},
{
"input": "6 1\n1 1 1 1 1 1",
"output": "6"
},
{
"input": "6 5\n7 6 8 9 10 5",
"output": "11"
},
{
"input": "10 420\n214 614 297 675 82 740 174 23 255 15",
"output": "13"
},
{
"input": "10 561\n657 23 1096 487 785 66 481 554 1000 821",
"output": "15"
},
{
"input": "100 342\n478 143 359 336 162 333 385 515 117 496 310 538 469 539 258 676 466 677 1 296 150 560 26 213 627 221 255 126 617 174 279 178 24 435 70 145 619 46 669 566 300 67 576 251 58 176 441 564 569 194 24 669 73 262 457 259 619 78 400 579 222 626 269 47 80 315 160 194 455 186 315 424 197 246 683 220 68 682 83 233 290 664 273 598 362 305 674 614 321 575 362 120 14 534 62 436 294 351 485 396",
"output": "144"
},
{
"input": "100 290\n244 49 276 77 449 261 468 458 201 424 9 131 300 88 432 394 104 77 13 289 435 259 111 453 168 394 156 412 351 576 178 530 81 271 228 564 125 328 42 372 205 61 180 471 33 360 567 331 222 318 241 117 529 169 188 484 202 202 299 268 246 343 44 364 333 494 59 236 84 485 50 8 428 8 571 227 205 310 210 9 324 472 368 490 114 84 296 305 411 351 569 393 283 120 510 171 232 151 134 366",
"output": "145"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 1\n2",
"output": "2"
},
{
"input": "46 71\n30 26 56 138 123 77 60 122 73 45 79 10 130 3 14 1 38 46 128 50 82 16 32 68 28 98 62 106 2 49 131 11 114 39 139 70 40 50 45 137 33 30 35 136 135 19",
"output": "63"
},
{
"input": "20 723\n212 602 293 591 754 91 1135 640 80 495 845 928 1399 498 926 1431 1226 869 814 1386",
"output": "31"
},
{
"input": "48 864\n843 1020 751 1694 18 1429 1395 1174 272 1158 1628 1233 1710 441 765 561 778 748 1501 1200 563 1263 1398 1687 1518 1640 1591 839 500 466 1603 1587 1201 1209 432 868 1159 639 649 628 9 91 1036 147 896 1557 941 518",
"output": "75"
},
{
"input": "26 708\n549 241 821 734 945 1161 566 1268 216 30 1142 730 529 1014 255 168 796 1148 89 113 1328 286 743 871 1259 1397",
"output": "41"
},
{
"input": "75 940\n1620 1745 1599 441 64 1466 1496 1239 1716 1475 778 106 1136 1212 1261 444 781 257 1071 747 626 232 609 1544 682 1326 469 1361 1460 1450 1207 1319 922 625 1737 1057 1698 592 692 80 1016 541 1254 201 682 1007 847 206 1066 809 259 109 240 1611 219 1455 1326 1377 1827 786 42 1002 1382 1592 543 1866 1198 334 1524 1760 340 1566 955 257 1118",
"output": "116"
}
] | 1,696,404,001 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 46 | 0 | n, h = map(int, input().split())
arr = list(map(int, input().split()))
ans = n
for each in arr:
if each > h:
ans += 1
print(ans) | Title: Vanya and Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
Input Specification:
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
Output Specification:
Print a single integer — the minimum possible valid width of the road.
Demo Input:
['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n']
Demo Output:
['4\n', '6\n', '11\n']
Note:
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | ```python
n, h = map(int, input().split())
arr = list(map(int, input().split()))
ans = n
for each in arr:
if each > h:
ans += 1
print(ans)
``` | 3 |
|
379 | A | New Year Candles | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number. | The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000). | Print a single integer — the number of hours Vasily can light up the room for. | [
"4 2\n",
"6 3\n"
] | [
"7\n",
"8\n"
] | Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours. | 500 | [
{
"input": "4 2",
"output": "7"
},
{
"input": "6 3",
"output": "8"
},
{
"input": "1000 1000",
"output": "1001"
},
{
"input": "123 5",
"output": "153"
},
{
"input": "1000 2",
"output": "1999"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "3 2",
"output": "5"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "999 2",
"output": "1997"
},
{
"input": "1000 3",
"output": "1499"
},
{
"input": "1000 4",
"output": "1333"
},
{
"input": "1 1000",
"output": "1"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "10 4",
"output": "13"
},
{
"input": "4 3",
"output": "5"
},
{
"input": "91 5",
"output": "113"
},
{
"input": "777 17",
"output": "825"
},
{
"input": "100 4",
"output": "133"
},
{
"input": "5 3",
"output": "7"
},
{
"input": "6 4",
"output": "7"
},
{
"input": "26 8",
"output": "29"
},
{
"input": "9 4",
"output": "11"
},
{
"input": "20 3",
"output": "29"
},
{
"input": "17 3",
"output": "25"
}
] | 1,691,069,839 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 31 | 0 | n , m = map(int,input().split())
count = n
while n >= m :
count += n // m
n //= m
print(count) | Title: New Year Candles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input Specification:
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
Output Specification:
Print a single integer — the number of hours Vasily can light up the room for.
Demo Input:
['4 2\n', '6 3\n']
Demo Output:
['7\n', '8\n']
Note:
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours. | ```python
n , m = map(int,input().split())
count = n
while n >= m :
count += n // m
n //= m
print(count)
``` | 0 |
|
525 | B | Pasha and String | PROGRAMMING | 1,400 | [
"constructive algorithms",
"greedy",
"math",
"strings"
] | null | null | Pasha got a very beautiful string *s* for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |*s*| from left to right, where |*s*| is the length of the given string.
Pasha didn't like his present very much so he decided to change it. After his birthday Pasha spent *m* days performing the following transformations on his string — each day he chose integer *a**i* and reversed a piece of string (a segment) from position *a**i* to position |*s*|<=-<=*a**i*<=+<=1. It is guaranteed that 2·*a**i*<=≤<=|*s*|.
You face the following task: determine what Pasha's string will look like after *m* days. | The first line of the input contains Pasha's string *s* of length from 2 to 2·105 characters, consisting of lowercase Latin letters.
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=105) — the number of days when Pasha changed his string.
The third line contains *m* space-separated elements *a**i* (1<=≤<=*a**i*; 2·*a**i*<=≤<=|*s*|) — the position from which Pasha started transforming the string on the *i*-th day. | In the first line of the output print what Pasha's string *s* will look like after *m* days. | [
"abcdef\n1\n2\n",
"vwxyz\n2\n2 2\n",
"abcdef\n3\n1 2 3\n"
] | [
"aedcbf\n",
"vwxyz\n",
"fbdcea\n"
] | none | 750 | [
{
"input": "abcdef\n1\n2",
"output": "aedcbf"
},
{
"input": "vwxyz\n2\n2 2",
"output": "vwxyz"
},
{
"input": "abcdef\n3\n1 2 3",
"output": "fbdcea"
},
{
"input": "jc\n5\n1 1 1 1 1",
"output": "cj"
},
{
"input": "wljqgdlxyc\n13\n3 4 3 3 5 4 4 2 4 4 5 3 3",
"output": "wyjldgqxlc"
},
{
"input": "keicnqmuqinhsmtudqcilocxkbqgzhbkitmqwttdyoyvcbxincwjryzknubpacsngorexaldfurondbednowemnnlphhboycfavs\n2\n5 12",
"output": "keiccyobhhphsmtudqcilocxkbqgzhbkitmqwttdyoyvcbxincwjryzknubpacsngorexaldfurondbednowemnnlniqumqnfavs"
},
{
"input": "xwcxggxvfqbdklewbxkjzibmufnaywuxsqvwakefxbbkfandvigasbhbatsxyqxicrosatfsfybedklsaztyyiuurfbrzmwumujy\n100\n14 43 30 13 8 19 33 7 8 14 15 35 5 18 44 1 35 1 18 7 50 47 9 49 28 29 39 37 27 17 19 12 5 24 37 42 37 23 35 31 10 26 5 38 40 34 42 47 2 40 43 34 16 25 14 45 35 38 46 48 49 27 49 38 10 49 5 7 3 3 41 25 24 34 37 33 17 50 48 11 40 43 48 10 9 50 18 39 32 13 26 40 37 16 45 50 27 3 7 31",
"output": "xjcxggxvfbbruliyyxkjzikdebnfyftxsorcaxqyxbtkfhbdvigasnababsxfekiwvqsauwsayfumblsaztbweukdfqrzmwumuwy"
}
] | 1,668,110,927 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | s = list(input())
m = int(input())
transformations = [int(i)-1 for i in input().split()]
transformations.sort()
index = [0 for char in s]
for i, trans in enumerate(transformations):
count = transformations.count(trans)
if i % 2 == 0 and count % 2 != 0:
index[trans: len(s)-trans] = [1 for k in index[trans: len(s)-trans]]
else:
index[trans: len(s)-trans] = [0 for k in index[trans: len(s)-trans]]
for position, i in enumerate(index[:len(s)//2]):
if i == 1:
s[position], s[len(s)-position-1] = s[len(s)-position-1], s[position]
print("".join(s))
| Title: Pasha and String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha got a very beautiful string *s* for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |*s*| from left to right, where |*s*| is the length of the given string.
Pasha didn't like his present very much so he decided to change it. After his birthday Pasha spent *m* days performing the following transformations on his string — each day he chose integer *a**i* and reversed a piece of string (a segment) from position *a**i* to position |*s*|<=-<=*a**i*<=+<=1. It is guaranteed that 2·*a**i*<=≤<=|*s*|.
You face the following task: determine what Pasha's string will look like after *m* days.
Input Specification:
The first line of the input contains Pasha's string *s* of length from 2 to 2·105 characters, consisting of lowercase Latin letters.
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=105) — the number of days when Pasha changed his string.
The third line contains *m* space-separated elements *a**i* (1<=≤<=*a**i*; 2·*a**i*<=≤<=|*s*|) — the position from which Pasha started transforming the string on the *i*-th day.
Output Specification:
In the first line of the output print what Pasha's string *s* will look like after *m* days.
Demo Input:
['abcdef\n1\n2\n', 'vwxyz\n2\n2 2\n', 'abcdef\n3\n1 2 3\n']
Demo Output:
['aedcbf\n', 'vwxyz\n', 'fbdcea\n']
Note:
none | ```python
s = list(input())
m = int(input())
transformations = [int(i)-1 for i in input().split()]
transformations.sort()
index = [0 for char in s]
for i, trans in enumerate(transformations):
count = transformations.count(trans)
if i % 2 == 0 and count % 2 != 0:
index[trans: len(s)-trans] = [1 for k in index[trans: len(s)-trans]]
else:
index[trans: len(s)-trans] = [0 for k in index[trans: len(s)-trans]]
for position, i in enumerate(index[:len(s)//2]):
if i == 1:
s[position], s[len(s)-position-1] = s[len(s)-position-1], s[position]
print("".join(s))
``` | 0 |
|
610 | A | Pasha and Stick | PROGRAMMING | 1,000 | [
"combinatorics",
"math"
] | null | null | Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*.
Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.
Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way. | The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick. | The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square. | [
"6\n",
"20\n"
] | [
"1\n",
"4\n"
] | There is only one way to divide the stick in the first sample {1, 1, 2, 2}.
Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work. | 500 | [
{
"input": "6",
"output": "1"
},
{
"input": "20",
"output": "4"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "2000000000",
"output": "499999999"
},
{
"input": "1924704072",
"output": "481176017"
},
{
"input": "73740586",
"output": "18435146"
},
{
"input": "1925088820",
"output": "481272204"
},
{
"input": "593070992",
"output": "148267747"
},
{
"input": "1925473570",
"output": "481368392"
},
{
"input": "629490186",
"output": "157372546"
},
{
"input": "1980649112",
"output": "495162277"
},
{
"input": "36661322",
"output": "9165330"
},
{
"input": "1943590793",
"output": "0"
},
{
"input": "71207034",
"output": "17801758"
},
{
"input": "1757577394",
"output": "439394348"
},
{
"input": "168305294",
"output": "42076323"
},
{
"input": "1934896224",
"output": "483724055"
},
{
"input": "297149088",
"output": "74287271"
},
{
"input": "1898001634",
"output": "474500408"
},
{
"input": "176409698",
"output": "44102424"
},
{
"input": "1873025522",
"output": "468256380"
},
{
"input": "5714762",
"output": "1428690"
},
{
"input": "1829551192",
"output": "457387797"
},
{
"input": "16269438",
"output": "4067359"
},
{
"input": "1663283390",
"output": "415820847"
},
{
"input": "42549941",
"output": "0"
},
{
"input": "1967345604",
"output": "491836400"
},
{
"input": "854000",
"output": "213499"
},
{
"input": "1995886626",
"output": "498971656"
},
{
"input": "10330019",
"output": "0"
},
{
"input": "1996193634",
"output": "499048408"
},
{
"input": "9605180",
"output": "2401294"
},
{
"input": "1996459740",
"output": "499114934"
},
{
"input": "32691948",
"output": "8172986"
},
{
"input": "1975903308",
"output": "493975826"
},
{
"input": "1976637136",
"output": "494159283"
},
{
"input": "29803038",
"output": "7450759"
},
{
"input": "1977979692",
"output": "494494922"
},
{
"input": "1978595336",
"output": "494648833"
},
{
"input": "27379344",
"output": "6844835"
},
{
"input": "1979729912",
"output": "494932477"
},
{
"input": "1980253780",
"output": "495063444"
},
{
"input": "1980751584",
"output": "495187895"
},
{
"input": "53224878",
"output": "13306219"
},
{
"input": "5",
"output": "0"
},
{
"input": "7",
"output": "0"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "0"
},
{
"input": "10",
"output": "2"
},
{
"input": "11",
"output": "0"
},
{
"input": "12",
"output": "2"
},
{
"input": "13",
"output": "0"
},
{
"input": "14",
"output": "3"
},
{
"input": "15",
"output": "0"
},
{
"input": "16",
"output": "3"
},
{
"input": "17",
"output": "0"
},
{
"input": "18",
"output": "4"
},
{
"input": "19",
"output": "0"
},
{
"input": "21",
"output": "0"
},
{
"input": "22",
"output": "5"
},
{
"input": "23",
"output": "0"
},
{
"input": "24",
"output": "5"
},
{
"input": "25",
"output": "0"
},
{
"input": "26",
"output": "6"
},
{
"input": "27",
"output": "0"
},
{
"input": "28",
"output": "6"
},
{
"input": "29",
"output": "0"
},
{
"input": "30",
"output": "7"
},
{
"input": "111",
"output": "0"
},
{
"input": "55",
"output": "0"
},
{
"input": "105",
"output": "0"
},
{
"input": "199",
"output": "0"
},
{
"input": "151",
"output": "0"
}
] | 1,565,659,992 | 2,147,483,647 | PyPy 3 | OK | TESTS | 76 | 140 | 0 | # @author Matheus Alves dos Santos
def divisible_by(value, n):
return ((value % n) == 0)
length = int(input())
ways = 0
if (divisible_by(length, 2)):
ways = (length // 4)
if (divisible_by(length, 4)):
ways -= 1
print(ways) | Title: Pasha and Stick
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has a wooden stick of some positive integer length *n*. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be *n*.
Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.
Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer *x*, such that the number of parts of length *x* in the first way differ from the number of parts of length *x* in the second way.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=2·109) — the length of Pasha's stick.
Output Specification:
The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.
Demo Input:
['6\n', '20\n']
Demo Output:
['1\n', '4\n']
Note:
There is only one way to divide the stick in the first sample {1, 1, 2, 2}.
Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work. | ```python
# @author Matheus Alves dos Santos
def divisible_by(value, n):
return ((value % n) == 0)
length = int(input())
ways = 0
if (divisible_by(length, 2)):
ways = (length // 4)
if (divisible_by(length, 4)):
ways -= 1
print(ways)
``` | 3 |
|
702 | A | Maximum Increase | PROGRAMMING | 800 | [
"dp",
"greedy",
"implementation"
] | null | null | You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array.
A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous. | The first line contains single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). | Print the maximum length of an increasing subarray of the given array. | [
"5\n1 7 2 11 15\n",
"6\n100 100 100 100 100 100\n",
"3\n1 2 3\n"
] | [
"3\n",
"1\n",
"3\n"
] | none | 0 | [
{
"input": "5\n1 7 2 11 15",
"output": "3"
},
{
"input": "6\n100 100 100 100 100 100",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "10\n802030518 598196518 640274071 983359971 71550121 96204862 799843967 446173607 796619138 402690754",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n1 2 3 3 4",
"output": "3"
},
{
"input": "4\n1 2 2 3",
"output": "2"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "3\n1 2 1",
"output": "2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n1 1 2",
"output": "2"
},
{
"input": "11\n1 2 3 1 2 3 2 1 2 3 4",
"output": "4"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "9"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "3\n3 2 1",
"output": "1"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "1\n1234394",
"output": "1"
}
] | 1,673,338,443 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | n = int(input())
a = [0]
for i in input().split():
a.append(int(i))
dp = []
x = 0
for i in range(1, len(a)):
if a[i] <= a[i - 1]:
dp.append(x)
x = 1
continue
x += 1
print(max(dp)) | Title: Maximum Increase
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array.
A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.
Input Specification:
The first line contains single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print the maximum length of an increasing subarray of the given array.
Demo Input:
['5\n1 7 2 11 15\n', '6\n100 100 100 100 100 100\n', '3\n1 2 3\n']
Demo Output:
['3\n', '1\n', '3\n']
Note:
none | ```python
n = int(input())
a = [0]
for i in input().split():
a.append(int(i))
dp = []
x = 0
for i in range(1, len(a)):
if a[i] <= a[i - 1]:
dp.append(x)
x = 1
continue
x += 1
print(max(dp))
``` | 0 |
|
938 | B | Run For Your Prize | PROGRAMMING | 1,100 | [
"brute force",
"greedy"
] | null | null | You and your friend are participating in a TV show "Run For Your Prize".
At the start of the show *n* prizes are located on a straight line. *i*-th prize is located at position *a**i*. Positions of all prizes are distinct. You start at position 1, your friend — at position 106 (and there is no prize in any of these two positions). You have to work as a team and collect all prizes in minimum possible time, in any order.
You know that it takes exactly 1 second to move from position *x* to position *x*<=+<=1 or *x*<=-<=1, both for you and your friend. You also have trained enough to instantly pick up any prize, if its position is equal to your current position (and the same is true for your friend). Carrying prizes does not affect your speed (or your friend's speed) at all.
Now you may discuss your strategy with your friend and decide who will pick up each prize. Remember that every prize must be picked up, either by you or by your friend.
What is the minimum number of seconds it will take to pick up all the prizes? | The first line contains one integer *n* (1<=≤<=*n*<=≤<=105) — the number of prizes.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (2<=≤<=*a**i*<=≤<=106<=-<=1) — the positions of the prizes. No two prizes are located at the same position. Positions are given in ascending order. | Print one integer — the minimum number of seconds it will take to collect all prizes. | [
"3\n2 3 9\n",
"2\n2 999995\n"
] | [
"8\n",
"5\n"
] | In the first example you take all the prizes: take the first at 1, the second at 2 and the third at 8.
In the second example you take the first prize in 1 second and your friend takes the other in 5 seconds, you do this simultaneously, so the total time is 5. | 0 | [
{
"input": "3\n2 3 9",
"output": "8"
},
{
"input": "2\n2 999995",
"output": "5"
},
{
"input": "1\n20",
"output": "19"
},
{
"input": "6\n2 3 500000 999997 999998 999999",
"output": "499999"
},
{
"input": "1\n999999",
"output": "1"
},
{
"input": "1\n510000",
"output": "490000"
},
{
"input": "3\n2 5 27",
"output": "26"
},
{
"input": "2\n600000 800000",
"output": "400000"
},
{
"input": "5\n2 5 6 27 29",
"output": "28"
},
{
"input": "1\n500001",
"output": "499999"
},
{
"input": "10\n3934 38497 42729 45023 51842 68393 77476 82414 91465 98055",
"output": "98054"
},
{
"input": "1\n900000",
"output": "100000"
},
{
"input": "1\n500000",
"output": "499999"
},
{
"input": "1\n999998",
"output": "2"
},
{
"input": "3\n999997 999998 999999",
"output": "3"
},
{
"input": "2\n999997 999999",
"output": "3"
},
{
"input": "2\n2 999998",
"output": "2"
},
{
"input": "2\n500000 500001",
"output": "499999"
},
{
"input": "1\n500002",
"output": "499998"
},
{
"input": "1\n700000",
"output": "300000"
},
{
"input": "2\n2 999999",
"output": "1"
},
{
"input": "2\n999998 999999",
"output": "2"
},
{
"input": "1\n999995",
"output": "5"
},
{
"input": "2\n499999 500001",
"output": "499999"
},
{
"input": "1\n499999",
"output": "499998"
},
{
"input": "2\n100 999900",
"output": "100"
},
{
"input": "2\n499999 500000",
"output": "499999"
},
{
"input": "2\n500001 999999",
"output": "499999"
},
{
"input": "3\n500000 500001 500002",
"output": "499999"
},
{
"input": "2\n2 500001",
"output": "499999"
},
{
"input": "2\n499999 999999",
"output": "499998"
},
{
"input": "2\n2 500000",
"output": "499999"
},
{
"input": "4\n2 3 4 999999",
"output": "3"
},
{
"input": "2\n100000 500001",
"output": "499999"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n800000",
"output": "200000"
},
{
"input": "1\n505050",
"output": "494950"
},
{
"input": "1\n753572",
"output": "246428"
},
{
"input": "2\n576696 760487",
"output": "423304"
},
{
"input": "10\n3 4 5 6 7 8 9 10 11 12",
"output": "11"
},
{
"input": "4\n2 3 4 5",
"output": "4"
},
{
"input": "4\n999996 999997 999998 999999",
"output": "4"
}
] | 1,520,034,937 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 93 | 13,004,800 | from bisect import bisect
def main():
n, l = int(input()), [*map(float, input().split()), 1e6]
i = bisect(l, 500000.5)
print(int(max(l[i - 1] - 1., 1e6 - l[i])))
if __name__ == '__main__':
main()
| Title: Run For Your Prize
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You and your friend are participating in a TV show "Run For Your Prize".
At the start of the show *n* prizes are located on a straight line. *i*-th prize is located at position *a**i*. Positions of all prizes are distinct. You start at position 1, your friend — at position 106 (and there is no prize in any of these two positions). You have to work as a team and collect all prizes in minimum possible time, in any order.
You know that it takes exactly 1 second to move from position *x* to position *x*<=+<=1 or *x*<=-<=1, both for you and your friend. You also have trained enough to instantly pick up any prize, if its position is equal to your current position (and the same is true for your friend). Carrying prizes does not affect your speed (or your friend's speed) at all.
Now you may discuss your strategy with your friend and decide who will pick up each prize. Remember that every prize must be picked up, either by you or by your friend.
What is the minimum number of seconds it will take to pick up all the prizes?
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=105) — the number of prizes.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (2<=≤<=*a**i*<=≤<=106<=-<=1) — the positions of the prizes. No two prizes are located at the same position. Positions are given in ascending order.
Output Specification:
Print one integer — the minimum number of seconds it will take to collect all prizes.
Demo Input:
['3\n2 3 9\n', '2\n2 999995\n']
Demo Output:
['8\n', '5\n']
Note:
In the first example you take all the prizes: take the first at 1, the second at 2 and the third at 8.
In the second example you take the first prize in 1 second and your friend takes the other in 5 seconds, you do this simultaneously, so the total time is 5. | ```python
from bisect import bisect
def main():
n, l = int(input()), [*map(float, input().split()), 1e6]
i = bisect(l, 500000.5)
print(int(max(l[i - 1] - 1., 1e6 - l[i])))
if __name__ == '__main__':
main()
``` | 0 |
|
115 | A | Party | PROGRAMMING | 900 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed? | The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles. | Print a single integer denoting the minimum number of groups that will be formed in the party. | [
"5\n-1\n1\n2\n1\n-1\n"
] | [
"3\n"
] | For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5 | 500 | [
{
"input": "5\n-1\n1\n2\n1\n-1",
"output": "3"
},
{
"input": "4\n-1\n1\n2\n3",
"output": "4"
},
{
"input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11",
"output": "4"
},
{
"input": "6\n-1\n-1\n2\n3\n1\n1",
"output": "3"
},
{
"input": "3\n-1\n1\n1",
"output": "2"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n2\n-1",
"output": "2"
},
{
"input": "2\n-1\n-1",
"output": "1"
},
{
"input": "3\n2\n-1\n1",
"output": "3"
},
{
"input": "3\n-1\n-1\n-1",
"output": "1"
},
{
"input": "5\n4\n5\n1\n-1\n4",
"output": "3"
},
{
"input": "12\n-1\n1\n1\n1\n1\n1\n3\n4\n3\n3\n4\n7",
"output": "4"
},
{
"input": "12\n-1\n-1\n1\n-1\n1\n1\n5\n11\n8\n6\n6\n4",
"output": "5"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n2\n-1\n-1\n-1",
"output": "2"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1",
"output": "1"
},
{
"input": "12\n3\n4\n2\n8\n7\n1\n10\n12\n5\n-1\n9\n11",
"output": "12"
},
{
"input": "12\n5\n6\n7\n1\n-1\n9\n12\n4\n8\n-1\n3\n2",
"output": "11"
},
{
"input": "12\n-1\n9\n11\n6\n6\n-1\n6\n3\n8\n6\n1\n6",
"output": "6"
},
{
"input": "12\n7\n8\n4\n12\n7\n9\n-1\n-1\n-1\n8\n6\n-1",
"output": "3"
},
{
"input": "12\n-1\n10\n-1\n1\n-1\n5\n9\n12\n-1\n-1\n3\n-1",
"output": "2"
},
{
"input": "12\n-1\n7\n9\n12\n1\n7\n-1\n-1\n8\n5\n4\n-1",
"output": "3"
},
{
"input": "12\n11\n11\n8\n9\n1\n1\n2\n-1\n10\n3\n-1\n8",
"output": "5"
},
{
"input": "12\n-1\n8\n9\n-1\n4\n2\n11\n1\n-1\n6\n-1\n10",
"output": "6"
},
{
"input": "12\n7\n4\n4\n-1\n6\n7\n9\n7\n-1\n7\n4\n1",
"output": "4"
},
{
"input": "12\n-1\n6\n10\n6\n1\n11\n5\n9\n11\n5\n1\n9",
"output": "4"
},
{
"input": "12\n8\n6\n6\n8\n6\n-1\n6\n6\n8\n6\n8\n8",
"output": "3"
},
{
"input": "12\n3\n8\n9\n12\n2\n8\n11\n12\n2\n9\n-1\n11",
"output": "7"
},
{
"input": "7\n7\n6\n-1\n-1\n4\n4\n-1",
"output": "3"
},
{
"input": "77\n53\n-1\n48\n-1\n51\n-1\n54\n43\n70\n19\n64\n31\n45\n58\n6\n61\n41\n56\n34\n-1\n54\n-1\n66\n65\n74\n16\n38\n1\n2\n61\n-1\n34\n-1\n-1\n7\n19\n57\n13\n61\n42\n-1\n34\n65\n16\n31\n61\n6\n5\n34\n15\n-1\n13\n4\n22\n74\n7\n-1\n70\n38\n61\n45\n53\n46\n22\n-1\n65\n34\n19\n43\n48\n33\n58\n36\n-1\n47\n24\n34",
"output": "6"
},
{
"input": "5\n2\n3\n4\n5\n-1",
"output": "5"
}
] | 1,695,155,446 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | n=int(input())
b=[]
c=[]
x=0
res=[]
for i in range (n):
a=int(input())
b.append(a)
for i in range(1,len(b)+1):
c.append(i)
for j in range(1,len(b)+1):
if i==b[j-1]:
c.append(j)
d=c.copy()
res.append(d)
c.clear()
for i in res:
if len(i)==1:
x=x+1
print(x)
| Title: Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
Output Specification:
Print a single integer denoting the minimum number of groups that will be formed in the party.
Demo Input:
['5\n-1\n1\n2\n1\n-1\n']
Demo Output:
['3\n']
Note:
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5 | ```python
n=int(input())
b=[]
c=[]
x=0
res=[]
for i in range (n):
a=int(input())
b.append(a)
for i in range(1,len(b)+1):
c.append(i)
for j in range(1,len(b)+1):
if i==b[j-1]:
c.append(j)
d=c.copy()
res.append(d)
c.clear()
for i in res:
if len(i)==1:
x=x+1
print(x)
``` | 0 |
|
721 | A | One-dimensional Japanese Crossword | PROGRAMMING | 800 | [
"implementation"
] | null | null | Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew). | The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right. | [
"3\nBBW\n",
"5\nBWBWB\n",
"4\nWWWW\n",
"4\nBBBB\n",
"13\nWBBBBWWBWBBBW\n"
] | [
"1\n2 ",
"3\n1 1 1 ",
"0\n",
"1\n4 ",
"3\n4 1 3 "
] | The last sample case correspond to the picture in the statement. | 500 | [
{
"input": "3\nBBW",
"output": "1\n2 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "4\nWWWW",
"output": "0"
},
{
"input": "4\nBBBB",
"output": "1\n4 "
},
{
"input": "13\nWBBBBWWBWBBBW",
"output": "3\n4 1 3 "
},
{
"input": "1\nB",
"output": "1\n1 "
},
{
"input": "2\nBB",
"output": "1\n2 "
},
{
"input": "100\nWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWB",
"output": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "1\nW",
"output": "0"
},
{
"input": "2\nWW",
"output": "0"
},
{
"input": "2\nWB",
"output": "1\n1 "
},
{
"input": "2\nBW",
"output": "1\n1 "
},
{
"input": "3\nBBB",
"output": "1\n3 "
},
{
"input": "3\nBWB",
"output": "2\n1 1 "
},
{
"input": "3\nWBB",
"output": "1\n2 "
},
{
"input": "3\nWWB",
"output": "1\n1 "
},
{
"input": "3\nWBW",
"output": "1\n1 "
},
{
"input": "3\nBWW",
"output": "1\n1 "
},
{
"input": "3\nWWW",
"output": "0"
},
{
"input": "100\nBBBWWWWWWBBWWBBWWWBBWBBBBBBBBBBBWBBBWBBWWWBBWWBBBWBWWBBBWWBBBWBBBBBWWWBWWBBWWWWWWBWBBWWBWWWBWBWWWWWB",
"output": "21\n3 2 2 2 11 3 2 2 3 1 3 3 5 1 2 1 2 1 1 1 1 "
},
{
"input": "5\nBBBWB",
"output": "2\n3 1 "
},
{
"input": "5\nBWWWB",
"output": "2\n1 1 "
},
{
"input": "5\nWWWWB",
"output": "1\n1 "
},
{
"input": "5\nBWWWW",
"output": "1\n1 "
},
{
"input": "5\nBBBWW",
"output": "1\n3 "
},
{
"input": "5\nWWBBB",
"output": "1\n3 "
},
{
"input": "10\nBBBBBWWBBB",
"output": "2\n5 3 "
},
{
"input": "10\nBBBBWBBWBB",
"output": "3\n4 2 2 "
},
{
"input": "20\nBBBBBWWBWBBWBWWBWBBB",
"output": "6\n5 1 2 1 1 3 "
},
{
"input": "20\nBBBWWWWBBWWWBWBWWBBB",
"output": "5\n3 2 1 1 3 "
},
{
"input": "20\nBBBBBBBBWBBBWBWBWBBB",
"output": "5\n8 3 1 1 3 "
},
{
"input": "20\nBBBWBWBWWWBBWWWWBWBB",
"output": "6\n3 1 1 2 1 2 "
},
{
"input": "40\nBBBBBBWWWWBWBWWWBWWWWWWWWWWWBBBBBBBBBBBB",
"output": "5\n6 1 1 1 12 "
},
{
"input": "40\nBBBBBWBWWWBBWWWBWBWWBBBBWWWWBWBWBBBBBBBB",
"output": "9\n5 1 2 1 1 4 1 1 8 "
},
{
"input": "50\nBBBBBBBBBBBWWWWBWBWWWWBBBBBBBBWWWWWWWBWWWWBWBBBBBB",
"output": "7\n11 1 1 8 1 1 6 "
},
{
"input": "50\nWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW",
"output": "0"
},
{
"input": "50\nBBBBBWWWWWBWWWBWWWWWBWWWBWWWWWWBBWBBWWWWBWWWWWWWBW",
"output": "9\n5 1 1 1 1 2 2 1 1 "
},
{
"input": "50\nWWWWBWWBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWBWWWWWWWBBBBB",
"output": "6\n1 1 1 1 1 5 "
},
{
"input": "50\nBBBBBWBWBWWBWBWWWWWWBWBWBWWWWWWWWWWWWWBWBWWWWBWWWB",
"output": "12\n5 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1\n50 "
},
{
"input": "100\nBBBBBBBBBBBWBWWWWBWWBBWBBWWWWWWWWWWBWBWWBWWWWWWWWWWWBBBWWBBWWWWWBWBWWWWBWWWWWWWWWWWBWWWWWBBBBBBBBBBB",
"output": "15\n11 1 1 2 2 1 1 1 3 2 1 1 1 1 11 "
},
{
"input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1\n100 "
},
{
"input": "100\nBBBBBBBBBBBBBBBBBBBBWBWBWWWWWBWWWWWWWWWWWWWWBBWWWBWWWWBWWBWWWWWWBWWWWWWWWWWWWWBWBBBBBBBBBBBBBBBBBBBB",
"output": "11\n20 1 1 1 2 1 1 1 1 1 20 "
},
{
"input": "100\nBBBBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWWWWWBWBWWWWWWBBWWWWWWWWWWWWBWWWWBWWWWWWWWWWWWBWWWWWWWBWWWWWWWBBBBBB",
"output": "11\n4 1 1 1 1 2 1 1 1 1 6 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "10\nWWBWWWBWBB",
"output": "3\n1 1 2 "
},
{
"input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1\n50 "
},
{
"input": "50\nBBBBBBBBBBBBBBBBBWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "2\n17 31 "
},
{
"input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "2\n24 42 "
},
{
"input": "90\nWWBWWBWBBWBBWWBWBWBBBWBWBBBWBWBWBWBWBWBWBWBBBBBWBBWWWWBWBBWBWWBBBWBWBWWBWBWBWBWWWWWWBWBBBB",
"output": "30\n1 1 2 2 1 1 3 1 3 1 1 1 1 1 1 1 5 2 1 2 1 3 1 1 1 1 1 1 1 4 "
},
{
"input": "100\nBWWWBWBWBBBBBWBWWBWBWWWBWBWBWWBBWWBBBWBBBWWBWBWWBBBBWBWBBBWBWBBWWWWWWBWWBBBBWBWBWWBWBWWWBWBWWBWBWWWB",
"output": "31\n1 1 1 5 1 1 1 1 1 1 2 3 3 1 1 4 1 3 1 2 1 4 1 1 1 1 1 1 1 1 1 "
},
{
"input": "90\nWBWBBBBBBWWWBBWWBWWWBBWWBWWWBWBBWBWBBWWWWBWBWBBWBBWBWWWBBWBBWWWWBWBBWWWBBBWBBWBWBBBBWWBWWB",
"output": "25\n1 6 2 1 2 1 1 2 1 2 1 1 2 2 1 2 2 1 2 3 2 1 4 1 1 "
},
{
"input": "80\nBBWWBBBWBBWWWWBBWBWBBWWWWWBWBBWWBWBWBWBWBWWBWWBWWWBWBBWBBWBBWBBBWWBBBBBBBWBBBWBB",
"output": "23\n2 3 2 2 1 2 1 2 1 1 1 1 1 1 1 1 2 2 2 3 7 3 2 "
},
{
"input": "65\nWWWWBWWWBBBBBWWWWWWBBBWWBBBBWWWWWWWWBBBWWWWBWBWWBBWWWWBWWWBBWBBBB",
"output": "11\n1 5 3 4 3 1 1 2 1 2 4 "
}
] | 1,619,044,359 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 | def printStrings(strings):
for element in strings:
print(element)
k=input()
n=input()
bcount=0
wcount=0
list=[]
for i in n:
if i=="B":
bcount=bcunt+1
elif i=="W":
if bcount>=1:
wcount=wcount+1
list.append(bcount)
bcount=0
if bcount>=1:
list.append(bcount)
wcount=wcount+1
print(wcount)
printStrings(list)
| Title: One-dimensional Japanese Crossword
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output Specification:
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Demo Input:
['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n']
Demo Output:
['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 ']
Note:
The last sample case correspond to the picture in the statement. | ```python
def printStrings(strings):
for element in strings:
print(element)
k=input()
n=input()
bcount=0
wcount=0
list=[]
for i in n:
if i=="B":
bcount=bcunt+1
elif i=="W":
if bcount>=1:
wcount=wcount+1
list.append(bcount)
bcount=0
if bcount>=1:
list.append(bcount)
wcount=wcount+1
print(wcount)
printStrings(list)
``` | -1 |
|
873 | D | Merge Sort | PROGRAMMING | 1,800 | [
"constructive algorithms",
"divide and conquer"
] | null | null | Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows:
1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=<<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions.
The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*).
The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted.
Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*.
Help Ivan to find an array he wants! | The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it. | If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them. | [
"3 3\n",
"4 1\n",
"5 6\n"
] | [
"2 1 3 ",
"1 2 3 4 ",
"-1\n"
] | none | 0 | [
{
"input": "3 3",
"output": "2 1 3 "
},
{
"input": "4 1",
"output": "1 2 3 4 "
},
{
"input": "5 6",
"output": "-1"
},
{
"input": "100 100",
"output": "-1"
},
{
"input": "10000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "10000 20001",
"output": "-1"
},
{
"input": "10000 30001",
"output": "-1"
},
{
"input": "20000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "20000 20001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "20000 30001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "30000 10001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "30000 20001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "30000 30001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "40000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "40000 20001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "40000 30001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "50000 10001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "50000 20001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "50000 30001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "60000 10001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "60000 20001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "60000 30001",
"output": "2 4 1 6 3 8 5 9 11 7 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 31 33 29 35 32 37 34 38 40 36 42 39 44 41 45 47 43 49 46 51 48 53 50 55 52 57 54 59 56 60 62 58 64 61 66 63 67 69 65 71 68 73 70 74 76 72 78 75 80 77 82 79 84 81 86 83 88 85 89 91 87 93 90 95 92 97 94 99 96 101 98 103 100 104 106 102 108 105 110 107 112 109 114 111 116 113 118 115 119 121 117 123 120 125 122 126 128 124 130 127 132 129 133 135 131 137 134 139 136 141 138 143 140 145 142 147 144 148 150 146 152 149 154 151 155 157..."
},
{
"input": "70000 10001",
"output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..."
},
{
"input": "70000 20001",
"output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..."
},
{
"input": "70000 30001",
"output": "3 1 5 2 7 4 9 6 11 8 13 10 15 12 16 18 14 20 17 22 19 24 21 26 23 28 25 30 27 32 29 33 35 31 37 34 39 36 41 38 43 40 45 42 47 44 49 46 50 52 48 54 51 56 53 58 55 60 57 62 59 64 61 66 63 67 69 65 71 68 73 70 75 72 77 74 79 76 81 78 83 80 84 86 82 88 85 90 87 92 89 94 91 96 93 98 95 100 97 101 103 99 105 102 107 104 109 106 111 108 113 110 115 112 117 114 118 120 116 122 119 124 121 126 123 128 125 130 127 132 129 134 131 135 137 133 139 136 141 138 143 140 145 142 147 144 149 146 151 148 152 154 150 156 153..."
},
{
"input": "80000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "80000 20001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "80000 30001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27 24 28 30 26 32 29 33 35 31 37 34 38 40 36 42 39 44 41 46 43 47 49 45 51 48 52 54 50 56 53 57 59 55 61 58 62 64 60 66 63 67 69 65 71 68 72 74 70 76 73 77 79 75 81 78 83 80 85 82 86 88 84 90 87 91 93 89 95 92 96 98 94 100 97 101 103 99 105 102 106 108 104 110 107 111 113 109 115 112 116 118 114 120 117 122 119 124 121 125 127 123 129 126 130 132 128 134 131 135 137 133 139 136 140 142 138 144 141 145 147 143 149 146 150 152 148 154 151 155 157..."
},
{
"input": "90000 10001",
"output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..."
},
{
"input": "90000 20001",
"output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..."
},
{
"input": "90000 30001",
"output": "3 1 4 6 2 8 5 9 11 7 13 10 14 16 12 17 19 15 20 22 18 24 21 25 27 23 28 30 26 31 33 29 35 32 36 38 34 39 41 37 42 44 40 46 43 47 49 45 50 52 48 53 55 51 57 54 58 60 56 61 63 59 64 66 62 68 65 69 71 67 72 74 70 75 77 73 79 76 80 82 78 83 85 81 86 88 84 90 87 91 93 89 94 96 92 97 99 95 101 98 102 104 100 105 107 103 108 110 106 112 109 113 115 111 116 118 114 119 121 117 123 120 124 126 122 127 129 125 130 132 128 134 131 135 137 133 138 140 136 141 143 139 145 142 146 148 144 149 151 147 152 154 150 156 153..."
},
{
"input": "100000 10001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "100000 20001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "100000 30001",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "100000 199999",
"output": "2 4 1 5 7 3 8 10 6 11 13 9 14 16 12 17 19 15 20 22 18 23 25 21 26 28 24 29 31 27 32 34 30 35 37 33 38 40 36 41 43 39 44 46 42 47 49 45 50 52 48 53 55 51 56 58 54 59 61 57 62 64 60 65 67 63 68 70 66 71 73 69 74 76 72 77 79 75 80 82 78 83 85 81 86 88 84 89 91 87 92 94 90 96 93 98 95 99 101 97 102 104 100 105 107 103 108 110 106 111 113 109 114 116 112 117 119 115 120 122 118 123 125 121 126 128 124 129 131 127 132 134 130 135 137 133 138 140 136 141 143 139 145 142 147 144 148 150 146 151 153 149 154 156 152..."
},
{
"input": "10 17",
"output": "3 1 4 6 2 8 5 9 7 10 "
}
] | 1,594,489,566 | 2,147,483,647 | PyPy 3 | OK | TESTS | 36 | 218 | 35,635,200 | n,k = [int(x) for x in input().split()]
if k%2 == 0:
print(-1)
else:
# a completely unsorted list of length n
# will have 2(n-1) child calls
if k > 2*n-1:
print(-1)
else:
arr = [n-x for x in range(n)]
sorts = 2*n-1-k
segs = [[(0,n,n//2)]]
layers = 0
nCopy = n
while nCopy:
layers += 1
nCopy //= 2
for x in range(layers):
newSeg = []
for l,r,mid in segs[x]:
if mid-l > 1:
newSeg.append((l,mid,(mid+l)//2))
if r-mid > 1:
newSeg.append((mid,r,(mid+r)//2))
segs.append(newSeg)
for x in range(sorts//2):
while not segs[-1]:
segs.pop()
l,r,mid = segs[-1].pop()
newPart = arr[mid:r] + arr[l:mid]
arr[l:r] = newPart
print(' '.join([str(x) for x in arr]))
| Title: Merge Sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows:
1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=<<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions.
The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*).
The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted.
Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*.
Help Ivan to find an array he wants!
Input Specification:
The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it.
Output Specification:
If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.
Demo Input:
['3 3\n', '4 1\n', '5 6\n']
Demo Output:
['2 1 3 ', '1 2 3 4 ', '-1\n']
Note:
none | ```python
n,k = [int(x) for x in input().split()]
if k%2 == 0:
print(-1)
else:
# a completely unsorted list of length n
# will have 2(n-1) child calls
if k > 2*n-1:
print(-1)
else:
arr = [n-x for x in range(n)]
sorts = 2*n-1-k
segs = [[(0,n,n//2)]]
layers = 0
nCopy = n
while nCopy:
layers += 1
nCopy //= 2
for x in range(layers):
newSeg = []
for l,r,mid in segs[x]:
if mid-l > 1:
newSeg.append((l,mid,(mid+l)//2))
if r-mid > 1:
newSeg.append((mid,r,(mid+r)//2))
segs.append(newSeg)
for x in range(sorts//2):
while not segs[-1]:
segs.pop()
l,r,mid = segs[-1].pop()
newPart = arr[mid:r] + arr[l:mid]
arr[l:r] = newPart
print(' '.join([str(x) for x in arr]))
``` | 3 |
|
312 | A | Whose sentence is it? | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100. | For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions. | [
"5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n"
] | [
"Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\n"
] | none | 500 | [
{
"input": "5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .",
"output": "Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!"
},
{
"input": "10\nLpAEKiHVJrzSZqBVSSyY\nYECGBlala.\nUZeGpeM.UCwiHmmA\nqt_,.b_.LSwJtJ.\nFAnXZtHlala.\nmiao.iapelala.\nCFPlbUgObrXLejPNu.F\nZSUfvisiHyrIMjMlala.\nmiao. lala.\nd,IWSeumytrVlala.",
"output": "OMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nFreda's"
},
{
"input": "10\nmiao.,taUvXPVlala.\nmiao.txEeId.X_lala.\nLZIeAEd JaeBVlala.\ncKPIsWpwIlala.\nfYp.eSvn,g\nKMx,nFEslala.\nmiao.QtMyxYqiajjuM\nDutxNkCqywgcnCYskcd\ngFLKACjeqfD\n,Ss UmY.wJvcX",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nFreda's\nOMG>.< I don't know!\nFreda's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nmiao.Plala.\nDVm,VYslala.\nmiao.rlala.\nmiao.,KQNL.fO_.QRc\nUBLCKEUePlala.\nIouS.Alala.\nmiao.lala.\nmiao.rlala.\nEJZwRJeKlala.\nmiao.Olala.",
"output": "OMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nRainbow's\nFreda's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!"
},
{
"input": "10\nmiao.grFTpju.jCLRnZ\ng.pVHYA_Usnm\nlloWONolcMFElala.\nAW,n.JJkOTe.Nd\n.bP.HvKlala.\nGziqPGQa,lala.\nmiao.,QkOCH.vFlala.\n.PUtOwImvUsoeh \nmiao.Z,KIds.R\nmiao.,_MDzoaAiJlala.",
"output": "Rainbow's\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nFreda's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!"
},
{
"input": "10\nmiao.xWfjV\nHFVrGCDQXyZ,Sbm\nLMDS.xVkTCAY.vm\nmiao.lLBglala.\nnl,jRPyClala.\nFYnHoXlala.\nmiao. oxaHE\n.WTrw_mNpOQCa\nHOk..wHYoyMhl\nQX,XpMuPIROM",
"output": "Rainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nFreda's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nJBQqiXlala.\npUNUWQRiMPCXv\nAiLnfNHWznwkC.lala.\nmiao.Dl_Oy\nxJJJkVkdfOzQBH_SmKh\nfgD_IHvdHiorE,W\nmiao.usBKixglala.\nwCpqPUzEtD\nmiao.rlala.\nmiao.JylcGvWlala.",
"output": "Freda's\nOMG>.< I don't know!\nFreda's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nmiao..FLhPl_Wjslala.\nmiao. tdEGtfdJlala.\nGAzEUlala.\nKCcmOa .aKBlZyYsdu.V\nmiao.lala.\njKylnM,FXK\nmiao.GBWqjGH.v\nmiao.RefxS Cni.\nOxaaEihuHQR_s,\nmiao.a,Axtlala.",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nNo.I_aTXlala.\nmiao.JKSCoRZS\nnOBMIlala.\nmiao.nlala.\nmiao._xqxoHIIlala.\nmiao.NJPy SWyiUDWc\nmiao.cCnahFaqqj.Xqp\nnreSMDeXPPYAQxI,W\nAktPajWimdd_qRn\nmiao.QHwKCYlala.",
"output": "Freda's\nRainbow's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\n \n,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ ,.._ \n \nmiao.miao.miao.\nlala.lala.lala.\nlala.miao.\nmiaolala. \nmiao.lala\nmiaolala_\n,.._ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nduClyjMIPsEuWmx_Ce.byVoizYlTM,sF\nuZHsNip_,Mwtg,FZjM_LzPC,_pSvEOyTHfAOvoZXvxCZdgYDTCDdCAoSVZWyxXGcLgWlala.\nEGtJFPAvTEcqjkhaGxdduaQ_rmUzF.WaU, EIuX B,aVzFFpFrxpwADXuayRD azDfj \n_tJqYzXyqc.,u.F,mUYukveBPWnPq,f,dJnPHuBazdnbRHfzwNUdRbheAIjcoaPcnLvocrzcioxCapb R\n.YUBeb_zmwUt.QQuUdQIiOXtqshcsycEe,HLytHlala.\ndJndLqGBHt.GfpN.BgvsbXoLh_DIzAJOtFDmLSCYEztvPcS_GHPxivzV,NPMmSAtfk.Mg.w,A UcCt_lCD.csEzyJJBYtSMkzqiA\nmiao.qlala.\nmiao.FmDlY\nmiao.UQI.aJmnanNvRLskuVaMybDMsOlala.\nmiao.lala.",
"output": "OMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nmiao.vyscfysAtWcPkpFHdwZqAQ,UPPcjhKQTlala.\nmiao.KESqus DybUuYFoWVpo..LWZh.UqEdUsTHFlKfzqkThAUPklala.\nUNoE vfZIAdxkiWKhsHPfsqRPTNQoHgAxooVLYxRzugHjo jaEHWQFF\nCCmdIwr.UkoiYWK.Z,,ZesMpISTXNgnpYnJaWquCyL,gO\n.JvOayhXK_bgoYbfAtnXg\nbvdSzRrXoGxVgWvdXnsjEnEfxDzIQo_aZVGDGrzwuAMtzVAHioMBx_DHuTxyieGbGuSRNUojOREqxBBxvCgqAOMzwIWT\nMBuaWduZmRaOGyIPzWOsBVeqtDrblAbXxmM_uRfqMvnVlLEuhVKlhidN_aigiXyq,ZEDqQAx\nmiao.wCHVCuVKNePKmIUFLL_lala.\nmiao.iAqstXHUv\n pMO yvPkNtnNwmUCao W,wW.OvIMVaEeVYHmqaniWq.ivlala.",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nFreda's"
},
{
"input": "10\nmiao.\nmiao.jrwLBCpNaDCjyoK.PFzbwWU.h.. wfQquG_P..lala.\nmiao.LGlYdKjw__.Chlala.\nW.wtr qG KDOHj.xWxPbXIXjD_,GJZDaAZ,JBHphsjWJwSKcZAIAi\nmiao.pHsGAZQDWPJQwKC.zHjJituLgp.eUrzObTI.wrpect.FMUJqu,Zuslala.\nmiao.YVlOpXccUA_YU igbsbZbhOVwyYTyOjnWqgiTmxwAuFa.flCHn.,MtVbqxZQl_BGHXWkwijGjuL, ,ezyNlala.\nmiao.xCrVSz.aMv UOSOroDlQxWeBmlWe.FA.ZfUmviMlala.\nxebAlala.\nmiao.qVSxqf vOTlala.\nD.oBUwsLQRgXAoNkQJhQN.w.oMhuvtujnmiwgQYMfjlNTSHh .lSKgI.OEp",
"output": "Rainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nZXXzYlTiQU\nkXE.DdcbOojSaSgjMcFBPubKHefEVAzbi,PDFgSZIz,lala.\nxEfrTCjKhhwBC.UNmJXgTGUdkQeVDlala.\nLfaEw.jvMmuOBWtfoiJNtDIlQAVWNU,xWK_efBBtfkM\nqtBpqKZMWZMX_NKrUAEKYyQcLZWQlqbM\nmiao.PrJEbUtInremuaKRItqXOrfQEjQcAak VQ\nMpGCq awvQaHRvDr uvtVMKsvZI\nmiao.A.RVGu.szCEp.pXQJwL EuTltlN.WradoTvWHJyhcNSoulala.\nmiao.rzlUHzUdxtDRpWRuc,QZwEBfsKKGHMLGtFymPPQdptLFlzZ_ORWqrlfOrlntuDkpXEvz.CxwAsFYUvpnOnFWG\nmiao.VXUoNBwlgBwcna_n.CgAAcKKUuiVA.doOJKHpMdwNwlHAcLpdfN.Awa SthrlEWpUcuOonUTxIQNszYcHDXxnhArrM..A",
"output": "OMG>.< I don't know!\nFreda's\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's"
},
{
"input": "10\nmiao.qbxBFzrjtWv.yOk\nDBgi,loApO AACrGnwssCHN\nmiao.LV.wbQEE_V.BSAtdTIHTQOJVJ_nGOthbL,nJvQ.UeWFpsa.GGsK_Uv,HQxHS,AN_bkrolala.\nmiao.tBEqk rIQuByGKhfq_iP.BW,nySZEfrfySEcqnnIzxC,lrjIiivbxlkoVXJFiegGFRn NO,txGPhVBcv.CVhMmNO zlala.\nmiao.aBZWDWxk.wkR ,NyCzGxJnJDqBZpetdUPAmmBZDXl_Tbflala.\nmiao. XN,uMwWm. VqloYr..jTLszlala.\n.rshcgfZ.eZOdMu_RMh\nmiao.ahiwpECEe.lala.\nLeoUSroTekQAMSO__M L_ZEeRD_tUihYvQETFB,RzJmFtFiKrU\nBtygQG_OoFEFBL.KsVWTYbtqtalXoStFCZ RINHda.NuLmlkRB.vAQJFvelbsfoJ.T,M sJn",
"output": "Rainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nYoYBCcaqhXLfvKKf.UYMODTHyPZlala.\ncxgWn J.Q\nmiao.nwH.IHntgKYDhdsjU DMTHXEVRyeJP ZaAecCIBJXuv.YjhEmtbjvjKnK.U,oc,x\nmiao.EcQ.FDtRJgmpAzxhq.RwXBLxjyC,IeMqaFoheMPFCGWBcwUAFnbiwlbz_fcsEGPfJaeryCtFocBNEWTlala.\nmiao.W\nmiao. ZQpIeyCXJSnFgAIzu.THfrmyoogYWQzFqblala.\nmiao.ifzdCwnTDcxpvdr OTC.YqPv.MKDp..utICtAsbfYyGlala.\nmiao.\nmiao.tS.U.wH.s,CxORZJsBAHLi,fXeoDJWVBH\nrcUMpeupOVRKrcIRAvU.rP kgUEfoeXcrFPQOBYG.BNvAQPg.XHMWizhLpZNljXc .LQmVXCi",
"output": "Freda's\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!"
},
{
"input": "10\nlala.\nmiao.milalala.lmmialamiao.la.o.iao.a.ao.\nmialala.o.\nmiao.millala.allala.amiao..miao.miao.lala.ao.miammiao.iao.o.\nmiao.miaomiao..\nlalmiao.amiao..\nmiao.lala.lamiamiaolala..o.lalala.miao..\nmlala.iao.lalamiao..\nlmlala.iao.alalamiao.lmialala.lala.miao.o.alala..lala..lalmiaomiao..lalmiao.a.lalamiao..miao.alala..\nlalllamiao.la.lala.alamiao.lalalala.lala..miao.lamiao.la.lallalamiao..a..a.",
"output": "Freda's\nRainbow's\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nlalllala.ala.lala.a.mmimiao.aomiao.lllala.ala.amiao.la.mialalala.la.o..imiao.miao.amlala.iao.o.\nmilala.alllala.ala.amiao.lamiao..o.\nlala.lalalala..lalalala..\nlala.miao.\nmimiao.ao.lala.\nlalmiao.amlala.iamialala.o.o..\nlalammlala.iaolammiao.imiao.ao.la..iao..\nmiao.mialala.omiao..mlala.iaolala..\nmiamiao.o.llallala.ala.la.miao.ala.miao.mimialmiao.ala.o.alala.miaomiao..olala..\nmialala.lamiao.la.lala.miao.ollala.allala.ala.lmiaommiao.imiao.ao.lallallala.a.miao.a..a..",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nFreda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "10\nlamiao.lamiao.mimiao.ao..\nllala.almiaomiao..lala.miao.a.\nlalala.lala.lalala.lala.lalala..la.\nlalala.la.miao.\nmiao.llalallala.miao.a.la.lala.almimiao.ao.a.\nmiao.mlala.ilala.aomilala.ao..\nmilala.ao.\nmmiao.iao.\nmlala.lala.ialalmiao.alala..olala.lala..\nlala.lmlalalalalala...iao.milala.aommiao.iao..alamiaolallala.milala.ao.a..lalalmiao.a..llala.amiao..",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nRainbow's\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "4\n \n miao. \n lala. \n ",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
},
{
"input": "1\nMiao.",
"output": "OMG>.< I don't know!"
},
{
"input": "1\nm",
"output": "OMG>.< I don't know!"
},
{
"input": "1\nMiAo.sdsdlala.",
"output": "Freda's"
},
{
"input": "1\nLalA.",
"output": "OMG>.< I don't know!"
},
{
"input": "3\n.\nm\nl",
"output": "OMG>.< I don't know!\nOMG>.< I don't know!\nOMG>.< I don't know!"
}
] | 1,474,321,717 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 61 | 0 | n = int(input())
for i in range(0, n):
text = input()
if text[0:4] == "miao.": print("Rainbow's")
elif text[-5:len(text)]: print("Freda's")
else: print("OMG>.< I don't know!") | Title: Whose sentence is it?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100.
Output Specification:
For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions.
Demo Input:
['5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n']
Demo Output:
["Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\n"]
Note:
none | ```python
n = int(input())
for i in range(0, n):
text = input()
if text[0:4] == "miao.": print("Rainbow's")
elif text[-5:len(text)]: print("Freda's")
else: print("OMG>.< I don't know!")
``` | 0 |
|
892 | A | Greed | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not! | The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans.
The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans. | Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower). | [
"2\n3 5\n3 6\n",
"3\n6 8 9\n6 10 12\n",
"5\n0 0 5 0 0\n1 1 8 10 5\n",
"4\n4 1 0 3\n5 2 2 3\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"YES\n"
] | In the first sample, there are already 2 cans, so the answer is "YES". | 500 | [
{
"input": "2\n3 5\n3 6",
"output": "YES"
},
{
"input": "3\n6 8 9\n6 10 12",
"output": "NO"
},
{
"input": "5\n0 0 5 0 0\n1 1 8 10 5",
"output": "YES"
},
{
"input": "4\n4 1 0 3\n5 2 2 3",
"output": "YES"
},
{
"input": "10\n9 10 24 11 1 7 8 3 28 14\n86 20 34 11 22 94 8 16 73 85",
"output": "YES"
},
{
"input": "4\n25 35 7 31\n70 37 43 35",
"output": "YES"
},
{
"input": "10\n15 26 15 14 14 39 40 4 25 39\n27 72 16 44 69 48 53 17 63 42",
"output": "NO"
},
{
"input": "5\n22 5 19 16 32\n26 10 43 38 37",
"output": "NO"
},
{
"input": "5\n32 4 22 40 26\n39 20 36 98 44",
"output": "YES"
},
{
"input": "6\n18 25 3 10 13 37\n38 73 19 35 24 37",
"output": "YES"
},
{
"input": "2\n2 2\n2 2",
"output": "YES"
},
{
"input": "2\n2 5\n2 5",
"output": "YES"
},
{
"input": "2\n1000 1008\n10000 2352",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5\n1 2 3 4 11",
"output": "YES"
},
{
"input": "4\n1 0 0 0\n2 0 0 0",
"output": "YES"
},
{
"input": "2\n0 0\n1 2",
"output": "YES"
},
{
"input": "3\n9 13 4\n10 14 5",
"output": "NO"
},
{
"input": "2\n0 0\n1 1",
"output": "YES"
},
{
"input": "5\n1 1 2 3 1\n1 1 2 3 4",
"output": "NO"
},
{
"input": "2\n0 0\n0 0",
"output": "YES"
},
{
"input": "3\n5 1 1\n5 5 5",
"output": "YES"
}
] | 1,510,943,030 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 46 | 0 | n = int(input())
a = list(map(int, input().split()))
b = list(map(input().split()))
m1 = max(b)
b.remove(m1)
m2 = max(b)
r = 0
for i in a:
r += i
if r < m1 + m2:
print("YES")
else:
print("NO")
| Title: Greed
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input Specification:
The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans.
The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans.
Output Specification:
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Demo Input:
['2\n3 5\n3 6\n', '3\n6 8 9\n6 10 12\n', '5\n0 0 5 0 0\n1 1 8 10 5\n', '4\n4 1 0 3\n5 2 2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'YES\n']
Note:
In the first sample, there are already 2 cans, so the answer is "YES". | ```python
n = int(input())
a = list(map(int, input().split()))
b = list(map(input().split()))
m1 = max(b)
b.remove(m1)
m2 = max(b)
r = 0
for i in a:
r += i
if r < m1 + m2:
print("YES")
else:
print("NO")
``` | -1 |
|
478 | C | Table Decorations | PROGRAMMING | 1,800 | [
"greedy"
] | null | null | You have *r* red, *g* green and *b* blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number *t* of tables can be decorated if we know number of balloons of each color?
Your task is to write a program that for given values *r*, *g* and *b* will find the maximum number *t* of tables, that can be decorated in the required manner. | The single line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space. | Print a single integer *t* — the maximum number of tables that can be decorated in the required manner. | [
"5 4 3\n",
"1 1 1\n",
"2 3 3\n"
] | [
"4\n",
"1\n",
"2\n"
] | In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively. | 1,500 | [
{
"input": "5 4 3",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 3 3",
"output": "2"
},
{
"input": "0 1 0",
"output": "0"
},
{
"input": "0 3 3",
"output": "2"
},
{
"input": "4 0 4",
"output": "2"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "100 99 56",
"output": "85"
},
{
"input": "1000 1000 1002",
"output": "1000"
},
{
"input": "0 1 1000000000",
"output": "1"
},
{
"input": "500000000 1000000000 500000000",
"output": "666666666"
},
{
"input": "1000000000 2000000000 1000000000",
"output": "1333333333"
},
{
"input": "2000000000 2000000000 2000000000",
"output": "2000000000"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "1 2000000000 1000000000",
"output": "1000000000"
},
{
"input": "1585222789 1889821127 2000000000",
"output": "1825014638"
},
{
"input": "10000 7500 7500",
"output": "8333"
},
{
"input": "150000 75000 75000",
"output": "100000"
},
{
"input": "999288131 55884921 109298382",
"output": "165183303"
},
{
"input": "100500 100500 3",
"output": "67001"
},
{
"input": "1463615122 1988383731 837331500",
"output": "1429776784"
},
{
"input": "1938 8999 1882",
"output": "3820"
},
{
"input": "45 33 76",
"output": "51"
},
{
"input": "100000 1 2",
"output": "3"
},
{
"input": "198488 50 18",
"output": "68"
},
{
"input": "82728372 939848 100139442",
"output": "61269220"
},
{
"input": "99 5747 5298",
"output": "3714"
},
{
"input": "3 5 2",
"output": "3"
},
{
"input": "7511 7512 7513",
"output": "7512"
},
{
"input": "1234567890 123456789 987654321",
"output": "781893000"
},
{
"input": "500000000 2000000000 500000000",
"output": "1000000000"
},
{
"input": "500000002 2000000000 500000001",
"output": "1000000001"
},
{
"input": "520000000 1000000033 501000000",
"output": "673666677"
},
{
"input": "10000 1000 100000",
"output": "11000"
},
{
"input": "2000000000 500000000 499999999",
"output": "999999999"
},
{
"input": "1999999999 500000000 500000000",
"output": "999999999"
},
{
"input": "1 1 9",
"output": "2"
},
{
"input": "3 0 0",
"output": "0"
},
{
"input": "6 1 1",
"output": "2"
},
{
"input": "2000000000 1999999999 1999999999",
"output": "1999999999"
},
{
"input": "3 4 9",
"output": "5"
},
{
"input": "3 3 6",
"output": "4"
}
] | 1,652,712,246 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 25 | 62 | 28,979,200 | a=list(map(int,input().split()))
b=max(a)
c=sum(a)-b
if b>=c:
if 2*c>=b:
d=min(b//3,c//3)
e=b-3*d
f=c-3*d
g=[e,f]
g.sort()
a.sort()
if a[1]*2==a[2]:
xx=a[1]
else:
xx=0
if 0 in g:
print(max(2*d,xx))
elif 2*g[0]<=g[1]:
print(max(2*d+g[0],xx))
else:
print(max(2*d+g[1]//2,xx))
else:
print(c)
else:
r=a[0]
gg=a[1]
bb=a[2]
aa=min(r,gg,bb)
cc=r-aa
d=gg-aa
e=bb-aa
z=[cc,d,e]
z.remove(0)
z.sort()
xr=0
while True:
diff=int(abs(z[0]-z[1]))
xr+=int(diff/2)
z[1]=z[1]-2*int(diff/2)
z[0]=z[0]-int(diff/2)
if diff<2:
break
y=min(min(z)//3,max(z)//3)
w=min(z)-3*y
x=max(z)-3*y
v=[w,x]
if 0 in v or v.count(1)==2:
print(aa+2*y+xr)
else:
print(aa+2*y+1+xr) | Title: Table Decorations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *r* red, *g* green and *b* blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number *t* of tables can be decorated if we know number of balloons of each color?
Your task is to write a program that for given values *r*, *g* and *b* will find the maximum number *t* of tables, that can be decorated in the required manner.
Input Specification:
The single line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space.
Output Specification:
Print a single integer *t* — the maximum number of tables that can be decorated in the required manner.
Demo Input:
['5 4 3\n', '1 1 1\n', '2 3 3\n']
Demo Output:
['4\n', '1\n', '2\n']
Note:
In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively. | ```python
a=list(map(int,input().split()))
b=max(a)
c=sum(a)-b
if b>=c:
if 2*c>=b:
d=min(b//3,c//3)
e=b-3*d
f=c-3*d
g=[e,f]
g.sort()
a.sort()
if a[1]*2==a[2]:
xx=a[1]
else:
xx=0
if 0 in g:
print(max(2*d,xx))
elif 2*g[0]<=g[1]:
print(max(2*d+g[0],xx))
else:
print(max(2*d+g[1]//2,xx))
else:
print(c)
else:
r=a[0]
gg=a[1]
bb=a[2]
aa=min(r,gg,bb)
cc=r-aa
d=gg-aa
e=bb-aa
z=[cc,d,e]
z.remove(0)
z.sort()
xr=0
while True:
diff=int(abs(z[0]-z[1]))
xr+=int(diff/2)
z[1]=z[1]-2*int(diff/2)
z[0]=z[0]-int(diff/2)
if diff<2:
break
y=min(min(z)//3,max(z)//3)
w=min(z)-3*y
x=max(z)-3*y
v=[w,x]
if 0 in v or v.count(1)==2:
print(aa+2*y+xr)
else:
print(aa+2*y+1+xr)
``` | 0 |
|
208 | A | Dubstep | PROGRAMMING | 900 | [
"strings"
] | null | null | Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. | The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. | Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. | [
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] | [
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] | In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | 500 | [
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,676,914,100 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 71 | 124 | 0 | arr = list(input().split('WUB'))
for e in arr:
if e == '':
continue
print(e, end=' ')
| Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | ```python
arr = list(input().split('WUB'))
for e in arr:
if e == '':
continue
print(e, end=' ')
``` | 3 |
|
260 | A | Adding Digits | PROGRAMMING | 1,400 | [
"implementation",
"math"
] | null | null | Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times. | The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105). | In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them. | [
"5 4 5\n",
"12 11 1\n",
"260 150 10\n"
] | [
"524848\n",
"121\n",
"-1\n"
] | none | 500 | [
{
"input": "5 4 5",
"output": "524848"
},
{
"input": "12 11 1",
"output": "121"
},
{
"input": "260 150 10",
"output": "-1"
},
{
"input": "78843 5684 42717",
"output": "-1"
},
{
"input": "93248 91435 1133",
"output": "-1"
},
{
"input": "100000 10 64479",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 21 73839",
"output": "9999990000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99991 623 36438",
"output": "9999150000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 334 94854",
"output": "9999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99252 9827 84849",
"output": "9925270000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99313 9833 10561",
"output": "9931330000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "94885 55815 11417",
"output": "9488550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99492 58525 53481",
"output": "9949250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99858 28531 79193",
"output": "9985850000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99136 47208 42607",
"output": "9913680000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "63270 19953 5555",
"output": "-1"
},
{
"input": "10240 128 100000",
"output": "1024000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "12 11 3",
"output": "12100"
},
{
"input": "14 12 99998",
"output": "1440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "1 11 3",
"output": "1100"
},
{
"input": "3 40 1",
"output": "-1"
},
{
"input": "150 100 10",
"output": "1500000000000"
},
{
"input": "5 10 1",
"output": "50"
},
{
"input": "1 15 10",
"output": "15000000000"
},
{
"input": "3 13 2",
"output": "390"
}
] | 1,647,451,327 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 717 | 307,200 | a,b,n=map(int,input().split(" "))
num=a
flag=True
for i in range(0,10):
if (num*10+i)%b==0:
num=num*10+i
a=num
flag=False
break
if flag==True:
print(-1)
else:
a*=pow(10,n-1)
print(a)
| Title: Adding Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
Input Specification:
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
Output Specification:
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Demo Input:
['5 4 5\n', '12 11 1\n', '260 150 10\n']
Demo Output:
['524848\n', '121\n', '-1\n']
Note:
none | ```python
a,b,n=map(int,input().split(" "))
num=a
flag=True
for i in range(0,10):
if (num*10+i)%b==0:
num=num*10+i
a=num
flag=False
break
if flag==True:
print(-1)
else:
a*=pow(10,n-1)
print(a)
``` | 3 |
|
263 | A | Beautiful Matrix | PROGRAMMING | 800 | [
"implementation"
] | null | null | You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful. | The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one. | Print a single integer — the minimum number of moves needed to make the matrix beautiful. | [
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] | [
"3\n",
"1\n"
] | none | 500 | [
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,697,379,472 | 2,147,483,647 | PyPy 3 | OK | TESTS | 25 | 154 | 0 | import math
l=[]
for i in range(5):
ll = list(map(int,input().split()))
l.append(ll)
m = 0
n = 0
x=0
for i in range(5):
a = l[i]
for j in range(5):
if a[j] == 1:
m=i
n=j
x=1
break
if x==1:
break
M=0
if m-2>=0:
M=m-2
else:
M=2-m
N=0
if n-2>=0:
N=n-2
else:
N=2-n
print(M+N) | Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
import math
l=[]
for i in range(5):
ll = list(map(int,input().split()))
l.append(ll)
m = 0
n = 0
x=0
for i in range(5):
a = l[i]
for j in range(5):
if a[j] == 1:
m=i
n=j
x=1
break
if x==1:
break
M=0
if m-2>=0:
M=m-2
else:
M=2-m
N=0
if n-2>=0:
N=n-2
else:
N=2-n
print(M+N)
``` | 3 |
|
262 | A | Roma and Lucky Numbers | PROGRAMMING | 800 | [
"implementation"
] | null | null | Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers.
Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem. | The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=100). The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the numbers that Roma has.
The numbers in the lines are separated by single spaces. | In a single line print a single integer — the answer to the problem. | [
"3 4\n1 2 4\n",
"3 2\n447 44 77\n"
] | [
"3\n",
"2\n"
] | In the first sample all numbers contain at most four lucky digits, so the answer is 3.
In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2. | 500 | [
{
"input": "3 4\n1 2 4",
"output": "3"
},
{
"input": "3 2\n447 44 77",
"output": "2"
},
{
"input": "2 2\n507978501 180480073",
"output": "2"
},
{
"input": "9 6\n655243746 167613748 1470546 57644035 176077477 56984809 44677 215706823 369042089",
"output": "9"
},
{
"input": "6 100\n170427799 37215529 675016434 168544291 683447134 950090227",
"output": "6"
},
{
"input": "4 2\n194041605 706221269 69909135 257655784",
"output": "3"
},
{
"input": "4 2\n9581849 67346651 530497 272158241",
"output": "4"
},
{
"input": "3 47\n378261451 163985731 230342101",
"output": "3"
},
{
"input": "2 3\n247776868 480572137",
"output": "1"
},
{
"input": "7 77\n366496749 549646417 278840199 119255907 33557677 379268590 150378796",
"output": "7"
},
{
"input": "40 31\n32230963 709031779 144328646 513494529 36547831 416998222 84161665 318773941 170724397 553666286 368402971 48581613 31452501 368026285 47903381 939151438 204145360 189920160 288159400 133145006 314295423 450219949 160203213 358403181 478734385 29331901 31051111 110710191 567314089 139695685 111511396 87708701 317333277 103301481 110400517 634446253 481551313 39202255 105948 738066085",
"output": "40"
},
{
"input": "1 8\n55521105",
"output": "1"
},
{
"input": "49 3\n34644511 150953622 136135827 144208961 359490601 86708232 719413689 188605873 64330753 488776302 104482891 63360106 437791390 46521319 70778345 339141601 136198441 292941209 299339510 582531183 555958105 437904637 74219097 439816011 236010407 122674666 438442529 186501223 63932449 407678041 596993853 92223251 849265278 480265849 30983497 330283357 186901672 20271344 794252593 123774176 27851201 52717531 479907210 196833889 149331196 82147847 255966471 278600081 899317843",
"output": "44"
},
{
"input": "26 2\n330381357 185218042 850474297 483015466 296129476 1205865 538807493 103205601 160403321 694220263 416255901 7245756 507755361 88187633 91426751 1917161 58276681 59540376 576539745 595950717 390256887 105690055 607818885 28976353 488947089 50643601",
"output": "22"
},
{
"input": "38 1\n194481717 126247087 815196361 106258801 381703249 283859137 15290101 40086151 213688513 577996947 513899717 371428417 107799271 11136651 5615081 323386401 381128815 34217126 17709913 520702093 201694245 570931849 169037023 417019726 282437316 7417126 271667553 11375851 185087449 410130883 383045677 5764771 905017051 328584026 215330671 299553233 15838255 234532105",
"output": "20"
},
{
"input": "44 9\n683216389 250581469 130029957 467020047 188395565 206237982 63257361 68314981 732878407 563579660 199133851 53045209 665723851 16273169 10806790 556633156 350593410 474645249 478790761 708234243 71841230 18090541 19836685 146373571 17947452 534010506 46933264 377035021 311636557 75193963 54321761 12759959 71120181 548816939 23608621 31876417 107672995 72575155 369667956 20574379 210596751 532163173 75726739 853719629",
"output": "44"
},
{
"input": "8 6\n204157376 10514197 65483881 347219841 263304577 296402721 11739011 229776191",
"output": "8"
},
{
"input": "38 29\n333702889 680931737 61137217 203030505 68728281 11414209 642645708 590904616 3042901 607198177 189041074 700764043 813035201 198341461 126403544 401436841 420826465 45046581 20249976 46978855 46397957 706610773 24701041 57954481 51603266 593109701 385569073 178982291 582152863 287317968 1474090 34825141 432421977 130257781 151516903 540852403 548392 117246529",
"output": "38"
},
{
"input": "19 3\n562569697 549131571 50676718 84501863 74567295 702372009 365895280 451459937 40378543 167666701 158635641 53639293 442332661 825055617 100109161 326616021 862332843 533271196 4791547",
"output": "18"
},
{
"input": "1 1\n44",
"output": "0"
},
{
"input": "1 1\n4",
"output": "1"
},
{
"input": "10 3\n444 447 774 777 7777 4447 4 7 7 4",
"output": "8"
}
] | 1,695,844,512 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 92 | 0 | n, k= map(int, input().split())
a = 0
b = 0
x = [str(y) for y in input().split()]
for i in x:
for j in range(len(i)):
if i[j] == "4" or i[j] == "7":
a+=1
if a <= k:
b+=1
a = 0
else:
a = 0
print(b)
| Title: Roma and Lucky Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers.
Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem.
Input Specification:
The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=100). The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the numbers that Roma has.
The numbers in the lines are separated by single spaces.
Output Specification:
In a single line print a single integer — the answer to the problem.
Demo Input:
['3 4\n1 2 4\n', '3 2\n447 44 77\n']
Demo Output:
['3\n', '2\n']
Note:
In the first sample all numbers contain at most four lucky digits, so the answer is 3.
In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2. | ```python
n, k= map(int, input().split())
a = 0
b = 0
x = [str(y) for y in input().split()]
for i in x:
for j in range(len(i)):
if i[j] == "4" or i[j] == "7":
a+=1
if a <= k:
b+=1
a = 0
else:
a = 0
print(b)
``` | 3 |
|
94 | A | Restoring Password | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Restoring Password | 2 | 256 | Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password".
Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address.
Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits.
Help Igor K. restore his ISQ account by the encrypted password and encryption specification. | The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9. | Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists. | [
"01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n",
"10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n"
] | [
"12345678\n",
"30234919\n"
] | none | 500 | [
{
"input": "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110",
"output": "12345678"
},
{
"input": "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000",
"output": "30234919"
},
{
"input": "00010101101110110101100110101100010101100010101111000101011010011010110010000011\n0101010110\n0001001101\n1001101011\n0000100011\n0010101111\n1110110101\n0001010110\n0110111000\n0000111110\n0010000011",
"output": "65264629"
},
{
"input": "10100100010010010011011001101000100100110110011010011001101011000100110110011010\n1111110011\n1001000111\n1001000100\n1100010011\n0110011010\n0010000001\n1110101110\n0010000110\n0010010011\n1010010001",
"output": "98484434"
},
{
"input": "00101100011111010001001000000110110000000110010011001111111010110010001011000000\n0010000001\n0110010011\n0010000010\n1011001000\n0011111110\n0110001000\n1111010001\n1011000000\n0000100110\n0010110001",
"output": "96071437"
},
{
"input": "10001110111110000001000010001010001110110000100010100010111101101101010000100010\n0000010110\n1101010111\n1000101111\n0001011110\n0011110101\n0101100100\n0110110101\n0000100010\n1000111011\n1110000001",
"output": "89787267"
},
{
"input": "10010100011001010001010101001101010100110100111011001010111100011001000010100000\n0011100000\n1001100100\n0001100100\n0010100000\n0101010011\n0010101110\n0010101111\n0100111011\n1001010001\n1111111110",
"output": "88447623"
},
{
"input": "01101100111000000101011011001110000001011111111000111111100001011010001001011001\n1000000101\n0101101000\n0101110101\n1101011110\n0000101100\n1111111000\n0001001101\n0110111011\n0110110011\n1001011001",
"output": "80805519"
},
{
"input": "11100011000100010110010011101010101010011110001100011010111110011000011010110111\n1110001100\n0110101111\n0100111010\n0101000000\n1001100001\n1010101001\n0000100010\n1010110111\n1100011100\n0100010110",
"output": "09250147"
},
{
"input": "10000110110000010100000010001000111101110110101011110111000100001101000000100010\n0000010100\n0000110001\n0110101011\n1101110001\n1000011011\n0000110100\n0011110111\n1000110010\n0000100010\n0000011011",
"output": "40862358"
},
{
"input": "01000000010000000110100101000110110000100100000001101100001000011111111001010001\n1011000010\n1111101010\n0111110011\n0000000110\n0000001001\n0001111111\n0110010010\n0100000001\n1011001000\n1001010001",
"output": "73907059"
},
{
"input": "01111000111110011001110101110011110000111110010001101100110110100111101011001101\n1110010001\n1001100000\n1100001000\n1010011110\n1011001101\n0111100011\n1101011100\n1110011001\n1111000011\n0010000101",
"output": "57680434"
},
{
"input": "01001100101000100010001011110001000101001001100010010000001001001100101001011111\n1001011111\n1110010111\n0111101011\n1000100010\n0011100101\n0100000010\n0010111100\n0100010100\n1001100010\n0100110010",
"output": "93678590"
},
{
"input": "01110111110000111011101010110110101011010100110111000011101101110101011101001000\n0110000101\n1010101101\n1101010111\n1101011100\n0100110111\n0111011111\n1100011001\n0111010101\n0000111011\n1101001000",
"output": "58114879"
},
{
"input": "11101001111100110101110011010100110011011110100111010110110011000111000011001101\n1100011100\n1100110101\n1011101000\n0011011110\n0011001101\n0100010001\n1110100111\n1010101100\n1110110100\n0101101100",
"output": "61146904"
},
{
"input": "10101010001011010001001001011000100101100001011011101010101110101010001010101000\n0010110101\n1010011010\n1010101000\n1011010001\n1010101011\n0010010110\n0110100010\n1010100101\n0001011011\n0110100001",
"output": "23558422"
},
{
"input": "11110101001100010000110100001110101011011111010100110001000001001010001001101111\n0101101100\n1001101111\n1010101101\n0100101000\n1111110000\n0101010010\n1100010000\n1111010100\n1101000011\n1011111111",
"output": "76827631"
},
{
"input": "10001100110000110111100011001101111110110011110101000011011100001101110000110111\n0011110101\n0101100011\n1000110011\n1011011001\n0111111011\n0101111011\n0000110111\n0100001110\n1000000111\n0110110111",
"output": "26240666"
},
{
"input": "10000100010000111101100100111101111011101000001001100001000110000010010000111101\n1001001111\n0000111101\n1000010001\n0110011101\n0110101000\n1011111001\n0111101110\n1000001001\n1101011111\n0001010100",
"output": "21067271"
},
{
"input": "01101111000110111100011011110001101111001010001100101000110001010101100100000010\n1010001100\n0011010011\n0101010110\n1111001100\n1100011000\n0100101100\n1001100101\n0110111100\n0011001101\n0100000010",
"output": "77770029"
},
{
"input": "10100111011010001011111000000111100000010101000011000010111101010000111010011101\n1010011101\n1010111111\n0110100110\n1111000100\n1110000001\n0000101111\n0011111000\n1000110001\n0101000011\n1010001011",
"output": "09448580"
},
{
"input": "10000111111000011111001010101010010011111001001111000010010100100011000010001100\n1101101110\n1001001111\n0000100101\n1100111010\n0010101010\n1110000110\n1100111101\n0010001100\n1110000001\n1000011111",
"output": "99411277"
},
{
"input": "10110110111011001111101100111100111111011011011011001111110110010011100010000111\n0111010011\n0111101100\n1001101010\n0101000101\n0010000111\n0011111101\n1011001111\n1101111000\n1011011011\n1001001110",
"output": "86658594"
},
{
"input": "01001001100101100011110110111100000110001111001000100000110111110010000000011000\n0100100110\n1000001011\n1000111110\n0000011000\n0101100011\n1101101111\n1111001000\n1011011001\n1000001101\n0010101000",
"output": "04536863"
},
{
"input": "10010100011101000011100100001100101111000010111100000010010000001001001101011101\n1001000011\n1101000011\n1001010001\n1101011101\n1000010110\n0011111101\n0010111100\n0000100100\n1010001000\n0101000110",
"output": "21066773"
},
{
"input": "01111111110101111111011111111111010010000001100000101000100100111001011010001001\n0111111111\n0101111111\n0100101101\n0001100000\n0011000101\n0011100101\n1101001000\n0010111110\n1010001001\n1111000111",
"output": "01063858"
},
{
"input": "00100011111001001010001111000011101000001110100000000100101011101000001001001010\n0010001111\n1001001010\n1010011001\n0011100111\n1000111000\n0011110000\n0000100010\n0001001010\n1111110111\n1110100000",
"output": "01599791"
},
{
"input": "11011101000100110100110011010101100011111010011010010011010010010010100110101111\n0100110100\n1001001010\n0001111101\n1101011010\n1101110100\n1100110101\n0110101111\n0110001111\n0001101000\n1010011010",
"output": "40579016"
},
{
"input": "10000010111101110110011000111110000011100110001111100100000111000011011000001011\n0111010100\n1010110110\n1000001110\n1110000100\n0110001111\n1101110110\n1100001101\n1000001011\n0000000101\n1001000001",
"output": "75424967"
},
{
"input": "11101100101110111110111011111010001111111111000001001001000010001111111110110010\n0101100001\n1111010011\n1110111110\n0100110100\n1110011111\n1000111111\n0010010000\n1110110010\n0011000010\n1111000001",
"output": "72259657"
},
{
"input": "01011110100101111010011000001001100000101001110011010111101011010000110110010101\n0100111100\n0101110011\n0101111010\n0110000010\n0101001111\n1101000011\n0110010101\n0111011010\n0001101110\n1001110011",
"output": "22339256"
},
{
"input": "01100000100101111000100001100010000110000010100100100001100000110011101001110000\n0101111000\n1001110000\n0001000101\n0110110111\n0010100100\n1000011000\n1101110110\n0110000010\n0001011010\n0011001110",
"output": "70554591"
},
{
"input": "11110011011000001001111100110101001000010100100000110011001110011111100100100001\n1010011000\n1111001101\n0100100001\n1111010011\n0100100000\n1001111110\n1010100111\n1000100111\n1000001001\n1100110011",
"output": "18124952"
},
{
"input": "10001001011000100101010110011101011001110010000001010110000101000100101111101010\n0101100001\n1100001100\n1111101010\n1000100101\n0010000001\n0100010010\n0010110110\n0101100111\n0000001110\n1101001110",
"output": "33774052"
},
{
"input": "00110010000111001001001100100010010111101011011110001011111100000101000100000001\n0100000001\n1011011110\n0010111111\n0111100111\n0100111001\n0000010100\n1001011110\n0111001001\n0100010011\n0011001000",
"output": "97961250"
},
{
"input": "01101100001000110101101100101111101110010011010111100011010100010001101000110101\n1001101001\n1000110101\n0110110000\n0111100100\n0011010111\n1110111001\n0001000110\n0000000100\n0001101001\n1011001011",
"output": "21954161"
},
{
"input": "10101110000011010110101011100000101101000110100000101101101101110101000011110010\n0110100000\n1011011011\n0011110010\n0001110110\n0010110100\n1100010010\n0001101011\n1010111000\n0011010110\n0111010100",
"output": "78740192"
},
{
"input": "11000101011100100111010000010001000001001100101100000011000000001100000101011010\n1100010101\n1111101011\n0101011010\n0100000100\n1000110111\n1100100111\n1100101100\n0111001000\n0000110000\n0110011111",
"output": "05336882"
},
{
"input": "11110100010000101110010110001000001011100101100010110011011011111110001100110110\n0101100010\n0100010001\n0000101110\n1100110110\n0101000101\n0011001011\n1111010001\n1000110010\n1111111000\n1010011111",
"output": "62020383"
},
{
"input": "00011001111110000011101011010001010111100110100101000110011111011001100000001100\n0111001101\n0101011110\n0001100111\n1101011111\n1110000011\n0000001100\n0111010001\n1101100110\n1010110100\n0110100101",
"output": "24819275"
},
{
"input": "10111110010011111001001111100101010111010011111001001110101000111110011001111101\n0011111001\n0101011101\n0100001010\n0001110010\n1001111101\n0011101010\n1111001001\n1100100001\n1001101000\n1011111001",
"output": "90010504"
},
{
"input": "01111101111100101010001001011110111001110111110111011111011110110111111011011111\n1111110111\n0010000101\n0110000100\n0111111011\n1011100111\n1100101010\n1011011111\n1100010001\n0111110111\n0010010111",
"output": "85948866"
},
{
"input": "01111100000111110000110010111001111100001001101010110010111010001000101001101010\n0100010101\n1011110101\n1010100100\n1010000001\n1001101010\n0101100110\n1000100010\n0111110000\n1100101110\n0110010110",
"output": "77874864"
},
{
"input": "11100011010000000010011110010111001011111001000111000000001000000000100111100101\n0000000010\n1110001101\n0011010101\n0111100101\n1001000111\n1101001111\n0111010110\n1100101111\n0110000000\n1101101011",
"output": "10374003"
},
{
"input": "01111011100111101110011001000110001111101000111110100100100001011111001011100010\n0110010100\n1100010001\n0111101110\n1001001000\n1010011011\n1000111110\n0010110101\n1011100010\n0101111100\n0110010001",
"output": "22955387"
},
{
"input": "11011010001100000011000100110011010101000110011110110000001100111100001000011111\n0000100010\n1000011111\n1101101000\n0110011110\n0011110000\n1100000011\n0010001100\n0101101000\n0001001100\n1101010100",
"output": "25893541"
},
{
"input": "01011001011111010010101111011001000011001100011101101111011011010011101011110110\n0100001100\n0101100101\n1111111011\n1111010010\n1111101100\n1100011101\n1011000011\n1101001110\n1011110110\n0110001010",
"output": "13805878"
},
{
"input": "11110011011000111111001100111110001111111100000010111100110100110011111111001101\n1111001101\n1001101010\n1100110010\n0011001111\n0001011110\n1000110011\n1000111111\n0110001010\n1001011101\n1100000010",
"output": "06369030"
},
{
"input": "01110011110010000011011001011000001000010110010110011001100001100110001100101000\n0000100001\n0110011000\n1010000010\n1110011101\n0111001111\n1100101000\n0010000011\n0110010000\n1100100101\n0110010110",
"output": "46909115"
},
{
"input": "00001011001111110111111111011111111101110101110100010111010010100101100001010110\n1111110111\n0001010110\n0111011011\n0111000001\n1010010110\n0101110100\n0001000101\n0000111000\n0110100001\n0000101100",
"output": "90005541"
}
] | 1,593,360,193 | 2,147,483,647 | PyPy 3 | OK | TESTS | 50 | 280 | 20,172,800 | s = input()
d = {}
for i in range(10):
x = input()
d[x] = str(i)
ans = ''
for i in range(8):
ans += d[s[i*10 : (i * 10 + 10)]]
print(ans) | Title: Restoring Password
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password".
Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address.
Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits.
Help Igor K. restore his ISQ account by the encrypted password and encryption specification.
Input Specification:
The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9.
Output Specification:
Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists.
Demo Input:
['01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n', '10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n']
Demo Output:
['12345678\n', '30234919\n']
Note:
none | ```python
s = input()
d = {}
for i in range(10):
x = input()
d[x] = str(i)
ans = ''
for i in range(8):
ans += d[s[i*10 : (i * 10 + 10)]]
print(ans)
``` | 3.892425 |
960 | B | Minimize the error | PROGRAMMING | 1,500 | [
"data structures",
"greedy",
"sortings"
] | null | null | You are given two arrays *A* and *B*, each of size *n*. The error, *E*, between these two arrays is defined . You have to perform exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*. In one operation, you have to choose one element of the array and increase or decrease it by 1.
Output the minimum possible value of error after *k*1 operations on array *A* and *k*2 operations on array *B* have been performed. | The first line contains three space-separated integers *n* (1<=≤<=*n*<=≤<=103), *k*1 and *k*2 (0<=≤<=*k*1<=+<=*k*2<=≤<=103, *k*1 and *k*2 are non-negative) — size of arrays and number of operations to perform on *A* and *B* respectively.
Second line contains *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — array *A*.
Third line contains *n* space separated integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=106<=≤<=*b**i*<=≤<=106)— array *B*. | Output a single integer — the minimum possible value of after doing exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*. | [
"2 0 0\n1 2\n2 3\n",
"2 1 0\n1 2\n2 2\n",
"2 5 7\n3 4\n14 4\n"
] | [
"2",
"0",
"1"
] | In the first sample case, we cannot perform any operations on *A* or *B*. Therefore the minimum possible error *E* = (1 - 2)<sup class="upper-index">2</sup> + (2 - 3)<sup class="upper-index">2</sup> = 2.
In the second sample case, we are required to perform exactly one operation on *A*. In order to minimize error, we increment the first element of *A* by 1. Now, *A* = [2, 2]. The error is now *E* = (2 - 2)<sup class="upper-index">2</sup> + (2 - 2)<sup class="upper-index">2</sup> = 0. This is the minimum possible error obtainable.
In the third sample case, we can increase the first element of *A* to 8, using the all of the 5 moves available to us. Also, the first element of *B* can be reduced to 8 using the 6 of the 7 available moves. Now *A* = [8, 4] and *B* = [8, 4]. The error is now *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 4)<sup class="upper-index">2</sup> = 0, but we are still left with 1 move for array *B*. Increasing the second element of *B* to 5 using the left move, we get *B* = [8, 5] and *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 5)<sup class="upper-index">2</sup> = 1. | 1,000 | [
{
"input": "2 0 0\n1 2\n2 3",
"output": "2"
},
{
"input": "2 1 0\n1 2\n2 2",
"output": "0"
},
{
"input": "2 5 7\n3 4\n14 4",
"output": "1"
},
{
"input": "2 0 1\n1 2\n2 2",
"output": "0"
},
{
"input": "2 1 1\n0 0\n1 1",
"output": "0"
},
{
"input": "5 5 5\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "3 4 5\n1 2 3\n3 2 1",
"output": "1"
},
{
"input": "3 1000 0\n1 2 3\n-1000 -1000 -1000",
"output": "1341346"
},
{
"input": "10 300 517\n-6 -2 6 5 -3 8 9 -10 8 6\n5 -9 -2 6 1 4 6 -2 5 -3",
"output": "1"
},
{
"input": "10 819 133\n87 22 30 89 82 -97 -52 25 76 -22\n-20 95 21 25 2 -3 45 -7 -98 -56",
"output": "0"
},
{
"input": "10 10 580\n302 -553 -281 -299 -270 -890 -989 -749 -418 486\n735 330 6 725 -984 209 -855 -786 -502 967",
"output": "2983082"
},
{
"input": "10 403 187\n9691 -3200 3016 3540 -9475 8840 -4705 7940 6293 -2631\n-2288 9129 4067 696 -6754 9869 -5747 701 3344 -3426",
"output": "361744892"
},
{
"input": "10 561 439\n76639 67839 10670 -23 -18393 65114 46538 67596 86615 90480\n50690 620 -33631 -75857 75634 91321 -81662 -93668 -98557 -43621",
"output": "116776723778"
},
{
"input": "10 765 62\n-929885 -995154 254071 -370672 -435272 584846 -301610 -234118 -82557 743536\n-36327 439149 -977780 -821019 -585558 953598 -151943 140715 -311253 -383103",
"output": "6216649853365"
},
{
"input": "22 334 246\n-462653 -618002 4973 -348485 366658 192390 274752 200590 138367 779540 -661269 642587 113645 -110388 -604418 -491231 -933401 -219332 -603140 836439 167007 210226\n357638 -646669 -558432 -434313 -285851 -119233 323088 -512237 -729293 215256 39316 -984201 -209814 715016 -271932 796550 988227 -89577 67202 462973 -942079 -823339",
"output": "15389604923763"
},
{
"input": "1 1000 0\n1000000\n-1000000",
"output": "3996001000000"
},
{
"input": "1 1000 0\n1000000\n1000000",
"output": "0"
},
{
"input": "1 0 1000\n-1000000\n1000000",
"output": "3996001000000"
},
{
"input": "1 0 1000\n1000000\n1000000",
"output": "0"
},
{
"input": "1 5 7\n1\n2",
"output": "1"
},
{
"input": "2 1 3\n2 2\n2 2",
"output": "0"
},
{
"input": "3 3 0\n1 1 1\n1 1 1",
"output": "1"
},
{
"input": "1 0 6\n0\n0",
"output": "0"
},
{
"input": "2 1 1\n2 2\n0 0",
"output": "2"
},
{
"input": "3 1 4\n0 0 0\n1 5 6",
"output": "19"
},
{
"input": "1 1 2\n0\n8",
"output": "25"
},
{
"input": "2 2 0\n3 3\n3 3",
"output": "0"
}
] | 1,687,425,660 | 1,260 | Python 3 | OK | TESTS | 80 | 46 | 512,000 | n, k1, k2 = map (int, input ().split ())
a = list (map (int, input ().split ()))
b = list (map (int, input ().split ()))
c = list ()
for i in range (n) : c.append (abs (a[i] - b[i]))
c.sort (reverse = True)
for i in range (k1 + k2) :
if c[0] <= 0 : c[0] += 1
else : c[0] -= 1
i = 1
while i < n and c[i] > c[0] : i += 1
if i == 1 : continue
else : c = c[1 : i] + [c[0]] + c[i :]
ans = 0
for i in c : ans = (ans + i * i)
print (ans)
| Title: Minimize the error
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays *A* and *B*, each of size *n*. The error, *E*, between these two arrays is defined . You have to perform exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*. In one operation, you have to choose one element of the array and increase or decrease it by 1.
Output the minimum possible value of error after *k*1 operations on array *A* and *k*2 operations on array *B* have been performed.
Input Specification:
The first line contains three space-separated integers *n* (1<=≤<=*n*<=≤<=103), *k*1 and *k*2 (0<=≤<=*k*1<=+<=*k*2<=≤<=103, *k*1 and *k*2 are non-negative) — size of arrays and number of operations to perform on *A* and *B* respectively.
Second line contains *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — array *A*.
Third line contains *n* space separated integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=106<=≤<=*b**i*<=≤<=106)— array *B*.
Output Specification:
Output a single integer — the minimum possible value of after doing exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*.
Demo Input:
['2 0 0\n1 2\n2 3\n', '2 1 0\n1 2\n2 2\n', '2 5 7\n3 4\n14 4\n']
Demo Output:
['2', '0', '1']
Note:
In the first sample case, we cannot perform any operations on *A* or *B*. Therefore the minimum possible error *E* = (1 - 2)<sup class="upper-index">2</sup> + (2 - 3)<sup class="upper-index">2</sup> = 2.
In the second sample case, we are required to perform exactly one operation on *A*. In order to minimize error, we increment the first element of *A* by 1. Now, *A* = [2, 2]. The error is now *E* = (2 - 2)<sup class="upper-index">2</sup> + (2 - 2)<sup class="upper-index">2</sup> = 0. This is the minimum possible error obtainable.
In the third sample case, we can increase the first element of *A* to 8, using the all of the 5 moves available to us. Also, the first element of *B* can be reduced to 8 using the 6 of the 7 available moves. Now *A* = [8, 4] and *B* = [8, 4]. The error is now *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 4)<sup class="upper-index">2</sup> = 0, but we are still left with 1 move for array *B*. Increasing the second element of *B* to 5 using the left move, we get *B* = [8, 5] and *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 5)<sup class="upper-index">2</sup> = 1. | ```python
n, k1, k2 = map (int, input ().split ())
a = list (map (int, input ().split ()))
b = list (map (int, input ().split ()))
c = list ()
for i in range (n) : c.append (abs (a[i] - b[i]))
c.sort (reverse = True)
for i in range (k1 + k2) :
if c[0] <= 0 : c[0] += 1
else : c[0] -= 1
i = 1
while i < n and c[i] > c[0] : i += 1
if i == 1 : continue
else : c = c[1 : i] + [c[0]] + c[i :]
ans = 0
for i in c : ans = (ans + i * i)
print (ans)
``` | 3 |
|
762 | A | k-th divisor | PROGRAMMING | 1,400 | [
"math",
"number theory"
] | null | null | You are given two integers *n* and *k*. Find *k*-th smallest divisor of *n*, or report that it doesn't exist.
Divisor of *n* is any such natural number, that *n* can be divided by it without remainder. | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=1015, 1<=≤<=*k*<=≤<=109). | If *n* has less than *k* divisors, output -1.
Otherwise, output the *k*-th smallest divisor of *n*. | [
"4 2\n",
"5 3\n",
"12 5\n"
] | [
"2\n",
"-1\n",
"6\n"
] | In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | 0 | [
{
"input": "4 2",
"output": "2"
},
{
"input": "5 3",
"output": "-1"
},
{
"input": "12 5",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "866421317361600 26880",
"output": "866421317361600"
},
{
"input": "866421317361600 26881",
"output": "-1"
},
{
"input": "1000000000000000 1000000000",
"output": "-1"
},
{
"input": "1000000000000000 100",
"output": "1953125"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "4 3",
"output": "4"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "9 3",
"output": "9"
},
{
"input": "21 3",
"output": "7"
},
{
"input": "67280421310721 1",
"output": "1"
},
{
"input": "6 3",
"output": "3"
},
{
"input": "3 3",
"output": "-1"
},
{
"input": "16 3",
"output": "4"
},
{
"input": "1 1000",
"output": "-1"
},
{
"input": "16 4",
"output": "8"
},
{
"input": "36 8",
"output": "18"
},
{
"input": "49 4",
"output": "-1"
},
{
"input": "9 4",
"output": "-1"
},
{
"input": "16 1",
"output": "1"
},
{
"input": "16 6",
"output": "-1"
},
{
"input": "16 5",
"output": "16"
},
{
"input": "25 4",
"output": "-1"
},
{
"input": "4010815561 2",
"output": "63331"
},
{
"input": "49 3",
"output": "49"
},
{
"input": "36 6",
"output": "9"
},
{
"input": "36 10",
"output": "-1"
},
{
"input": "25 3",
"output": "25"
},
{
"input": "22876792454961 28",
"output": "7625597484987"
},
{
"input": "1234 2",
"output": "2"
},
{
"input": "179458711 2",
"output": "179458711"
},
{
"input": "900104343024121 100000",
"output": "-1"
},
{
"input": "8 3",
"output": "4"
},
{
"input": "100 6",
"output": "20"
},
{
"input": "15500 26",
"output": "-1"
},
{
"input": "111111 1",
"output": "1"
},
{
"input": "100000000000000 200",
"output": "160000000000"
},
{
"input": "1000000000000 100",
"output": "6400000"
},
{
"input": "100 10",
"output": "-1"
},
{
"input": "1000000000039 2",
"output": "1000000000039"
},
{
"input": "64 5",
"output": "16"
},
{
"input": "999999961946176 33",
"output": "63245552"
},
{
"input": "376219076689 3",
"output": "376219076689"
},
{
"input": "999999961946176 63",
"output": "999999961946176"
},
{
"input": "1048576 12",
"output": "2048"
},
{
"input": "745 21",
"output": "-1"
},
{
"input": "748 6",
"output": "22"
},
{
"input": "999999961946176 50",
"output": "161082468097"
},
{
"input": "10 3",
"output": "5"
},
{
"input": "1099511627776 22",
"output": "2097152"
},
{
"input": "1000000007 100010",
"output": "-1"
},
{
"input": "3 1",
"output": "1"
},
{
"input": "100 8",
"output": "50"
},
{
"input": "100 7",
"output": "25"
},
{
"input": "7 2",
"output": "7"
},
{
"input": "999999961946176 64",
"output": "-1"
},
{
"input": "20 5",
"output": "10"
},
{
"input": "999999999999989 2",
"output": "999999999999989"
},
{
"input": "100000000000000 114",
"output": "10240000"
},
{
"input": "99999640000243 3",
"output": "9999991"
},
{
"input": "999998000001 566",
"output": "333332666667"
},
{
"input": "99999820000081 2",
"output": "9999991"
},
{
"input": "49000042000009 3",
"output": "49000042000009"
},
{
"input": "151491429961 4",
"output": "-1"
},
{
"input": "32416190071 2",
"output": "32416190071"
},
{
"input": "1000 8",
"output": "25"
},
{
"input": "1999967841 15",
"output": "1999967841"
},
{
"input": "26880 26880",
"output": "-1"
},
{
"input": "151491429961 3",
"output": "151491429961"
},
{
"input": "90000000000 300",
"output": "100000000"
},
{
"input": "98765004361 10",
"output": "-1"
},
{
"input": "15 2",
"output": "3"
},
{
"input": "16 2",
"output": "2"
},
{
"input": "1996 2",
"output": "2"
},
{
"input": "1997 2",
"output": "1997"
},
{
"input": "1999 2",
"output": "1999"
},
{
"input": "1998 2",
"output": "2"
},
{
"input": "1998 1",
"output": "1"
},
{
"input": "1998 7",
"output": "27"
},
{
"input": "1998 8",
"output": "37"
},
{
"input": "100000380000361 2",
"output": "10000019"
},
{
"input": "15 1",
"output": "1"
},
{
"input": "100000000000000 226",
"output": "-1"
},
{
"input": "844030857550613 517",
"output": "-1"
},
{
"input": "4567890 14",
"output": "430"
},
{
"input": "123123123 123123123",
"output": "-1"
},
{
"input": "24 4",
"output": "4"
},
{
"input": "999999993568952 17",
"output": "31622777"
},
{
"input": "99999999994190 9",
"output": "241656799"
},
{
"input": "999997874844049 4",
"output": "-1"
},
{
"input": "99999999999931 2",
"output": "99999999999931"
},
{
"input": "2 3",
"output": "-1"
},
{
"input": "67280421310721 2",
"output": "67280421310721"
}
] | 1,512,659,473 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 4 | 2,000 | 5,529,600 | a=list(map(int,input().split()))
c=0
i=1
l=0
divisor=0
while i<=a[0]+1 :
if c==a[1] :
break
else :
if a[0]%i==0 :
c+=1
divisor=i
i+=1
elif a[0]%i==a[0] :
l+=1
break
else :
i+=1
if l==0 :
print(divisor)
else :
print(-1)
| Title: k-th divisor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two integers *n* and *k*. Find *k*-th smallest divisor of *n*, or report that it doesn't exist.
Divisor of *n* is any such natural number, that *n* can be divided by it without remainder.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=1015, 1<=≤<=*k*<=≤<=109).
Output Specification:
If *n* has less than *k* divisors, output -1.
Otherwise, output the *k*-th smallest divisor of *n*.
Demo Input:
['4 2\n', '5 3\n', '12 5\n']
Demo Output:
['2\n', '-1\n', '6\n']
Note:
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.
In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1. | ```python
a=list(map(int,input().split()))
c=0
i=1
l=0
divisor=0
while i<=a[0]+1 :
if c==a[1] :
break
else :
if a[0]%i==0 :
c+=1
divisor=i
i+=1
elif a[0]%i==a[0] :
l+=1
break
else :
i+=1
if l==0 :
print(divisor)
else :
print(-1)
``` | 0 |
|
709 | A | Juicer | PROGRAMMING | 900 | [
"implementation"
] | null | null | Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section? | The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer. | Print one integer — the number of times Kolya will have to empty the waste section. | [
"2 7 10\n5 6\n",
"1 5 10\n7\n",
"3 10 10\n5 7 7\n",
"1 1 1\n1\n"
] | [
"1\n",
"0\n",
"1\n",
"0\n"
] | In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all. | 500 | [
{
"input": "2 7 10\n5 6",
"output": "1"
},
{
"input": "1 5 10\n7",
"output": "0"
},
{
"input": "3 10 10\n5 7 7",
"output": "1"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "2 951637 951638\n44069 951637",
"output": "1"
},
{
"input": "50 100 129\n55 130 91 19 116 3 63 52 104 76 75 27 151 99 149 147 39 148 84 9 132 49 40 112 124 141 144 93 36 32 146 74 48 38 150 55 94 32 107 69 77 81 33 57 62 98 78 127 154 126",
"output": "12"
},
{
"input": "100 1000 1083\n992 616 818 359 609 783 263 989 501 929 362 394 919 1081 870 830 1097 975 62 346 531 367 323 457 707 360 949 334 867 116 478 417 961 963 1029 114 867 1008 988 916 983 1077 959 942 572 961 579 318 721 337 488 717 111 70 416 685 987 130 353 107 61 191 827 849 106 815 211 953 111 398 889 860 801 71 375 320 395 1059 116 222 931 444 582 74 677 655 88 173 686 491 661 186 114 832 615 814 791 464 517 850",
"output": "36"
},
{
"input": "2 6 8\n2 1",
"output": "0"
},
{
"input": "5 15 16\n7 11 5 12 8",
"output": "2"
},
{
"input": "15 759966 759967\n890397 182209 878577 548548 759966 812923 759966 860479 200595 381358 299175 339368 759966 907668 69574",
"output": "4"
},
{
"input": "5 234613 716125\n642626 494941 234613 234613 234613",
"output": "0"
},
{
"input": "50 48547 567054\n529808 597004 242355 559114 78865 537318 631455 733020 655072 645093 309010 855034 306058 625046 524574 834944 27330 664392 443637 821584 338013 490702 289520 675471 885846 258814 134220 571301 84875 94132 200425 928833 375166 521232 317961 175315 947093 89971 322071 174033 48547 998535 954205 704114 943163 438900 48547 538422 48547 48547",
"output": "0"
},
{
"input": "5 10 20\n10 10 10 10 1",
"output": "1"
},
{
"input": "5 10 11\n10 10 10 10 1",
"output": "2"
},
{
"input": "3 10 10\n4 3 3",
"output": "0"
},
{
"input": "3 5 5\n5 5 5",
"output": "1"
},
{
"input": "3 4 14\n5 5 5",
"output": "0"
},
{
"input": "2 7 10\n1234 1234",
"output": "0"
},
{
"input": "1 5 6\n10",
"output": "0"
},
{
"input": "3 4 6\n1 2 3",
"output": "0"
},
{
"input": "5 10 12\n13 13 13 13 13",
"output": "0"
},
{
"input": "3 4 5\n5 7 9",
"output": "0"
},
{
"input": "3 10 100\n5 5 5",
"output": "0"
},
{
"input": "5 1 2\n2 2 2 2 2",
"output": "0"
},
{
"input": "5 5 5\n5 5 5 5 5",
"output": "2"
},
{
"input": "2 2 5\n5 5",
"output": "0"
},
{
"input": "3 1 4\n2 2 2",
"output": "0"
}
] | 1,685,849,308 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 10,956,800 | number_of_oranges, orange_limit_size, waste_limit = map(int, input().split())
amount_waste = 0
oranges = map(int, input().split())
for orange in oranges:
if orange < orange_limit_size:
amount_waste += orange
print(amount_waste // waste_limit) | Title: Juicer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?
Input Specification:
The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.
Output Specification:
Print one integer — the number of times Kolya will have to empty the waste section.
Demo Input:
['2 7 10\n5 6\n', '1 5 10\n7\n', '3 10 10\n5 7 7\n', '1 1 1\n1\n']
Demo Output:
['1\n', '0\n', '1\n', '0\n']
Note:
In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all. | ```python
number_of_oranges, orange_limit_size, waste_limit = map(int, input().split())
amount_waste = 0
oranges = map(int, input().split())
for orange in oranges:
if orange < orange_limit_size:
amount_waste += orange
print(amount_waste // waste_limit)
``` | 0 |
|
78 | A | Haiku | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Haiku | 2 | 256 | Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not. | The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification. | Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes). | [
"on codeforces \nbeta round is running\n a rustling of keys \n",
"how many gallons\nof edo s rain did you drink\n cuckoo\n"
] | [
"YES",
"NO"
] | none | 500 | [
{
"input": "on codeforces \nbeta round is running\n a rustling of keys ",
"output": "YES"
},
{
"input": "how many gallons\nof edo s rain did you drink\n cuckoo",
"output": "NO"
},
{
"input": " hatsu shigure\n saru mo komino wo\nhoshige nari",
"output": "YES"
},
{
"input": "o vetus stagnum\n rana de ripa salit\n ac sonant aquae",
"output": "NO"
},
{
"input": " furuike ya\nkawazu tobikomu\nmizu no oto ",
"output": "YES"
},
{
"input": " noch da leich\na stamperl zum aufwaerma\n da pfarrer kimmt a ",
"output": "NO"
},
{
"input": " sommerfuglene \n hvorfor bruge mange ord\n et kan gore det",
"output": "YES"
},
{
"input": " ab der mittagszeit\n ist es etwas schattiger\n ein wolkenhimmel",
"output": "NO"
},
{
"input": "tornando a vederli\ni fiori di ciliegio la sera\nson divenuti frutti",
"output": "NO"
},
{
"input": "kutaburete\nyado karu koro ya\nfuji no hana",
"output": "YES"
},
{
"input": " beginnings of poetry\n the rice planting songs \n of the interior",
"output": "NO"
},
{
"input": " door zomerregens\n zijn de kraanvogelpoten\n korter geworden",
"output": "NO"
},
{
"input": " derevo na srub\na ptitsi bezzabotno\n gnezdishko tam vyut",
"output": "YES"
},
{
"input": "writing in the dark\nunaware that my pen\nhas run out of ink",
"output": "NO"
},
{
"input": "kusaaiu\nuieueua\nuo efaa",
"output": "YES"
},
{
"input": "v\nh\np",
"output": "NO"
},
{
"input": "i\ni\nu",
"output": "NO"
},
{
"input": "awmio eoj\nabdoolceegood\nwaadeuoy",
"output": "YES"
},
{
"input": "xzpnhhnqsjpxdboqojixmofawhdjcfbscq\nfoparnxnbzbveycoltwdrfbwwsuobyoz hfbrszy\nimtqryscsahrxpic agfjh wvpmczjjdrnwj mcggxcdo",
"output": "YES"
},
{
"input": "wxjcvccp cppwsjpzbd dhizbcnnllckybrnfyamhgkvkjtxxfzzzuyczmhedhztugpbgpvgh\nmdewztdoycbpxtp bsiw hknggnggykdkrlihvsaykzfiiw\ndewdztnngpsnn lfwfbvnwwmxoojknygqb hfe ibsrxsxr",
"output": "YES"
},
{
"input": "nbmtgyyfuxdvrhuhuhpcfywzrbclp znvxw synxmzymyxcntmhrjriqgdjh xkjckydbzjbvtjurnf\nhhnhxdknvamywhsrkprofnyzlcgtdyzzjdsfxyddvilnzjziz qmwfdvzckgcbrrxplxnxf mpxwxyrpesnewjrx ajxlfj\nvcczq hddzd cvefmhxwxxyqcwkr fdsndckmesqeq zyjbwbnbyhybd cta nsxzidl jpcvtzkldwd",
"output": "YES"
},
{
"input": "rvwdsgdsrutgjwscxz pkd qtpmfbqsmctuevxdj kjzknzghdvxzlaljcntg jxhvzn yciktbsbyscfypx x xhkxnfpdp\nwdfhvqgxbcts mnrwbr iqttsvigwdgvlxwhsmnyxnttedonxcfrtmdjjmacvqtkbmsnwwvvrlxwvtggeowtgsqld qj\nvsxcdhbzktrxbywpdvstr meykarwtkbm pkkbhvwvelclfmpngzxdmblhcvf qmabmweldplmczgbqgzbqnhvcdpnpjtch ",
"output": "YES"
},
{
"input": "brydyfsmtzzkpdsqvvztmprhqzbzqvgsblnz naait tdtiprjsttwusdykndwcccxfmzmrmfmzjywkpgbfnjpypgcbcfpsyfj k\nucwdfkfyxxxht lxvnovqnnsqutjsyagrplb jhvtwdptrwcqrovncdvqljjlrpxcfbxqgsfylbgmcjpvpl ccbcybmigpmjrxpu\nfgwtpcjeywgnxgbttgx htntpbk tkkpwbgxwtbxvcpkqbzetjdkcwad tftnjdxxjdvbpfibvxuglvx llyhgjvggtw jtjyphs",
"output": "YES"
},
{
"input": "nyc aqgqzjjlj mswgmjfcxlqdscheskchlzljlsbhyn iobxymwzykrsnljj\nnnebeaoiraga\nqpjximoqzswhyyszhzzrhfwhf iyxysdtcpmikkwpugwlxlhqfkn",
"output": "NO"
},
{
"input": "lzrkztgfe mlcnq ay ydmdzxh cdgcghxnkdgmgfzgahdjjmqkpdbskreswpnblnrc fmkwziiqrbskp\np oukeaz gvvy kghtrjlczyl qeqhgfgfej\nwfolhkmktvsjnrpzfxcxzqmfidtlzmuhxac wsncjgmkckrywvxmnjdpjpfydhk qlmdwphcvyngansqhl",
"output": "NO"
},
{
"input": "yxcboqmpwoevrdhvpxfzqmammak\njmhphkxppkqkszhqqtkvflarsxzla pbxlnnnafqbsnmznfj qmhoktgzix qpmrgzxqvmjxhskkksrtryehfnmrt dtzcvnvwp\nscwymuecjxhw rdgsffqywwhjpjbfcvcrnisfqllnbplpadfklayjguyvtrzhwblftclfmsr",
"output": "NO"
},
{
"input": "qfdwsr jsbrpfmn znplcx nhlselflytndzmgxqpgwhpi ghvbbxrkjdirfghcybhkkqdzmyacvrrcgsneyjlgzfvdmxyjmph\nylxlyrzs drbktzsniwcbahjkgohcghoaczsmtzhuwdryjwdijmxkmbmxv yyfrokdnsx\nyw xtwyzqlfxwxghugoyscqlx pljtz aldfskvxlsxqgbihzndhxkswkxqpwnfcxzfyvncstfpqf",
"output": "NO"
},
{
"input": "g rguhqhcrzmuqthtmwzhfyhpmqzzosa\nmhjimzvchkhejh irvzejhtjgaujkqfxhpdqjnxr dvqallgssktqvsxi\npcwbliftjcvuzrsqiswohi",
"output": "NO"
},
{
"input": " ngxtlq iehiise vgffqcpnmsoqzyseuqqtggokymol zn\nvjdjljazeujwoubkcvtsbepooxqzrueaauokhepiquuopfild\ngoabauauaeotoieufueeknudiilupouaiaexcoapapu",
"output": "NO"
},
{
"input": "ycnvnnqk mhrmhctpkfbc qbyvtjznmndqjzgbcxmvrpkfcll zwspfptmbxgrdv dsgkk nfytsqjrnfbhh pzdldzymvkdxxwh\nvnhjfwgdnyjptsmblyxmpzylsbjlmtkkwjcbqwjctqvrlqqkdsrktxlnslspvnn mdgsmzblhbnvpczmqkcffwhwljqkzmk hxcm\nrghnjvzcpprrgmtgytpkzyc mrdnnhpkwypwqbtzjyfwvrdwyjltbzxtbstzs xdjzdmx yjsqtzlrnvyssvglsdjrmsrfrcdpqt",
"output": "NO"
},
{
"input": "ioeeaioeiuoeaeieuuieooaouiuouiioaueeaiaiuoaoiioeeaauooiuuieeuaeeoauieeaiuoieiaieuoauaaoioooieueueuai\nuooaoeeaoiuuoeioaoouaououoeioiaeueoioaiouaeaoioiuuaueeuaiuoiueoiuaoeeieeouaeeaeeieioeoiiieuuueuuieuo\naeeouieeieoueaioeoioooiouaeeeiaaioueauaueiouuuaieuuioiaeiueauueaoieauauoeueuaiueuuaueeoueauaeaoieeoo",
"output": "NO"
},
{
"input": "mkgycgmrqtmlwj vddlbsgdptyfrcj nmrddjchydvbsyhgzytfzvwbhqmmhkpsyrcnknpsqrr wcnf wfvgskrnsmnrcqgcnc q\npn cclcyjjdp rqkpgyrvyyd hhynrochfa pthyffqvskazfshvzcmvhmsrxvquzkdvivlbfypwbgltfbjmhzljvgzmxjlrjdjq\nauuouieuaaauoooeoiuoaueuueauiuuuiaeuwuoiaeoeaieauieouoaeiiieiioiuuuiuauioiiuoeauueieaueiaioeuaoaaau",
"output": "NO"
},
{
"input": "ab\na\na",
"output": "NO"
},
{
"input": "on codeforcs\nbeta round is runnineg\na rustling of keys",
"output": "NO"
},
{
"input": "coooooooooooooooood\nd\nd",
"output": "NO"
},
{
"input": "on codeforces \nbeta round is runninog\n a rustling of keys",
"output": "NO"
},
{
"input": "aaaaaaaaaaaaaaa\na\na",
"output": "NO"
},
{
"input": "aaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc",
"output": "YES"
},
{
"input": "cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\ncccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\ncccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc",
"output": "NO"
},
{
"input": "aaaaaa\naaaaaa\naaaaa",
"output": "NO"
}
] | 1,689,868,242 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | s1 = input()
s2 = input()
s3 = input()
vowels = 'aeiou'
count = 0
for ch in s1:
for ch in s2:
for ch in s3:
if ch in vowels:
count += 1
if s1.count==5 and s2.count==7 and s3.count==5::
print("YES")
else:
print("NO") | Title: Haiku
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input Specification:
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output Specification:
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Demo Input:
['on codeforces \nbeta round is running\n a rustling of keys \n', 'how many gallons\nof edo s rain did you drink\n cuckoo\n']
Demo Output:
['YES', 'NO']
Note:
none | ```python
s1 = input()
s2 = input()
s3 = input()
vowels = 'aeiou'
count = 0
for ch in s1:
for ch in s2:
for ch in s3:
if ch in vowels:
count += 1
if s1.count==5 and s2.count==7 and s3.count==5::
print("YES")
else:
print("NO")
``` | -1 |
31 | A | Worms Evolution | PROGRAMMING | 1,200 | [
"implementation"
] | A. Worms Evolution | 2 | 256 | Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are *n* forms of worms. Worms of these forms have lengths *a*1, *a*2, ..., *a**n*. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of worm's forms. The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000) — lengths of worms of each form. | Output 3 distinct integers *i* *j* *k* (1<=≤<=*i*,<=*j*,<=*k*<=≤<=*n*) — such indexes of worm's forms that *a**i*<==<=*a**j*<=+<=*a**k*. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that *a**j*<==<=*a**k*. | [
"5\n1 2 3 5 7\n",
"5\n1 8 1 5 1\n"
] | [
"3 2 1\n",
"-1\n"
] | none | 500 | [
{
"input": "5\n1 2 3 5 7",
"output": "3 2 1"
},
{
"input": "5\n1 8 1 5 1",
"output": "-1"
},
{
"input": "4\n303 872 764 401",
"output": "-1"
},
{
"input": "6\n86 402 133 524 405 610",
"output": "6 4 1"
},
{
"input": "8\n217 779 418 895 996 473 3 22",
"output": "5 2 1"
},
{
"input": "10\n858 972 670 15 662 114 33 273 53 310",
"output": "2 6 1"
},
{
"input": "100\n611 697 572 770 603 870 128 245 49 904 468 982 788 943 549 288 668 796 803 515 999 735 912 49 298 80 412 841 494 434 543 298 17 571 271 105 70 313 178 755 194 279 585 766 412 164 907 841 776 556 731 268 735 880 176 267 287 65 239 588 155 658 821 47 783 595 585 69 226 906 429 161 999 148 7 484 362 585 952 365 92 749 904 525 307 626 883 367 450 755 564 950 728 724 69 106 119 157 96 290",
"output": "1 38 25"
},
{
"input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438",
"output": "1 63 61"
},
{
"input": "100\n156 822 179 298 981 82 610 345 373 378 895 734 768 15 78 335 764 608 932 297 717 553 916 367 425 447 361 195 66 70 901 236 905 744 919 564 296 610 963 628 840 52 100 750 345 308 37 687 192 704 101 815 10 990 216 358 823 546 578 821 706 148 182 582 421 482 829 425 121 337 500 301 402 868 66 935 625 527 746 585 308 523 488 914 608 709 875 252 151 781 447 2 756 176 976 302 450 35 680 791",
"output": "1 98 69"
},
{
"input": "100\n54 947 785 838 359 647 92 445 48 465 323 486 101 86 607 31 860 420 709 432 435 372 272 37 903 814 309 197 638 58 259 822 793 564 309 22 522 907 101 853 486 824 614 734 630 452 166 532 256 499 470 9 933 452 256 450 7 26 916 406 257 285 895 117 59 369 424 133 16 417 352 440 806 236 478 34 889 469 540 806 172 296 73 655 261 792 868 380 204 454 330 53 136 629 236 850 134 560 264 291",
"output": "2 29 27"
},
{
"input": "99\n175 269 828 129 499 890 127 263 995 807 508 289 996 226 437 320 365 642 757 22 190 8 345 499 834 713 962 889 336 171 608 492 320 257 472 801 176 325 301 306 198 729 933 4 640 322 226 317 567 586 249 237 202 633 287 128 911 654 719 988 420 855 361 574 716 899 317 356 581 440 284 982 541 111 439 29 37 560 961 224 478 906 319 416 736 603 808 87 762 697 392 713 19 459 262 238 239 599 997",
"output": "1 44 30"
},
{
"input": "98\n443 719 559 672 16 69 529 632 953 999 725 431 54 22 346 968 558 696 48 669 963 129 257 712 39 870 498 595 45 821 344 925 179 388 792 346 755 213 423 365 344 659 824 356 773 637 628 897 841 155 243 536 951 361 192 105 418 431 635 596 150 162 145 548 473 531 750 306 377 354 450 975 79 743 656 733 440 940 19 139 237 346 276 227 64 799 479 633 199 17 796 362 517 234 729 62 995 535",
"output": "2 70 40"
},
{
"input": "97\n359 522 938 862 181 600 283 1000 910 191 590 220 761 818 903 264 751 751 987 316 737 898 168 925 244 674 34 950 754 472 81 6 37 520 112 891 981 454 897 424 489 238 363 709 906 951 677 828 114 373 589 835 52 89 97 435 277 560 551 204 879 469 928 523 231 163 183 609 821 915 615 969 616 23 874 437 844 321 78 53 643 786 585 38 744 347 150 179 988 985 200 11 15 9 547 886 752",
"output": "1 23 10"
},
{
"input": "4\n303 872 764 401",
"output": "-1"
},
{
"input": "100\n328 397 235 453 188 254 879 225 423 36 384 296 486 592 231 849 856 255 213 898 234 800 701 529 951 693 507 326 15 905 618 348 967 927 28 979 752 850 343 35 84 302 36 390 482 826 249 918 91 289 973 457 557 348 365 239 709 565 320 560 153 130 647 708 483 469 788 473 322 844 830 562 611 961 397 673 69 960 74 703 369 968 382 451 328 160 211 230 566 208 7 545 293 73 806 375 157 410 303 58",
"output": "1 79 6"
},
{
"input": "33\n52 145 137 734 180 847 178 286 716 134 181 630 358 764 593 762 785 28 1 468 189 540 764 485 165 656 114 58 628 108 605 584 257",
"output": "8 30 7"
},
{
"input": "57\n75 291 309 68 444 654 985 158 514 204 116 918 374 806 176 31 49 455 269 66 722 713 164 818 317 295 546 564 134 641 28 13 987 478 146 219 213 940 289 173 157 666 168 391 392 71 870 477 446 988 414 568 964 684 409 671 454",
"output": "2 41 29"
},
{
"input": "88\n327 644 942 738 84 118 981 686 530 404 137 197 434 16 693 183 423 325 410 345 941 329 7 106 79 867 584 358 533 675 192 718 641 329 900 768 404 301 101 538 954 590 401 954 447 14 559 337 756 586 934 367 538 928 945 936 770 641 488 579 206 869 902 139 216 446 723 150 829 205 373 578 357 368 960 40 121 206 503 385 521 161 501 694 138 370 709 308",
"output": "1 77 61"
},
{
"input": "100\n804 510 266 304 788 625 862 888 408 82 414 470 777 991 729 229 933 406 601 1 596 720 608 706 432 361 527 548 59 548 474 515 4 991 263 568 681 24 117 563 576 587 281 643 904 521 891 106 842 884 943 54 605 815 504 757 311 374 335 192 447 652 633 410 455 402 382 150 432 836 413 819 669 875 638 925 217 805 632 520 605 266 728 795 162 222 603 159 284 790 914 443 775 97 789 606 859 13 851 47",
"output": "1 77 42"
},
{
"input": "100\n449 649 615 713 64 385 927 466 138 126 143 886 80 199 208 43 196 694 92 89 264 180 617 970 191 196 910 150 275 89 693 190 191 99 542 342 45 592 114 56 451 170 64 589 176 102 308 92 402 153 414 675 352 157 69 150 91 288 163 121 816 184 20 234 836 12 593 150 793 439 540 93 99 663 186 125 349 247 476 106 77 523 215 7 363 278 441 745 337 25 148 384 15 915 108 211 240 58 23 408",
"output": "1 6 5"
},
{
"input": "90\n881 436 52 308 97 261 153 931 670 538 702 156 114 445 154 685 452 76 966 790 93 42 547 65 736 364 136 489 719 322 239 628 696 735 55 703 622 375 100 188 804 341 546 474 484 446 729 290 974 301 602 225 996 244 488 983 882 460 962 754 395 617 61 640 534 292 158 375 632 902 420 979 379 38 100 67 963 928 190 456 545 571 45 716 153 68 844 2 102 116",
"output": "1 14 2"
},
{
"input": "80\n313 674 262 240 697 146 391 221 793 504 896 818 92 899 86 370 341 339 306 887 937 570 830 683 729 519 240 833 656 847 427 958 435 704 853 230 758 347 660 575 843 293 649 396 437 787 654 599 35 103 779 783 447 379 444 585 902 713 791 150 851 228 306 721 996 471 617 403 102 168 197 741 877 481 968 545 331 715 236 654",
"output": "1 13 8"
},
{
"input": "70\n745 264 471 171 946 32 277 511 269 469 89 831 69 2 369 407 583 602 646 633 429 747 113 302 722 321 344 824 241 372 263 287 822 24 652 758 246 967 219 313 882 597 752 965 389 775 227 556 95 904 308 340 899 514 400 187 275 318 621 546 659 488 199 154 811 1 725 79 925 82",
"output": "1 63 60"
},
{
"input": "60\n176 502 680 102 546 917 516 801 392 435 635 492 398 456 653 444 472 513 634 378 273 276 44 920 68 124 800 167 825 250 452 264 561 344 98 933 381 939 426 51 568 548 206 887 342 763 151 514 156 354 486 546 998 649 356 438 295 570 450 589",
"output": "2 26 20"
},
{
"input": "50\n608 92 889 33 146 803 402 91 868 400 828 505 375 558 584 129 361 776 974 123 765 804 326 186 61 927 904 511 762 775 640 593 300 664 897 461 869 911 986 789 607 500 309 457 294 104 724 471 216 155",
"output": "3 25 11"
},
{
"input": "40\n40 330 98 612 747 336 640 381 991 366 22 167 352 12 868 166 603 40 313 869 609 981 609 804 54 729 8 854 347 300 828 922 39 633 695 988 4 530 545 176",
"output": "5 10 8"
},
{
"input": "30\n471 920 308 544 347 222 878 671 467 332 215 180 681 114 151 203 492 951 653 614 453 510 540 422 399 532 113 198 932 825",
"output": "2 21 9"
},
{
"input": "20\n551 158 517 475 595 108 764 961 590 297 761 841 659 568 82 888 733 214 993 359",
"output": "3 20 2"
},
{
"input": "10\n983 748 726 406 196 993 2 251 66 263",
"output": "-1"
},
{
"input": "9\n933 266 457 863 768 257 594 136 145",
"output": "-1"
},
{
"input": "8\n537 198 48 771 944 868 700 163",
"output": "7 8 1"
},
{
"input": "7\n140 779 639 679 768 479 158",
"output": "2 3 1"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "-1"
},
{
"input": "5\n700 939 173 494 120",
"output": "-1"
},
{
"input": "4\n303 872 764 401",
"output": "-1"
},
{
"input": "3\n907 452 355",
"output": "-1"
},
{
"input": "3\n963 630 333",
"output": "1 3 2"
},
{
"input": "3\n2 2 4",
"output": "3 2 1"
},
{
"input": "3\n2 4 100",
"output": "-1"
}
] | 1,610,080,730 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 186 | 0 | x=int(input())
f=1
y=sorted(list(map(int,input().split())))
for i in range(x-1):
if(y[i]+y[i+1] in y[i+2:]):
f=0
print(i+2+1,i+1+1,i+1)
break
if(f):
print("-1") | Title: Worms Evolution
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are *n* forms of worms. Worms of these forms have lengths *a*1, *a*2, ..., *a**n*. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of worm's forms. The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000) — lengths of worms of each form.
Output Specification:
Output 3 distinct integers *i* *j* *k* (1<=≤<=*i*,<=*j*,<=*k*<=≤<=*n*) — such indexes of worm's forms that *a**i*<==<=*a**j*<=+<=*a**k*. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that *a**j*<==<=*a**k*.
Demo Input:
['5\n1 2 3 5 7\n', '5\n1 8 1 5 1\n']
Demo Output:
['3 2 1\n', '-1\n']
Note:
none | ```python
x=int(input())
f=1
y=sorted(list(map(int,input().split())))
for i in range(x-1):
if(y[i]+y[i+1] in y[i+2:]):
f=0
print(i+2+1,i+1+1,i+1)
break
if(f):
print("-1")
``` | 0 |
472 | A | Design Tutorial: Learn from Math | PROGRAMMING | 800 | [
"math",
"number theory"
] | null | null | One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers. | The only line contains an integer *n* (12<=≤<=*n*<=≤<=106). | Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them. | [
"12\n",
"15\n",
"23\n",
"1000000\n"
] | [
"4 8\n",
"6 9\n",
"8 15\n",
"500000 500000\n"
] | In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number. | 500 | [
{
"input": "12",
"output": "4 8"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "23",
"output": "8 15"
},
{
"input": "1000000",
"output": "500000 500000"
},
{
"input": "63874",
"output": "4 63870"
},
{
"input": "14568",
"output": "4 14564"
},
{
"input": "192",
"output": "4 188"
},
{
"input": "86",
"output": "4 82"
},
{
"input": "46220",
"output": "4 46216"
},
{
"input": "57114",
"output": "4 57110"
},
{
"input": "869",
"output": "4 865"
},
{
"input": "738457",
"output": "4 738453"
},
{
"input": "58113",
"output": "6 58107"
},
{
"input": "4864",
"output": "4 4860"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "74752",
"output": "4 74748"
},
{
"input": "6073",
"output": "4 6069"
},
{
"input": "1289",
"output": "4 1285"
},
{
"input": "20",
"output": "4 16"
},
{
"input": "58134",
"output": "4 58130"
},
{
"input": "57756",
"output": "4 57752"
},
{
"input": "765",
"output": "6 759"
},
{
"input": "59",
"output": "4 55"
},
{
"input": "991666",
"output": "4 991662"
},
{
"input": "70761",
"output": "4 70757"
},
{
"input": "13",
"output": "4 9"
},
{
"input": "999999",
"output": "4 999995"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "21",
"output": "6 15"
},
{
"input": "19",
"output": "4 15"
},
{
"input": "100007",
"output": "6 100001"
},
{
"input": "999987",
"output": "6 999981"
},
{
"input": "22",
"output": "4 18"
}
] | 1,699,130,267 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | n=input();m=(n%2*5+4);print m,n-m
| Title: Design Tutorial: Learn from Math
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
Input Specification:
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
Output Specification:
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
Demo Input:
['12\n', '15\n', '23\n', '1000000\n']
Demo Output:
['4 8\n', '6 9\n', '8 15\n', '500000 500000\n']
Note:
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number. | ```python
n=input();m=(n%2*5+4);print m,n-m
``` | -1 |
|
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,637,602,495 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | s=input()
UC=0
LC=0
for i in s:
if(i.islower()):
LC=LC+1
elif(i.isupper()):
UC=UC+1
if(UC>LC):
print(s.upper())
elif(LC>UC):
print(s.lower())
elif(LC==UC):
print(s.lower()) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
s=input()
UC=0
LC=0
for i in s:
if(i.islower()):
LC=LC+1
elif(i.isupper()):
UC=UC+1
if(UC>LC):
print(s.upper())
elif(LC>UC):
print(s.lower())
elif(LC==UC):
print(s.lower())
``` | 3.977 |
2 | A | Winner | PROGRAMMING | 1,500 | [
"hashing",
"implementation"
] | A. Winner | 1 | 64 | The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points. | The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive. | Print the name of the winner. | [
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] | [
"andrew\n",
"andrew\n"
] | none | 0 | [
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},
{
"input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734",
"output": "ksjuuerbnlklcfdjeyq"
},
{
"input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879",
"output": "fcgslzkicjrpbqaifgweyzreajjfdo"
},
{
"input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735",
"output": "aawtvezfntstrcpgbzjbf"
},
{
"input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158",
"output": "ivhgbxiv"
}
] | 1,665,039,025 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | dic={}
dic_1={}
dict={}
n=int(input())
for j in range(n):
dict[j]=input()
if dict[j][:-2] not in dic.keys():
dic[dict[j][:-2]]=int(dict[j][-1])
else:
dic[dict[j][:-2]]+=int(dict[j][-1])
s=[a for a in dic.values()]
m=max(s)
while s.count(m)!=1 and j in range(n):
x=dict[n-j-1]
dic[x[:-2]]-=x[-1]
j+=1
print(key) | Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none | ```python
dic={}
dic_1={}
dict={}
n=int(input())
for j in range(n):
dict[j]=input()
if dict[j][:-2] not in dic.keys():
dic[dict[j][:-2]]=int(dict[j][-1])
else:
dic[dict[j][:-2]]+=int(dict[j][-1])
s=[a for a in dic.values()]
m=max(s)
while s.count(m)!=1 and j in range(n):
x=dict[n-j-1]
dic[x[:-2]]-=x[-1]
j+=1
print(key)
``` | -1 |
455 | A | Boredom | PROGRAMMING | 1,500 | [
"dp"
] | null | null | Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105). | Print a single integer — the maximum number of points that Alex can earn. | [
"2\n1 2\n",
"3\n1 2 3\n",
"9\n1 2 1 3 2 2 2 2 3\n"
] | [
"2\n",
"4\n",
"10\n"
] | Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points. | 500 | [
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n1 2 3",
"output": "4"
},
{
"input": "9\n1 2 1 3 2 2 2 2 3",
"output": "10"
},
{
"input": "5\n3 3 4 5 4",
"output": "11"
},
{
"input": "5\n5 3 5 3 4",
"output": "16"
},
{
"input": "5\n4 2 3 2 5",
"output": "9"
},
{
"input": "10\n10 5 8 9 5 6 8 7 2 8",
"output": "46"
},
{
"input": "10\n1 1 1 1 1 1 2 3 4 4",
"output": "14"
},
{
"input": "100\n6 6 8 9 7 9 6 9 5 7 7 4 5 3 9 1 10 3 4 5 8 9 6 5 6 4 10 9 1 4 1 7 1 4 9 10 8 2 9 9 10 5 8 9 5 6 8 7 2 8 7 6 2 6 10 8 6 2 5 5 3 2 8 8 5 3 6 2 1 4 7 2 7 3 7 4 10 10 7 5 4 7 5 10 7 1 1 10 7 7 7 2 3 4 2 8 4 7 4 4",
"output": "296"
},
{
"input": "100\n6 1 5 7 10 10 2 7 3 7 2 10 7 6 3 5 5 5 3 7 2 4 2 7 7 4 2 8 2 10 4 7 9 1 1 7 9 7 1 10 10 9 5 6 10 1 7 5 8 1 1 5 3 10 2 4 3 5 2 7 4 9 5 10 1 3 7 6 6 9 3 6 6 10 1 10 6 1 10 3 4 1 7 9 2 7 8 9 3 3 2 4 6 6 1 2 9 4 1 2",
"output": "313"
},
{
"input": "100\n7 6 3 8 8 3 10 5 3 8 6 4 6 9 6 7 3 9 10 7 5 5 9 10 7 2 3 8 9 5 4 7 9 3 6 4 9 10 7 6 8 7 6 6 10 3 7 4 5 7 7 5 1 5 4 8 7 3 3 4 7 8 5 9 2 2 3 1 6 4 6 6 6 1 7 10 7 4 5 3 9 2 4 1 5 10 9 3 9 6 8 5 2 1 10 4 8 5 10 9",
"output": "298"
},
{
"input": "100\n2 10 9 1 2 6 7 2 2 8 9 9 9 5 6 2 5 1 1 10 7 4 5 5 8 1 9 4 10 1 9 3 1 8 4 10 8 8 2 4 6 5 1 4 2 2 1 2 8 5 3 9 4 10 10 7 8 6 1 8 2 6 7 1 6 7 3 10 10 3 7 7 6 9 6 8 8 10 4 6 4 3 3 3 2 3 10 6 8 5 5 10 3 7 3 1 1 1 5 5",
"output": "312"
},
{
"input": "100\n4 9 7 10 4 7 2 6 1 9 1 8 7 5 5 7 6 7 9 8 10 5 3 5 7 10 3 2 1 3 8 9 4 10 4 7 6 4 9 6 7 1 9 4 3 5 8 9 2 7 10 5 7 5 3 8 10 3 8 9 3 4 3 10 6 5 1 8 3 2 5 8 4 7 5 3 3 2 6 9 9 8 2 7 6 3 2 2 8 8 4 5 6 9 2 3 2 2 5 2",
"output": "287"
},
{
"input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8",
"output": "380"
},
{
"input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8",
"output": "380"
},
{
"input": "100\n10 5 8 4 4 4 1 4 5 8 3 10 2 4 1 10 8 1 1 6 8 4 2 9 1 3 1 7 7 9 3 5 5 8 6 9 9 4 8 1 3 3 2 6 1 5 4 5 3 5 5 6 7 5 7 9 3 5 4 9 2 6 8 1 1 7 7 3 8 9 8 7 3 2 4 1 6 1 3 9 4 2 2 8 5 10 1 8 8 5 1 5 6 9 4 5 6 5 10 2",
"output": "265"
},
{
"input": "100\n7 5 1 8 5 6 6 2 6 2 7 7 3 6 2 4 4 2 10 2 2 2 10 6 6 1 5 10 9 1 5 9 8 9 4 1 10 5 7 5 7 6 4 8 8 1 7 8 3 8 2 1 8 4 10 3 5 6 6 10 9 6 5 1 10 7 6 9 9 2 10 10 9 1 2 1 7 7 4 10 1 10 5 5 3 8 9 8 1 4 10 2 4 5 4 4 1 6 2 9",
"output": "328"
},
{
"input": "100\n5 6 10 7 1 7 10 1 9 1 5 1 4 1 3 3 7 9 1 6 1 6 5 7 1 6 3 1 3 6 3 8 2 4 1 5 2 10 7 3 10 4 10 1 5 4 2 9 7 9 5 7 10 4 1 4 8 9 3 1 3 7 7 4 3 7 7 10 6 9 5 5 6 5 3 9 8 8 5 5 4 10 9 4 10 4 1 8 3 5 4 10 9 3 10 4 10 7 10 9",
"output": "324"
},
{
"input": "10\n7 4 5 3 9 1 10 3 4 5",
"output": "34"
},
{
"input": "10\n8 9 6 5 6 4 10 9 1 4",
"output": "39"
},
{
"input": "10\n1 7 1 4 9 10 8 2 9 9",
"output": "40"
},
{
"input": "1\n100000",
"output": "100000"
}
] | 1,668,614,721 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 810 | 13,209,600 | n = int(input())
list1 = list(map(int,input().split()))
list2 = [0]*1000001
for i in list1:
list2[i] += 1
list3 = [0]*1000001
for q in range(1,1000001):
list3[q] = max(list3[q-1],list3[q-2]+q*list2[q])
print(list3[max(list1)]) | Title: Boredom
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Print a single integer — the maximum number of points that Alex can earn.
Demo Input:
['2\n1 2\n', '3\n1 2 3\n', '9\n1 2 1 3 2 2 2 2 3\n']
Demo Output:
['2\n', '4\n', '10\n']
Note:
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points. | ```python
n = int(input())
list1 = list(map(int,input().split()))
list2 = [0]*1000001
for i in list1:
list2[i] += 1
list3 = [0]*1000001
for q in range(1,1000001):
list3[q] = max(list3[q-1],list3[q-2]+q*list2[q])
print(list3[max(list1)])
``` | 3 |
|
960 | A | Check the string | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes). | The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'. | Print "YES" or "NO", according to the condition. | [
"aaabccc\n",
"bbacc\n",
"aabc\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | 500 | [
{
"input": "aaabccc",
"output": "YES"
},
{
"input": "bbacc",
"output": "NO"
},
{
"input": "aabc",
"output": "YES"
},
{
"input": "aabbcc",
"output": "YES"
},
{
"input": "aaacccbb",
"output": "NO"
},
{
"input": "abc",
"output": "YES"
},
{
"input": "acba",
"output": "NO"
},
{
"input": "bbabbc",
"output": "NO"
},
{
"input": "bbbabacca",
"output": "NO"
},
{
"input": "aabcbcaca",
"output": "NO"
},
{
"input": "aaaaabbbbbb",
"output": "NO"
},
{
"input": "c",
"output": "NO"
},
{
"input": "cc",
"output": "NO"
},
{
"input": "bbb",
"output": "NO"
},
{
"input": "bc",
"output": "NO"
},
{
"input": "ccbcc",
"output": "NO"
},
{
"input": "aaa",
"output": "NO"
},
{
"input": "aaccaa",
"output": "NO"
},
{
"input": "a",
"output": "NO"
},
{
"input": "b",
"output": "NO"
},
{
"input": "abca",
"output": "NO"
},
{
"input": "aabbcccc",
"output": "NO"
},
{
"input": "abac",
"output": "NO"
},
{
"input": "abcc",
"output": "NO"
},
{
"input": "abcb",
"output": "NO"
},
{
"input": "aacc",
"output": "NO"
},
{
"input": "aabbaacccc",
"output": "NO"
},
{
"input": "aabb",
"output": "NO"
},
{
"input": "ac",
"output": "NO"
},
{
"input": "abbacc",
"output": "NO"
},
{
"input": "abacc",
"output": "NO"
},
{
"input": "ababc",
"output": "NO"
},
{
"input": "aa",
"output": "NO"
},
{
"input": "aabaccc",
"output": "NO"
},
{
"input": "bbcc",
"output": "NO"
},
{
"input": "aaabcbc",
"output": "NO"
},
{
"input": "acbbc",
"output": "NO"
},
{
"input": "babc",
"output": "NO"
},
{
"input": "bbbcc",
"output": "NO"
},
{
"input": "bbc",
"output": "NO"
},
{
"input": "abababccc",
"output": "NO"
},
{
"input": "ccbbaa",
"output": "NO"
}
] | 1,540,108,759 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 16 | 140 | 204,800 | s = input()
exist = True
for prev, nex in zip(list(s)[:-1], list(s)[1:]):
if prev > nex:
exist = False
anum = s.count("a")
bnum = s.count("b")
cnum = s.count("c")
if anum != cnum and bnum != cnum:
exist = False
if exist:
print("YES")
else:
print("NO") | Title: Check the string
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.
B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.
You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).
Input Specification:
The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.
Output Specification:
Print "YES" or "NO", according to the condition.
Demo Input:
['aaabccc\n', 'bbacc\n', 'aabc\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
Consider first example: the number of 'c' is equal to the number of 'a'.
Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.
Consider third example: the number of 'c' is equal to the number of 'b'. | ```python
s = input()
exist = True
for prev, nex in zip(list(s)[:-1], list(s)[1:]):
if prev > nex:
exist = False
anum = s.count("a")
bnum = s.count("b")
cnum = s.count("c")
if anum != cnum and bnum != cnum:
exist = False
if exist:
print("YES")
else:
print("NO")
``` | 0 |
|
295 | B | Greg and Graph | PROGRAMMING | 1,700 | [
"dp",
"graphs",
"shortest paths"
] | null | null | Greg has a weighed directed graph, consisting of *n* vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:
- The game consists of *n* steps. - On the *i*-th step Greg removes vertex number *x**i* from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. - Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that *d*(*i*,<=*v*,<=*u*) is the shortest path between vertices *v* and *u* in the graph that formed before deleting vertex *x**i*, then Greg wants to know the value of the following sum: .
Help Greg, print the value of the required sum before each step. | The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the number of vertices in the graph.
Next *n* lines contain *n* integers each — the graph adjacency matrix: the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*a**ij*<=≤<=105,<=*a**ii*<==<=0) represents the weight of the edge that goes from vertex *i* to vertex *j*.
The next line contains *n* distinct integers: *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=*n*) — the vertices that Greg deletes. | Print *n* integers — the *i*-th number equals the required sum before the *i*-th step.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier. | [
"1\n0\n1\n",
"2\n0 5\n4 0\n1 2\n",
"4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3\n"
] | [
"0 ",
"9 0 ",
"17 23 404 0 "
] | none | 1,000 | [
{
"input": "1\n0\n1",
"output": "0 "
},
{
"input": "2\n0 5\n4 0\n1 2",
"output": "9 0 "
},
{
"input": "4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3",
"output": "17 23 404 0 "
},
{
"input": "4\n0 57148 51001 13357\n71125 0 98369 67226\n49388 90852 0 66291\n39573 38165 97007 0\n2 3 1 4",
"output": "723897 306638 52930 0 "
},
{
"input": "5\n0 27799 15529 16434 44291\n47134 0 90227 26873 52252\n41605 21269 0 9135 55784\n70744 17563 79061 0 73981\n70529 35681 91073 52031 0\n5 2 3 1 4",
"output": "896203 429762 232508 87178 0 "
},
{
"input": "6\n0 72137 71041 29217 96749 46417\n40199 0 55907 57677 68590 78796\n83463 50721 0 30963 31779 28646\n94529 47831 98222 0 61665 73941\n24397 66286 2971 81613 0 52501\n26285 3381 51438 45360 20160 0\n6 3 2 4 5 1",
"output": "1321441 1030477 698557 345837 121146 0 "
},
{
"input": "7\n0 34385 31901 51111 10191 14089 95685\n11396 0 8701 33277 1481 517 46253\n51313 2255 0 5948 66085 37201 65310\n21105 60985 10748 0 89271 42883 77345\n34686 29401 73565 47795 0 13793 66997\n70279 49576 62900 40002 70943 0 89601\n65045 1681 28239 12023 40414 89585 0\n3 5 7 6 1 2 4",
"output": "1108867 1016339 729930 407114 206764 94262 0 "
},
{
"input": "8\n0 74961 47889 4733 72876 21399 63105 48239\n15623 0 9680 89133 57989 63401 26001 29608\n42369 82390 0 32866 46171 11871 67489 54070\n23425 80027 18270 0 28105 42657 40876 29267\n78793 18701 7655 94798 0 88885 71424 86914\n44835 76636 11553 46031 13617 0 16971 51915\n33037 53719 43116 52806 56897 71241 0 11629\n2119 62373 93265 69513 5770 90751 36619 0\n3 7 6 5 8 1 2 4",
"output": "1450303 1188349 900316 531281 383344 219125 169160 0 "
},
{
"input": "9\n0 85236 27579 82251 69479 24737 87917 15149 52311\n59640 0 74687 34711 3685 30121 4961 7552 83399\n33376 68733 0 81357 18042 74297 15466 29476 5865\n7493 5601 3321 0 20263 55901 45756 55361 87633\n26751 17161 76681 40376 0 39745 50717 56887 90055\n18885 76353 47089 43601 21561 0 60571 33551 53753\n74595 877 71853 93156 97499 70876 0 22713 63961\n67725 25309 56358 92376 40641 35433 39781 0 97482\n81818 12561 85961 81445 3941 76799 31701 43725 0\n6 2 9 3 5 7 1 4 8",
"output": "2106523 1533575 1645151 1255230 946667 618567 287636 147737 0 "
}
] | 1,670,860,333 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 3,000 | 1,228,800 | #Lee la entrada
A = []
n = int(input())
for i in range(n):
arcos = []
arcos = [int(item) for item in input().split()]
A.append(arcos)
delnode = [int(item) for item in input().split()]
#print(A)
sumas = []
for w in range(n-1,-1,-1):
#Usamos Floyd-Warshall
j = delnode[w] -1
for i in range(len(A)):
for k in range(len(A)):
#print(A[i])
if A[i][j] + A[j][k] <A[i][k]:
A[i][k] = A[i][j] + A[j][k]
#Calcula e imprime la suma de los caminos minimos
sumas.append(sum(sum(A[delnode[i]-1][delnode[j]-1] for j in range(w,len(A)))for i in range(w,len(A))))
for i in range(len(sumas)-1,-1,-1):
print(sumas[i])
| Title: Greg and Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg has a weighed directed graph, consisting of *n* vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:
- The game consists of *n* steps. - On the *i*-th step Greg removes vertex number *x**i* from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. - Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that *d*(*i*,<=*v*,<=*u*) is the shortest path between vertices *v* and *u* in the graph that formed before deleting vertex *x**i*, then Greg wants to know the value of the following sum: .
Help Greg, print the value of the required sum before each step.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the number of vertices in the graph.
Next *n* lines contain *n* integers each — the graph adjacency matrix: the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*a**ij*<=≤<=105,<=*a**ii*<==<=0) represents the weight of the edge that goes from vertex *i* to vertex *j*.
The next line contains *n* distinct integers: *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=*n*) — the vertices that Greg deletes.
Output Specification:
Print *n* integers — the *i*-th number equals the required sum before the *i*-th step.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Demo Input:
['1\n0\n1\n', '2\n0 5\n4 0\n1 2\n', '4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3\n']
Demo Output:
['0 ', '9 0 ', '17 23 404 0 ']
Note:
none | ```python
#Lee la entrada
A = []
n = int(input())
for i in range(n):
arcos = []
arcos = [int(item) for item in input().split()]
A.append(arcos)
delnode = [int(item) for item in input().split()]
#print(A)
sumas = []
for w in range(n-1,-1,-1):
#Usamos Floyd-Warshall
j = delnode[w] -1
for i in range(len(A)):
for k in range(len(A)):
#print(A[i])
if A[i][j] + A[j][k] <A[i][k]:
A[i][k] = A[i][j] + A[j][k]
#Calcula e imprime la suma de los caminos minimos
sumas.append(sum(sum(A[delnode[i]-1][delnode[j]-1] for j in range(w,len(A)))for i in range(w,len(A))))
for i in range(len(sumas)-1,-1,-1):
print(sumas[i])
``` | 0 |
|
408 | A | Line to Cashier | PROGRAMMING | 900 | [
"implementation"
] | null | null | Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.
There are *n* cashiers at the exit from the supermarket. At the moment the queue for the *i*-th cashier already has *k**i* people. The *j*-th person standing in the queue to the *i*-th cashier has *m**i*,<=*j* items in the basket. Vasya knows that:
- the cashier needs 5 seconds to scan one item; - after the cashier scans each item of some customer, he needs 15 seconds to take the customer's money and give him the change.
Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of cashes in the shop. The second line contains *n* space-separated integers: *k*1,<=*k*2,<=...,<=*k**n* (1<=≤<=*k**i*<=≤<=100), where *k**i* is the number of people in the queue to the *i*-th cashier.
The *i*-th of the next *n* lines contains *k**i* space-separated integers: *m**i*,<=1,<=*m**i*,<=2,<=...,<=*m**i*,<=*k**i* (1<=≤<=*m**i*,<=*j*<=≤<=100) — the number of products the *j*-th person in the queue for the *i*-th cash has. | Print a single integer — the minimum number of seconds Vasya needs to get to the cashier. | [
"1\n1\n1\n",
"4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8\n"
] | [
"20\n",
"100\n"
] | In the second test sample, if Vasya goes to the first queue, he gets to the cashier in 100·5 + 15 = 515 seconds. But if he chooses the second queue, he will need 1·5 + 2·5 + 2·5 + 3·5 + 4·15 = 100 seconds. He will need 1·5 + 9·5 + 1·5 + 3·15 = 100 seconds for the third one and 7·5 + 8·5 + 2·15 = 105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue. | 500 | [
{
"input": "1\n1\n1",
"output": "20"
},
{
"input": "4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8",
"output": "100"
},
{
"input": "4\n5 4 5 5\n3 1 3 1 2\n3 1 1 3\n1 1 1 2 2\n2 2 1 1 3",
"output": "100"
},
{
"input": "5\n5 3 6 6 4\n7 5 3 3 9\n6 8 2\n1 10 8 5 9 2\n9 7 8 5 9 10\n9 8 3 3",
"output": "125"
},
{
"input": "5\n10 10 10 10 10\n6 7 8 6 8 5 9 8 10 5\n9 6 9 8 7 8 8 10 8 5\n8 7 7 8 7 5 6 8 9 5\n6 5 10 5 5 10 7 8 5 5\n10 9 8 7 6 9 7 9 6 5",
"output": "480"
},
{
"input": "10\n9 10 10 10 9 5 9 7 8 7\n11 6 10 4 4 15 7 15 5\n3 9 11 12 11 1 13 13 1 5\n6 15 9 12 3 2 8 12 11 10\n7 1 1 6 10 2 6 1 14 2\n8 14 2 3 6 1 14 1 12\n6 10 9 3 5\n13 12 12 7 13 4 4 8 10\n5 6 4 3 14 9 13\n8 12 1 5 7 4 13 1\n1 9 5 3 5 1 4",
"output": "240"
},
{
"input": "10\n5 5 5 5 5 5 5 5 5 5\n5 5 4 5 4\n6 5 7 7 6\n5 4 4 5 5\n4 4 5 5 5\n7 6 4 5 7\n4 6 5 4 5\n6 6 7 6 6\n4 5 4 4 7\n7 5 4 4 5\n6 6 7 4 4",
"output": "190"
},
{
"input": "1\n1\n100",
"output": "515"
},
{
"input": "1\n90\n90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90",
"output": "41850"
}
] | 1,667,661,790 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 46 | 0 | n = int(input())
input()
times = []
for i in range(n):
x = [int(i) for i in input().split()]
times.append((sum(x)*5) + len(x)*15)
print(min(times))
| Title: Line to Cashier
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.
There are *n* cashiers at the exit from the supermarket. At the moment the queue for the *i*-th cashier already has *k**i* people. The *j*-th person standing in the queue to the *i*-th cashier has *m**i*,<=*j* items in the basket. Vasya knows that:
- the cashier needs 5 seconds to scan one item; - after the cashier scans each item of some customer, he needs 15 seconds to take the customer's money and give him the change.
Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of cashes in the shop. The second line contains *n* space-separated integers: *k*1,<=*k*2,<=...,<=*k**n* (1<=≤<=*k**i*<=≤<=100), where *k**i* is the number of people in the queue to the *i*-th cashier.
The *i*-th of the next *n* lines contains *k**i* space-separated integers: *m**i*,<=1,<=*m**i*,<=2,<=...,<=*m**i*,<=*k**i* (1<=≤<=*m**i*,<=*j*<=≤<=100) — the number of products the *j*-th person in the queue for the *i*-th cash has.
Output Specification:
Print a single integer — the minimum number of seconds Vasya needs to get to the cashier.
Demo Input:
['1\n1\n1\n', '4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8\n']
Demo Output:
['20\n', '100\n']
Note:
In the second test sample, if Vasya goes to the first queue, he gets to the cashier in 100·5 + 15 = 515 seconds. But if he chooses the second queue, he will need 1·5 + 2·5 + 2·5 + 3·5 + 4·15 = 100 seconds. He will need 1·5 + 9·5 + 1·5 + 3·15 = 100 seconds for the third one and 7·5 + 8·5 + 2·15 = 105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue. | ```python
n = int(input())
input()
times = []
for i in range(n):
x = [int(i) for i in input().split()]
times.append((sum(x)*5) + len(x)*15)
print(min(times))
``` | 3 |
|
818 | A | Diplomas and Certificates | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners. | The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas. | Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners. | [
"18 2\n",
"9 10\n",
"1000000000000 5\n",
"1000000000000 499999999999\n"
] | [
"3 6 9\n",
"0 0 9\n",
"83333333333 416666666665 500000000002\n",
"1 499999999999 500000000000\n"
] | none | 0 | [
{
"input": "18 2",
"output": "3 6 9"
},
{
"input": "9 10",
"output": "0 0 9"
},
{
"input": "1000000000000 5",
"output": "83333333333 416666666665 500000000002"
},
{
"input": "1000000000000 499999999999",
"output": "1 499999999999 500000000000"
},
{
"input": "1 1",
"output": "0 0 1"
},
{
"input": "5 3",
"output": "0 0 5"
},
{
"input": "42 6",
"output": "3 18 21"
},
{
"input": "1000000000000 1000",
"output": "499500499 499500499000 500000000501"
},
{
"input": "999999999999 999999",
"output": "499999 499998500001 500000999999"
},
{
"input": "732577309725 132613",
"output": "2762066 366285858458 366288689201"
},
{
"input": "152326362626 15",
"output": "4760198832 71402982480 76163181314"
},
{
"input": "2 1",
"output": "0 0 2"
},
{
"input": "1000000000000 500000000000",
"output": "0 0 1000000000000"
},
{
"input": "100000000000 50000000011",
"output": "0 0 100000000000"
},
{
"input": "1000000000000 32416187567",
"output": "15 486242813505 513757186480"
},
{
"input": "1000000000000 7777777777",
"output": "64 497777777728 502222222208"
},
{
"input": "1000000000000 77777777777",
"output": "6 466666666662 533333333332"
},
{
"input": "100000000000 578485652",
"output": "86 49749766072 50250233842"
},
{
"input": "999999999999 10000000000",
"output": "49 490000000000 509999999950"
},
{
"input": "7 2",
"output": "1 2 4"
},
{
"input": "420506530901 752346673804",
"output": "0 0 420506530901"
},
{
"input": "960375521135 321688347872",
"output": "1 321688347872 638687173262"
},
{
"input": "1000000000000 1000000000000",
"output": "0 0 1000000000000"
},
{
"input": "99999999999 15253636363",
"output": "3 45760909089 54239090907"
},
{
"input": "19 2",
"output": "3 6 10"
},
{
"input": "999999999999 1000000000000",
"output": "0 0 999999999999"
},
{
"input": "1000000000000 5915587276",
"output": "84 496909331184 503090668732"
},
{
"input": "1000000000000 1000000006",
"output": "499 499000002994 500999996507"
},
{
"input": "549755813888 134217728",
"output": "2047 274743689216 275012122625"
},
{
"input": "99999999999 3333333",
"output": "14999 49996661667 50003323333"
},
{
"input": "9 1",
"output": "2 2 5"
},
{
"input": "1000000000000 250000000001",
"output": "1 250000000001 749999999998"
},
{
"input": "5 1",
"output": "1 1 3"
},
{
"input": "3107038133 596040207",
"output": "2 1192080414 1914957717"
},
{
"input": "1000000000000 73786977",
"output": "6776 499980556152 500019437072"
},
{
"input": "1000000000000 73786976",
"output": "6776 499980549376 500019443848"
},
{
"input": "1000000000000 25000000000",
"output": "19 475000000000 524999999981"
},
{
"input": "216929598879 768233755932",
"output": "0 0 216929598879"
},
{
"input": "1000000000000 250000000000",
"output": "1 250000000000 749999999999"
},
{
"input": "1000000000000 100000000001",
"output": "4 400000000004 599999999992"
},
{
"input": "100000000000 100000000001",
"output": "0 0 100000000000"
},
{
"input": "900000000000 100281800001",
"output": "4 401127200004 498872799992"
},
{
"input": "906028900004 109123020071",
"output": "4 436492080284 469536819716"
},
{
"input": "1000000000000 1",
"output": "250000000000 250000000000 500000000000"
}
] | 1,554,141,229 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 93 | 0 | n,k=map(int,input().split())
f=int(n/2)
f1=int(f/(k+1))
print(f1,f1*2,n-f1*3) | Title: Diplomas and Certificates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners.
Input Specification:
The first (and the only) line of input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas.
Output Specification:
Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners.
Demo Input:
['18 2\n', '9 10\n', '1000000000000 5\n', '1000000000000 499999999999\n']
Demo Output:
['3 6 9\n', '0 0 9\n', '83333333333 416666666665 500000000002\n', '1 499999999999 500000000000\n']
Note:
none | ```python
n,k=map(int,input().split())
f=int(n/2)
f1=int(f/(k+1))
print(f1,f1*2,n-f1*3)
``` | 0 |
|
334 | B | Eight Point Sets | PROGRAMMING | 1,400 | [
"sortings"
] | null | null | Gerald is very particular to eight point sets. He thinks that any decent eight point set must consist of all pairwise intersections of three distinct integer vertical straight lines and three distinct integer horizontal straight lines, except for the average of these nine points. In other words, there must be three integers *x*1,<=*x*2,<=*x*3 and three more integers *y*1,<=*y*2,<=*y*3, such that *x*1<=<<=*x*2<=<<=*x*3, *y*1<=<<=*y*2<=<<=*y*3 and the eight point set consists of all points (*x**i*,<=*y**j*) (1<=≤<=*i*,<=*j*<=≤<=3), except for point (*x*2,<=*y*2).
You have a set of eight points. Find out if Gerald can use this set? | The input consists of eight lines, the *i*-th line contains two space-separated integers *x**i* and *y**i* (0<=≤<=*x**i*,<=*y**i*<=≤<=106). You do not have any other conditions for these points. | In a single line print word "respectable", if the given set of points corresponds to Gerald's decency rules, and "ugly" otherwise. | [
"0 0\n0 1\n0 2\n1 0\n1 2\n2 0\n2 1\n2 2\n",
"0 0\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0\n",
"1 1\n1 2\n1 3\n2 1\n2 2\n2 3\n3 1\n3 2\n"
] | [
"respectable\n",
"ugly\n",
"ugly\n"
] | none | 1,000 | [
{
"input": "0 0\n0 1\n0 2\n1 0\n1 2\n2 0\n2 1\n2 2",
"output": "respectable"
},
{
"input": "0 0\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0",
"output": "ugly"
},
{
"input": "1 1\n1 2\n1 3\n2 1\n2 2\n2 3\n3 1\n3 2",
"output": "ugly"
},
{
"input": "0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "ugly"
},
{
"input": "1000000 1000000\n1000000 999999\n1000000 999998\n999999 1000000\n999999 999998\n999998 1000000\n999998 999999\n999998 999998",
"output": "respectable"
},
{
"input": "0 0\n1 0\n0 1\n1 1\n0 2\n1 2\n0 3\n1 3",
"output": "ugly"
},
{
"input": "0 0\n2 1\n1 0\n0 2\n2 2\n1 0\n2 1\n0 2",
"output": "ugly"
},
{
"input": "0 0\n2 1\n1 0\n0 2\n2 2\n1 0\n2 1\n0 2",
"output": "ugly"
},
{
"input": "791649 383826\n10864 260573\n504506 185571\n899991 511500\n503197 876976\n688727 569035\n343255 961333\n439355 759581",
"output": "ugly"
},
{
"input": "750592 335292\n226387 434036\n299976 154633\n593197 600998\n62014 689355\n566268 571630\n381455 222817\n50555 288617",
"output": "ugly"
},
{
"input": "716334 42808\n211710 645370\n515258 96837\n14392 766713\n439265 939607\n430602 918570\n845044 187545\n957977 441674",
"output": "ugly"
},
{
"input": "337873 813442\n995185 863182\n375545 263618\n310042 130019\n358572 560779\n305725 729179\n377381 267545\n41376 312626",
"output": "ugly"
},
{
"input": "803784 428886\n995691 328351\n211844 386054\n375491 74073\n692402 660275\n366073 536431\n485832 941417\n96032 356022",
"output": "ugly"
},
{
"input": "999231 584954\n246553 267441\n697080 920011\n173593 403511\n58535 101909\n131124 924182\n779830 204560\n684576 533111",
"output": "ugly"
},
{
"input": "666888 741208\n685852 578759\n211123 826453\n244759 601804\n670436 748132\n976425 387060\n587850 804554\n430242 805528",
"output": "ugly"
},
{
"input": "71768 834717\n13140 834717\n13140 991083\n880763 386898\n71768 386898\n880763 991083\n880763 834717\n13140 386898",
"output": "ugly"
},
{
"input": "941532 913025\n941532 862399\n686271 913025\n686271 862399\n686271 461004\n941532 461004\n908398 862399\n908398 913025",
"output": "ugly"
},
{
"input": "251515 680236\n761697 669947\n251515 669947\n761697 680236\n251515 476629\n761697 476629\n453296 669947\n453296 476629",
"output": "ugly"
},
{
"input": "612573 554036\n195039 655769\n472305 655769\n612573 655769\n195039 160740\n472305 160740\n472305 554036\n612573 160740",
"output": "ugly"
},
{
"input": "343395 788566\n171702 674699\n171702 788566\n971214 788566\n343395 9278\n971214 9278\n343395 674699\n971214 674699",
"output": "ugly"
},
{
"input": "38184 589856\n281207 447136\n281207 42438\n38184 42438\n38184 447136\n880488 589856\n281207 589856\n880488 42438",
"output": "ugly"
},
{
"input": "337499 89260\n337499 565883\n603778 89260\n603778 565883\n234246 89260\n603778 17841\n337499 17841\n234246 17841",
"output": "ugly"
},
{
"input": "180952 311537\n180952 918548\n126568 918548\n180952 268810\n732313 918548\n126568 311537\n126568 268810\n732313 311537",
"output": "ugly"
},
{
"input": "323728 724794\n265581 165113\n323728 146453\n265581 146453\n591097 146453\n265581 724794\n323728 165113\n591097 165113",
"output": "ugly"
},
{
"input": "642921 597358\n922979 597358\n127181 616833\n642921 828316\n922979 828316\n127181 597358\n922979 616833\n127181 828316",
"output": "respectable"
},
{
"input": "69586 260253\n74916 203798\n985457 203798\n74916 943932\n985457 943932\n69586 943932\n985457 260253\n69586 203798",
"output": "respectable"
},
{
"input": "57930 637387\n883991 573\n57930 573\n57930 499963\n399327 573\n399327 637387\n883991 637387\n883991 499963",
"output": "respectable"
},
{
"input": "52820 216139\n52820 999248\n290345 216139\n290345 999248\n308639 216139\n308639 999248\n52820 477113\n308639 477113",
"output": "respectable"
},
{
"input": "581646 464672\n493402 649074\n581646 649074\n214619 649074\n581646 252709\n214619 252709\n214619 464672\n493402 252709",
"output": "respectable"
},
{
"input": "787948 77797\n421941 615742\n421941 77797\n400523 77797\n400523 111679\n787948 615742\n400523 615742\n787948 111679",
"output": "respectable"
},
{
"input": "583956 366985\n759621 567609\n756846 567609\n759621 176020\n583956 567609\n583956 176020\n759621 366985\n756846 176020",
"output": "respectable"
},
{
"input": "0 50000\n0 0\n0 1000000\n50000 0\n50000 1000000\n1000000 0\n1000000 50000\n1000000 1000000",
"output": "respectable"
},
{
"input": "0 8\n0 9\n0 10\n1 8\n3 8\n3 8\n3 9\n3 10",
"output": "ugly"
},
{
"input": "0 1\n0 1\n0 2\n1 1\n1 2\n2 1\n2 1\n2 2",
"output": "ugly"
},
{
"input": "1 2\n1 3\n1 4\n2 2\n2 4\n4 2\n4 2\n4 4",
"output": "ugly"
},
{
"input": "0 0\n0 1\n0 2\n0 0\n1 2\n2 0\n2 1\n2 2",
"output": "ugly"
},
{
"input": "0 0\n0 0\n0 0\n1 1\n1 1\n2 2\n2 2\n2 2",
"output": "ugly"
},
{
"input": "0 0\n0 0\n0 2\n1 1\n1 2\n2 0\n2 1\n2 2",
"output": "ugly"
},
{
"input": "0 0\n0 1\n0 3\n1 0\n1 3\n2 0\n2 2\n2 3",
"output": "ugly"
},
{
"input": "0 0\n0 1\n0 2\n1 0\n1 2\n3 0\n3 1\n3 2",
"output": "respectable"
},
{
"input": "1 1\n1 2\n1 5\n2 1\n2 5\n5 1\n5 2\n5 5",
"output": "respectable"
},
{
"input": "1 1\n1 2\n1 2\n2 3\n2 1\n3 3\n3 1\n3 3",
"output": "ugly"
},
{
"input": "0 0\n0 0\n1 0\n0 1\n2 1\n1 2\n2 2\n2 2",
"output": "ugly"
},
{
"input": "1 1\n1 1\n1 3\n2 1\n2 3\n3 2\n3 2\n3 3",
"output": "ugly"
},
{
"input": "1 0\n1 0\n1 0\n2 3\n2 3\n3 4\n3 4\n3 4",
"output": "ugly"
}
] | 1,457,188,910 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
#include <set>
int main()
{
std::set<int> set1, set2;
for(int i = 0; i < 8; i++)
{
int x, y;
std::cin >> x >> y;
set1.insert(x);
set2.insert(y);
}
if(set1.size() == 3 && set2.size() == 3) std::cout << "respectable" << std::endl; else std::cout << "ugly" << std::endl;
return 0;
} | Title: Eight Point Sets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gerald is very particular to eight point sets. He thinks that any decent eight point set must consist of all pairwise intersections of three distinct integer vertical straight lines and three distinct integer horizontal straight lines, except for the average of these nine points. In other words, there must be three integers *x*1,<=*x*2,<=*x*3 and three more integers *y*1,<=*y*2,<=*y*3, such that *x*1<=<<=*x*2<=<<=*x*3, *y*1<=<<=*y*2<=<<=*y*3 and the eight point set consists of all points (*x**i*,<=*y**j*) (1<=≤<=*i*,<=*j*<=≤<=3), except for point (*x*2,<=*y*2).
You have a set of eight points. Find out if Gerald can use this set?
Input Specification:
The input consists of eight lines, the *i*-th line contains two space-separated integers *x**i* and *y**i* (0<=≤<=*x**i*,<=*y**i*<=≤<=106). You do not have any other conditions for these points.
Output Specification:
In a single line print word "respectable", if the given set of points corresponds to Gerald's decency rules, and "ugly" otherwise.
Demo Input:
['0 0\n0 1\n0 2\n1 0\n1 2\n2 0\n2 1\n2 2\n', '0 0\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0\n', '1 1\n1 2\n1 3\n2 1\n2 2\n2 3\n3 1\n3 2\n']
Demo Output:
['respectable\n', 'ugly\n', 'ugly\n']
Note:
none | ```python
#include <iostream>
#include <set>
int main()
{
std::set<int> set1, set2;
for(int i = 0; i < 8; i++)
{
int x, y;
std::cin >> x >> y;
set1.insert(x);
set2.insert(y);
}
if(set1.size() == 3 && set2.size() == 3) std::cout << "respectable" << std::endl; else std::cout << "ugly" << std::endl;
return 0;
}
``` | -1 |
|
776 | B | Sherlock and his girlfriend | PROGRAMMING | 1,200 | [
"constructive algorithms",
"number theory"
] | null | null | Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.
He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1.
Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.
Help Sherlock complete this trivial task. | The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces. | The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.
The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price.
If there are multiple ways to color the pieces using *k* colors, you can output any of them. | [
"3\n",
"4\n"
] | [
"2\n1 1 2 ",
"2\n2 1 1 2\n"
] | In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct. | 1,000 | [
{
"input": "3",
"output": "2\n1 1 2 "
},
{
"input": "4",
"output": "2\n1 1 2 1 "
},
{
"input": "17",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 "
},
{
"input": "25",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 "
},
{
"input": "85",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 "
},
{
"input": "105",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 "
},
{
"input": "123",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "452",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "641",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "293",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "733",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "10",
"output": "2\n1 1 2 1 2 1 2 2 2 1 "
},
{
"input": "287",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "3202",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "728",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "3509",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "5137",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "2023",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "4890",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "8507",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "1796",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "3466",
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"input": "1098",
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"input": "11226",
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"input": "11731",
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"input": "11644",
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{
"input": "14553",
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{
"input": "17307",
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{
"input": "23189",
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"input": "6818",
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"input": "1054",
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"input": "28163",
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"input": "30885",
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"input": "27673",
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{
"input": "11656",
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{
"input": "36325",
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{
"input": "31205",
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{
"input": "29958",
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},
{
"input": "1696",
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},
{
"input": "44907",
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},
{
"input": "13736",
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},
{
"input": "29594",
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},
{
"input": "19283",
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},
{
"input": "15346",
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},
{
"input": "41794",
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},
{
"input": "99998",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "100000",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "2",
"output": "1\n1 1 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "2",
"output": "1\n1 1 "
},
{
"input": "100000",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "99971",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
}
] | 1,644,469,331 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 233 | 307,200 | n = int(input())
sieve = [False]*(n+2)
two = False
for i in range(2,n+2):
if not sieve[i]:
for j in range(2,(n+1)//i + 1): # To hit one below
sieve[i*j] = True
two = True
# paint prime numbers with different and others with different colors
if two:
print(2)
else:
print(1)
for i in range(2, n+2):
print((sieve[i] == True)*2 + (sieve[i] == False)*1, end=' ') | Title: Sherlock and his girlfriend
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.
He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1.
Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.
Help Sherlock complete this trivial task.
Input Specification:
The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces.
Output Specification:
The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.
The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price.
If there are multiple ways to color the pieces using *k* colors, you can output any of them.
Demo Input:
['3\n', '4\n']
Demo Output:
['2\n1 1 2 ', '2\n2 1 1 2\n']
Note:
In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct. | ```python
n = int(input())
sieve = [False]*(n+2)
two = False
for i in range(2,n+2):
if not sieve[i]:
for j in range(2,(n+1)//i + 1): # To hit one below
sieve[i*j] = True
two = True
# paint prime numbers with different and others with different colors
if two:
print(2)
else:
print(1)
for i in range(2, n+2):
print((sieve[i] == True)*2 + (sieve[i] == False)*1, end=' ')
``` | 3 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | A positive integer is called a 2-3-integer, if it is equal to 2*x*·3*y* for some non-negative integers *x* and *y*. In other words, these integers are such integers that only have 2 and 3 among their prime divisors. For example, integers 1, 6, 9, 16 and 108 — are 2-3 integers, while 5, 10, 21 and 120 are not.
Print the number of 2-3-integers on the given segment [*l*,<=*r*], i. e. the number of sich 2-3-integers *t* that *l*<=≤<=*t*<=≤<=*r*. | The only line contains two integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=2·109). | Print a single integer the number of 2-3-integers on the segment [*l*,<=*r*]. | [
"1 10\n",
"100 200\n",
"1 2000000000\n"
] | [
"7\n",
"5\n",
"326\n"
] | In the first example the 2-3-integers are 1, 2, 3, 4, 6, 8 and 9.
In the second example the 2-3-integers are 108, 128, 144, 162 and 192. | 0 | [
{
"input": "1 10",
"output": "7"
},
{
"input": "100 200",
"output": "5"
},
{
"input": "1 2000000000",
"output": "326"
},
{
"input": "1088391168 1934917632",
"output": "17"
},
{
"input": "1088391167 1934917632",
"output": "17"
},
{
"input": "1088391169 1934917632",
"output": "16"
},
{
"input": "1088391168 1934917631",
"output": "16"
},
{
"input": "1088391168 1934917633",
"output": "17"
},
{
"input": "4 134217728",
"output": "250"
},
{
"input": "209952 43046722",
"output": "112"
},
{
"input": "25165825 43046719",
"output": "13"
},
{
"input": "5183 25165825",
"output": "153"
},
{
"input": "388645 455910",
"output": "3"
},
{
"input": "472069 972050",
"output": "14"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "12 1999999931",
"output": "319"
},
{
"input": "1999999999 1999999999",
"output": "0"
},
{
"input": "2000000000 2000000000",
"output": "0"
},
{
"input": "1934917632 1934917632",
"output": "1"
},
{
"input": "1836660096 1836660096",
"output": "1"
},
{
"input": "1811939328 1811939328",
"output": "1"
},
{
"input": "1719926784 1719926784",
"output": "1"
},
{
"input": "1632586752 1632586752",
"output": "1"
},
{
"input": "1610612736 1610612736",
"output": "1"
},
{
"input": "1207959552 1207959552",
"output": "1"
},
{
"input": "129140163 129140163",
"output": "1"
},
{
"input": "12345 54321",
"output": "21"
},
{
"input": "1889569 25165826",
"output": "56"
},
{
"input": "73 7077888",
"output": "165"
},
{
"input": "7077888 45349631",
"output": "43"
},
{
"input": "6144 7077886",
"output": "122"
},
{
"input": "3779135 4194302",
"output": "2"
},
{
"input": "214 161243134",
"output": "232"
},
{
"input": "214 18874369",
"output": "180"
},
{
"input": "28 863",
"output": "26"
},
{
"input": "1417174 172186886",
"output": "112"
},
{
"input": "27 102036671",
"output": "234"
},
{
"input": "147458 102036672",
"output": "138"
},
{
"input": "33554431 102036671",
"output": "28"
},
{
"input": "1572863 33554433",
"output": "68"
},
{
"input": "33554434 181398528",
"output": "43"
},
{
"input": "373247 17915905",
"output": "81"
},
{
"input": "4251526 68024450",
"output": "65"
},
{
"input": "553599 555493",
"output": "0"
},
{
"input": "69739 671621",
"output": "40"
},
{
"input": "618583 755045",
"output": "4"
},
{
"input": "838469 943236",
"output": "2"
},
{
"input": "509607936 1836660096",
"output": "38"
},
{
"input": "509607935 1836660096",
"output": "38"
},
{
"input": "509607937 1836660096",
"output": "37"
},
{
"input": "509607936 1836660095",
"output": "37"
},
{
"input": "509607936 1836660097",
"output": "38"
}
] | 1,521,303,638 | 2,738 | Python 3 | WRONG_ANSWER | TESTS | 26 | 93 | 7,065,600 | from math import log
def solve(l, r):
s = set()
for i in range(int(log(r, 2) + 1)):
for j in range(int(log(r, 3) + 1)):
x = 2**i * 3**j
if l <= x <= r:
s.add(x)
return len(s)
def test():
assert solve(1, 10) == 7
assert solve(100, 200) == 5
assert solve(1, 2000000000) == 326
if __name__ == '__main__':
l, r = [int(x) for x in input().split()]
print(solve(l, r))
# test()
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A positive integer is called a 2-3-integer, if it is equal to 2*x*·3*y* for some non-negative integers *x* and *y*. In other words, these integers are such integers that only have 2 and 3 among their prime divisors. For example, integers 1, 6, 9, 16 and 108 — are 2-3 integers, while 5, 10, 21 and 120 are not.
Print the number of 2-3-integers on the given segment [*l*,<=*r*], i. e. the number of sich 2-3-integers *t* that *l*<=≤<=*t*<=≤<=*r*.
Input Specification:
The only line contains two integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=2·109).
Output Specification:
Print a single integer the number of 2-3-integers on the segment [*l*,<=*r*].
Demo Input:
['1 10\n', '100 200\n', '1 2000000000\n']
Demo Output:
['7\n', '5\n', '326\n']
Note:
In the first example the 2-3-integers are 1, 2, 3, 4, 6, 8 and 9.
In the second example the 2-3-integers are 108, 128, 144, 162 and 192. | ```python
from math import log
def solve(l, r):
s = set()
for i in range(int(log(r, 2) + 1)):
for j in range(int(log(r, 3) + 1)):
x = 2**i * 3**j
if l <= x <= r:
s.add(x)
return len(s)
def test():
assert solve(1, 10) == 7
assert solve(100, 200) == 5
assert solve(1, 2000000000) == 326
if __name__ == '__main__':
l, r = [int(x) for x in input().split()]
print(solve(l, r))
# test()
``` | 0 |
|
716 | A | Crazy Computer | PROGRAMMING | 800 | [
"implementation"
] | null | null | ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything. | The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word. | Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*. | [
"6 5\n1 3 8 14 19 20\n",
"6 1\n1 3 5 7 9 10\n"
] | [
"3",
"2"
] | The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1. | 500 | [
{
"input": "6 5\n1 3 8 14 19 20",
"output": "3"
},
{
"input": "6 1\n1 3 5 7 9 10",
"output": "2"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "5 5\n1 7 12 13 14",
"output": "4"
},
{
"input": "2 1000000000\n1 1000000000",
"output": "2"
},
{
"input": "3 5\n1 10 20",
"output": "1"
},
{
"input": "3 10\n1 2 3",
"output": "3"
},
{
"input": "2 1\n1 100",
"output": "1"
},
{
"input": "3 1\n1 2 10",
"output": "1"
},
{
"input": "2 1\n1 2",
"output": "2"
}
] | 1,629,310,530 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 124 | 15,155,200 | #loser707
n,c=map(int,input().split())
a=list(map(int,input().split()))
l=[]
ans=0
for x,y in zip(a,a[1:]):
l.append(y-x)
for i in l:
if i<=c:
ans=ans+1
else:
ans=0
print(ans+1) | Title: Crazy Computer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.
Input Specification:
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word.
Output Specification:
Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*.
Demo Input:
['6 5\n1 3 8 14 19 20\n', '6 1\n1 3 5 7 9 10\n']
Demo Output:
['3', '2']
Note:
The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1. | ```python
#loser707
n,c=map(int,input().split())
a=list(map(int,input().split()))
l=[]
ans=0
for x,y in zip(a,a[1:]):
l.append(y-x)
for i in l:
if i<=c:
ans=ans+1
else:
ans=0
print(ans+1)
``` | 3 |
|
265 | A | Colorful Stones (Simplified Edition) | PROGRAMMING | 800 | [
"implementation"
] | null | null | There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence. | The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence. | Print the final 1-based position of Liss in a single line. | [
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] | [
"2\n",
"3\n",
"15\n"
] | none | 500 | [
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
{
"input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB",
"output": "1"
},
{
"input": "RRGGBRGRBG\nBRRGGBBGGR",
"output": "7"
},
{
"input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG",
"output": "15"
},
{
"input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR",
"output": "8"
},
{
"input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR",
"output": "16"
},
{
"input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR",
"output": "18"
},
{
"input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG",
"output": "23"
},
{
"input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR",
"output": "14"
},
{
"input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG",
"output": "12"
},
{
"input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB",
"output": "12"
},
{
"input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR",
"output": "2"
},
{
"input": "R\nB",
"output": "1"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "50"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
},
{
"input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
}
] | 1,618,691,464 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 124 | 0 | s = input()
t = list(input())
move = 1
i = 0
while t:
if t[0] == s[i]:
move += 1
i += 1
t.pop(0)
print(move) | Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none | ```python
s = input()
t = list(input())
move = 1
i = 0
while t:
if t[0] == s[i]:
move += 1
i += 1
t.pop(0)
print(move)
``` | 3 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | This is an interactive problem.
Natasha is going to fly to Mars. Finally, Natasha sat in the rocket. She flies, flies... but gets bored. She wishes to arrive to Mars already! So she decides to find something to occupy herself. She couldn't think of anything better to do than to calculate the distance to the red planet.
Let's define $x$ as the distance to Mars. Unfortunately, Natasha does not know $x$. But it is known that $1 \le x \le m$, where Natasha knows the number $m$. Besides, $x$ and $m$ are positive integers.
Natasha can ask the rocket questions. Every question is an integer $y$ ($1 \le y \le m$). The correct answer to the question is $-1$, if $x<y$, $0$, if $x=y$, and $1$, if $x>y$. But the rocket is broken — it does not always answer correctly. Precisely: let the correct answer to the current question be equal to $t$, then, if the rocket answers this question correctly, then it will answer $t$, otherwise it will answer $-t$.
In addition, the rocket has a sequence $p$ of length $n$. Each element of the sequence is either $0$ or $1$. The rocket processes this sequence in the cyclic order, that is $1$-st element, $2$-nd, $3$-rd, $\ldots$, $(n-1)$-th, $n$-th, $1$-st, $2$-nd, $3$-rd, $\ldots$, $(n-1)$-th, $n$-th, $\ldots$. If the current element is $1$, the rocket answers correctly, if $0$ — lies. Natasha doesn't know the sequence $p$, but she knows its length — $n$.
You can ask the rocket no more than $60$ questions.
Help Natasha find the distance to Mars. Assume, that the distance to Mars does not change while Natasha is asking questions.
Your solution will not be accepted, if it does not receive an answer $0$ from the rocket (even if the distance to Mars is uniquely determined by the already received rocket's answers). | The first line contains two integers $m$ and $n$ ($1 \le m \le 10^9$, $1 \le n \le 30$) — the maximum distance to Mars and the number of elements in the sequence $p$. | none | [
"5 2\n1\n-1\n-1\n1\n0\n"
] | [
"1\n2\n4\n5\n3\n"
] | In the example, hacking would look like this:
5 2 3
1 0
This means that the current distance to Mars is equal to $3$, Natasha knows that it does not exceed $5$, and the rocket answers in order: correctly, incorrectly, correctly, incorrectly ...
Really:
on the first query ($1$) the correct answer is $1$, the rocket answered correctly: $1$;
on the second query ($2$) the correct answer is $1$, the rocket answered incorrectly: $-1$;
on the third query ($4$) the correct answer is $-1$, the rocket answered correctly: $-1$;
on the fourth query ($5$) the correct answer is $-1$, the rocket answered incorrectly: $1$;
on the fifth query ($3$) the correct and incorrect answer is $0$. | 0 | [
{
"input": "5 2 3\n1 0",
"output": "3 queries, x=3"
},
{
"input": "1 1 1\n1",
"output": "1 queries, x=1"
},
{
"input": "3 2 3\n1 0",
"output": "4 queries, x=3"
},
{
"input": "6 3 5\n1 1 1",
"output": "5 queries, x=5"
},
{
"input": "10 4 3\n0 0 1 0",
"output": "6 queries, x=3"
},
{
"input": "30 5 16\n0 1 1 1 0",
"output": "6 queries, x=16"
},
{
"input": "60 6 21\n1 0 0 1 0 1",
"output": "11 queries, x=21"
},
{
"input": "100 7 73\n0 0 0 1 0 1 1",
"output": "14 queries, x=73"
},
{
"input": "1000000000 29 958572235\n1 1 0 1 1 1 1 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0",
"output": "58 queries, x=958572235"
},
{
"input": "738009704 30 116044407\n0 0 1 1 0 0 1 0 0 0 1 0 0 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1",
"output": "59 queries, x=116044407"
},
{
"input": "300 8 165\n1 1 1 0 0 1 1 0",
"output": "16 queries, x=165"
},
{
"input": "600 9 150\n0 0 1 0 1 0 1 0 1",
"output": "19 queries, x=150"
},
{
"input": "1000 10 140\n0 0 0 0 1 0 0 0 0 0",
"output": "20 queries, x=140"
},
{
"input": "3000 11 1896\n1 0 1 1 0 0 0 0 1 1 1",
"output": "21 queries, x=1896"
},
{
"input": "6000 12 4679\n1 0 1 1 1 1 1 0 0 0 0 1",
"output": "23 queries, x=4679"
},
{
"input": "10000 13 4977\n1 0 1 1 0 0 0 1 0 0 1 1 0",
"output": "26 queries, x=4977"
},
{
"input": "30000 14 60\n1 1 1 0 0 1 0 1 0 0 1 0 0 0",
"output": "28 queries, x=60"
},
{
"input": "60000 15 58813\n0 1 1 0 1 1 0 0 0 1 1 1 1 0 1",
"output": "27 queries, x=58813"
},
{
"input": "100000 16 79154\n1 1 1 0 0 0 0 0 1 1 0 1 0 1 0 1",
"output": "32 queries, x=79154"
},
{
"input": "300000 17 11107\n1 0 0 0 1 0 0 0 1 1 1 0 0 1 1 1 0",
"output": "34 queries, x=11107"
},
{
"input": "600000 18 146716\n0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 1",
"output": "37 queries, x=146716"
},
{
"input": "1000000 19 418016\n1 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 0 0",
"output": "38 queries, x=418016"
},
{
"input": "3000000 20 642518\n1 0 0 1 0 1 1 1 1 1 0 0 0 1 0 1 0 1 0 1",
"output": "41 queries, x=642518"
},
{
"input": "6000000 21 3516807\n0 0 0 1 0 1 0 1 1 0 0 0 0 1 0 1 1 1 0 0 0",
"output": "43 queries, x=3516807"
},
{
"input": "10000000 22 8115129\n1 0 1 0 0 0 0 0 0 1 1 0 1 0 0 0 0 1 1 0 0 1",
"output": "42 queries, x=8115129"
},
{
"input": "30000000 23 10362635\n0 1 0 0 1 1 1 1 0 0 0 1 0 1 1 0 1 1 1 1 0 0 0",
"output": "48 queries, x=10362635"
},
{
"input": "60000000 24 52208533\n1 1 1 0 1 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 1 1 0",
"output": "46 queries, x=52208533"
},
{
"input": "100000000 25 51744320\n0 1 1 1 1 0 1 1 1 1 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1",
"output": "50 queries, x=51744320"
},
{
"input": "300000000 26 264009490\n1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 1 0 1 1 1 0 1 1",
"output": "54 queries, x=264009490"
},
{
"input": "600000000 27 415720732\n1 1 1 1 1 1 0 0 1 1 1 0 1 0 1 1 0 0 1 1 1 1 0 1 0 1 0",
"output": "56 queries, x=415720732"
},
{
"input": "1000000000 28 946835863\n0 0 1 0 1 1 1 0 1 0 1 1 0 1 0 1 1 0 0 0 1 0 1 0 1 1 0 0",
"output": "58 queries, x=946835863"
},
{
"input": "1000000000 29 124919287\n0 0 1 0 0 0 1 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 1 1 0 1 0 0",
"output": "59 queries, x=124919287"
},
{
"input": "1000000000 30 202669473\n1 1 0 1 1 1 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 1 1 1 0 0 0 0 0",
"output": "58 queries, x=202669473"
},
{
"input": "1000000000 13 532121080\n1 1 1 0 1 1 0 0 0 0 1 0 1",
"output": "42 queries, x=532121080"
},
{
"input": "1000000000 27 105669924\n0 1 1 1 0 1 0 1 0 0 0 1 0 0 1 1 0 1 1 0 0 1 0 1 1 1 1",
"output": "57 queries, x=105669924"
},
{
"input": "1000000000 11 533994576\n0 0 1 0 1 1 1 1 0 1 0",
"output": "38 queries, x=533994576"
},
{
"input": "1000000000 9 107543421\n1 0 0 1 1 1 1 1 1",
"output": "39 queries, x=107543421"
},
{
"input": "1000000000 23 976059561\n1 0 0 0 0 1 0 0 1 1 1 1 1 1 1 0 0 0 0 1 0 0 1",
"output": "53 queries, x=976059561"
},
{
"input": "1000000000 7 549608406\n1 1 1 0 1 1 1",
"output": "36 queries, x=549608406"
},
{
"input": "1000000000 21 123157250\n0 1 1 1 1 1 1 0 0 0 1 1 1 1 0 0 0 1 0 0 1",
"output": "49 queries, x=123157250"
},
{
"input": "1000000000 19 696706094\n0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0",
"output": "47 queries, x=696706094"
},
{
"input": "1000000000 3 125030747\n0 0 0",
"output": "33 queries, x=125030747"
},
{
"input": "1000000000 17 993546887\n1 0 1 0 1 1 1 1 1 1 0 1 0 1 1 1 1",
"output": "46 queries, x=993546887"
},
{
"input": "1000000000 15 567095731\n1 1 1 0 0 1 1 1 0 1 0 0 1 0 0",
"output": "45 queries, x=567095731"
},
{
"input": "1000000000 29 140644576\n1 1 1 1 1 1 0 1 0 0 0 1 0 1 0 1 1 1 0 1 1 1 0 1 1 0 1 0 0",
"output": "58 queries, x=140644576"
},
{
"input": "1000000000 13 714193420\n0 1 0 0 0 1 0 0 0 0 1 1 1",
"output": "43 queries, x=714193420"
},
{
"input": "1000000000 27 142518072\n0 0 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 1 0 0",
"output": "52 queries, x=142518072"
},
{
"input": "1000000000 25 11034213\n0 0 1 0 1 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0",
"output": "54 queries, x=11034213"
},
{
"input": "1000000000 9 584583057\n1 1 1 0 0 1 0 0 0",
"output": "35 queries, x=584583057"
},
{
"input": "1000000000 23 863164606\n1 1 0 1 0 1 0 1 0 1 0 0 1 0 0 0 1 1 1 0 0 1 1",
"output": "53 queries, x=863164606"
},
{
"input": "1000000000 21 731680746\n1 1 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 1 0 1",
"output": "51 queries, x=731680746"
},
{
"input": "1000000000 5 305229590\n0 0 1 1 0",
"output": "35 queries, x=305229590"
},
{
"input": "1000000000 3 28521539\n0 0 1",
"output": "31 queries, x=28521539"
},
{
"input": "1000000000 3 602070383\n0 1 1",
"output": "32 queries, x=602070383"
},
{
"input": "1000000000 2 880651931\n1 1",
"output": "30 queries, x=880651931"
},
{
"input": "1000000000 16 749168072\n1 1 0 0 0 1 0 0 1 1 1 1 1 1 1 0",
"output": "46 queries, x=749168072"
},
{
"input": "1000000000 30 322716916\n1 0 1 1 1 1 0 1 1 0 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 1 1 0 0",
"output": "58 queries, x=322716916"
},
{
"input": "1000000000 14 191233057\n0 0 1 0 0 1 1 1 1 0 0 0 1 1",
"output": "43 queries, x=191233057"
},
{
"input": "1000000000 30 1\n1 1 0 1 1 0 0 1 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 0 0 1 0 0 0 0",
"output": "1 queries, x=1"
},
{
"input": "1000000000 30 1\n1 0 1 1 1 1 1 1 0 1 0 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1",
"output": "1 queries, x=1"
},
{
"input": "1000000000 30 1\n1 0 1 0 0 0 0 1 1 0 0 1 1 0 1 1 1 0 1 0 1 1 1 0 0 0 1 0 1 1",
"output": "1 queries, x=1"
},
{
"input": "1000000000 30 1\n1 0 1 0 0 0 1 1 1 0 1 1 1 1 0 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1",
"output": "1 queries, x=1"
},
{
"input": "1000000000 30 1\n1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0",
"output": "1 queries, x=1"
},
{
"input": "1000000000 30 1000000000\n1 1 1 0 0 0 1 1 1 1 0 1 0 0 0 1 1 0 1 1 0 0 0 1 0 0 0 0 1 0",
"output": "60 queries, x=1000000000"
},
{
"input": "1000000000 30 1000000000\n1 1 1 0 0 1 1 1 0 0 1 1 0 0 1 1 1 0 1 0 1 1 0 0 1 0 1 1 1 0",
"output": "60 queries, x=1000000000"
},
{
"input": "1000000000 30 1000000000\n0 0 1 1 1 0 0 1 0 0 1 1 0 1 0 0 1 0 1 1 1 1 1 1 0 0 1 0 1 1",
"output": "60 queries, x=1000000000"
},
{
"input": "1000000000 30 1000000000\n0 0 0 1 1 1 1 1 1 0 1 0 1 0 1 1 0 1 1 0 0 1 0 0 1 0 0 1 0 1",
"output": "60 queries, x=1000000000"
},
{
"input": "1000000000 30 1000000000\n0 0 0 1 1 1 0 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 1 0 0 1 1 0 0 1",
"output": "60 queries, x=1000000000"
},
{
"input": "1 30 1\n1 1 1 0 1 0 0 0 0 1 1 0 1 0 0 1 0 1 0 1 0 1 0 1 1 0 0 0 1 1",
"output": "1 queries, x=1"
},
{
"input": "1 30 1\n1 1 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 0",
"output": "1 queries, x=1"
},
{
"input": "1 30 1\n1 0 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 0 1 1 0 0 1 1 0 1 0 0 0",
"output": "1 queries, x=1"
},
{
"input": "1 30 1\n1 0 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 0 1 0 1 0 0",
"output": "1 queries, x=1"
},
{
"input": "1 30 1\n1 0 1 1 1 1 1 0 0 0 1 1 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 0 1 1",
"output": "1 queries, x=1"
},
{
"input": "2 1 2\n1",
"output": "2 queries, x=2"
},
{
"input": "1000000000 1 1000000000\n1",
"output": "31 queries, x=1000000000"
},
{
"input": "10000 1 10000\n1",
"output": "15 queries, x=10000"
},
{
"input": "1000000000 1 999999999\n1",
"output": "30 queries, x=999999999"
},
{
"input": "100000 2 15\n1 0",
"output": "19 queries, x=15"
},
{
"input": "200000 1 110000\n1",
"output": "17 queries, x=110000"
},
{
"input": "123456789 1 42\n1",
"output": "27 queries, x=42"
},
{
"input": "1000000000 1 9\n1",
"output": "30 queries, x=9"
},
{
"input": "200000 2 100002\n1 0",
"output": "19 queries, x=100002"
},
{
"input": "1000000000 3 234567890\n0 1 0",
"output": "31 queries, x=234567890"
},
{
"input": "1000000000 5 321732193\n1 1 0 1 0",
"output": "35 queries, x=321732193"
},
{
"input": "1000000000 1 804289384\n1",
"output": "27 queries, x=804289384"
},
{
"input": "1000000000 2 999999998\n1 0",
"output": "32 queries, x=999999998"
},
{
"input": "1000000000 5 384618761\n0 1 1 0 1",
"output": "33 queries, x=384618761"
},
{
"input": "100000000 1 100\n0",
"output": "28 queries, x=100"
},
{
"input": "1000000000 1 804289384\n0",
"output": "27 queries, x=804289384"
},
{
"input": "100000000 1 100000000\n1",
"output": "28 queries, x=100000000"
},
{
"input": "40 1 4\n0",
"output": "6 queries, x=4"
},
{
"input": "1000000000 2 999999998\n0 1",
"output": "32 queries, x=999999998"
},
{
"input": "1000000000 1 1000000000\n0",
"output": "31 queries, x=1000000000"
},
{
"input": "1000000000 2 255555555\n1 0",
"output": "31 queries, x=255555555"
},
{
"input": "1000000000 2 1000000000\n0 1",
"output": "32 queries, x=1000000000"
},
{
"input": "1000000000 1 999999999\n0",
"output": "30 queries, x=999999999"
},
{
"input": "1000000000 2 888888888\n0 1",
"output": "31 queries, x=888888888"
},
{
"input": "1000000000 1 77000000\n1",
"output": "31 queries, x=77000000"
},
{
"input": "1000000000 1 123456789\n1",
"output": "27 queries, x=123456789"
},
{
"input": "10000 1 228\n0",
"output": "14 queries, x=228"
},
{
"input": "1000000000 1 12345\n1",
"output": "31 queries, x=12345"
},
{
"input": "1000000000 1 77000000\n0",
"output": "31 queries, x=77000000"
},
{
"input": "1000000000 1 23333\n0",
"output": "31 queries, x=23333"
},
{
"input": "1000000000 4 100\n0 1 0 1",
"output": "34 queries, x=100"
},
{
"input": "100000000 1 200\n1",
"output": "27 queries, x=200"
},
{
"input": "1000000000 3 5\n0 1 0",
"output": "33 queries, x=5"
},
{
"input": "1000000000 12 2\n1 1 1 1 1 1 0 0 1 1 1 1",
"output": "41 queries, x=2"
},
{
"input": "1000000000 1 5\n0",
"output": "31 queries, x=5"
},
{
"input": "100000 2 99999\n0 0",
"output": "18 queries, x=99999"
},
{
"input": "100000 2 2\n0 1",
"output": "18 queries, x=2"
},
{
"input": "1000000 1 91923\n0",
"output": "21 queries, x=91923"
},
{
"input": "1000000 2 1235\n0 1",
"output": "22 queries, x=1235"
},
{
"input": "1000000000 1 5\n1",
"output": "31 queries, x=5"
},
{
"input": "100000000 2 1234567\n0 1",
"output": "28 queries, x=1234567"
},
{
"input": "1000000000 1 1\n1",
"output": "1 queries, x=1"
},
{
"input": "1000000000 4 999999999\n1 0 0 1",
"output": "33 queries, x=999999999"
},
{
"input": "1000000000 4 1000000000\n1 0 1 0",
"output": "34 queries, x=1000000000"
},
{
"input": "1000000000 4 1\n1 0 1 0",
"output": "1 queries, x=1"
},
{
"input": "1000000000 5 500\n1 1 0 0 1",
"output": "34 queries, x=500"
},
{
"input": "1000 1 13\n1",
"output": "11 queries, x=13"
},
{
"input": "999999999 2 123456789\n1 0",
"output": "31 queries, x=123456789"
},
{
"input": "1000000000 5 1000000000\n1 1 1 1 1",
"output": "35 queries, x=1000000000"
},
{
"input": "1000000000 3 123456789\n1 0 1",
"output": "29 queries, x=123456789"
},
{
"input": "1000000000 1 123\n1",
"output": "29 queries, x=123"
},
{
"input": "10 1 9\n1",
"output": "4 queries, x=9"
},
{
"input": "1000000000 1 1\n0",
"output": "1 queries, x=1"
},
{
"input": "100000000 2 12345678\n0 1",
"output": "28 queries, x=12345678"
},
{
"input": "1000000000 30 1000000000\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "60 queries, x=1000000000"
},
{
"input": "805306369 30 805306369\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "60 queries, x=805306369"
},
{
"input": "1000000000 2 123\n0 1",
"output": "30 queries, x=123"
},
{
"input": "1000000000 1 9\n0",
"output": "30 queries, x=9"
},
{
"input": "100000 2 90000\n1 0",
"output": "18 queries, x=90000"
},
{
"input": "1000000000 2 123456789\n1 0",
"output": "28 queries, x=123456789"
},
{
"input": "1000000000 2 804289384\n1 0",
"output": "28 queries, x=804289384"
},
{
"input": "1000000000 30 999999970\n1 0 0 1 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 1 0 0 0 1 0 1 0 0 0 1",
"output": "60 queries, x=999999970"
}
] | 1,532,716,016 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 155 | 1,024,000 | import sys, math
from fractions import gcd
from itertools import permutations
debug = 0
if debug:
f = open("input.txt", "r")
input = f.readline
else:
input = sys.stdin.readline
def mp():
# global input
return list(map(int, input().split()))
def _main():
m, n = mp()
p = [0] * n
for i in range(n):
print(1)
stdout.flush()
ans = int(input())
if not ans:
return
if ans == 1:
p[i] = 1
else:
p[i] = 0
l, r = 1, m
i = 0
while True:
mid = (l + r) // 2
print(mid)
stdout.flush()
ans = int(input())
if not ans:
return
if p[i]:
if ans == 1:
l = mid
else:
r = mid
else:
if ans == 1:
r = mid
else:
l = mid
i += 1
i %= n
_main()
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This is an interactive problem.
Natasha is going to fly to Mars. Finally, Natasha sat in the rocket. She flies, flies... but gets bored. She wishes to arrive to Mars already! So she decides to find something to occupy herself. She couldn't think of anything better to do than to calculate the distance to the red planet.
Let's define $x$ as the distance to Mars. Unfortunately, Natasha does not know $x$. But it is known that $1 \le x \le m$, where Natasha knows the number $m$. Besides, $x$ and $m$ are positive integers.
Natasha can ask the rocket questions. Every question is an integer $y$ ($1 \le y \le m$). The correct answer to the question is $-1$, if $x<y$, $0$, if $x=y$, and $1$, if $x>y$. But the rocket is broken — it does not always answer correctly. Precisely: let the correct answer to the current question be equal to $t$, then, if the rocket answers this question correctly, then it will answer $t$, otherwise it will answer $-t$.
In addition, the rocket has a sequence $p$ of length $n$. Each element of the sequence is either $0$ or $1$. The rocket processes this sequence in the cyclic order, that is $1$-st element, $2$-nd, $3$-rd, $\ldots$, $(n-1)$-th, $n$-th, $1$-st, $2$-nd, $3$-rd, $\ldots$, $(n-1)$-th, $n$-th, $\ldots$. If the current element is $1$, the rocket answers correctly, if $0$ — lies. Natasha doesn't know the sequence $p$, but she knows its length — $n$.
You can ask the rocket no more than $60$ questions.
Help Natasha find the distance to Mars. Assume, that the distance to Mars does not change while Natasha is asking questions.
Your solution will not be accepted, if it does not receive an answer $0$ from the rocket (even if the distance to Mars is uniquely determined by the already received rocket's answers).
Input Specification:
The first line contains two integers $m$ and $n$ ($1 \le m \le 10^9$, $1 \le n \le 30$) — the maximum distance to Mars and the number of elements in the sequence $p$.
Output Specification:
none
Demo Input:
['5 2\n1\n-1\n-1\n1\n0\n']
Demo Output:
['1\n2\n4\n5\n3\n']
Note:
In the example, hacking would look like this:
5 2 3
1 0
This means that the current distance to Mars is equal to $3$, Natasha knows that it does not exceed $5$, and the rocket answers in order: correctly, incorrectly, correctly, incorrectly ...
Really:
on the first query ($1$) the correct answer is $1$, the rocket answered correctly: $1$;
on the second query ($2$) the correct answer is $1$, the rocket answered incorrectly: $-1$;
on the third query ($4$) the correct answer is $-1$, the rocket answered correctly: $-1$;
on the fourth query ($5$) the correct answer is $-1$, the rocket answered incorrectly: $1$;
on the fifth query ($3$) the correct and incorrect answer is $0$. | ```python
import sys, math
from fractions import gcd
from itertools import permutations
debug = 0
if debug:
f = open("input.txt", "r")
input = f.readline
else:
input = sys.stdin.readline
def mp():
# global input
return list(map(int, input().split()))
def _main():
m, n = mp()
p = [0] * n
for i in range(n):
print(1)
stdout.flush()
ans = int(input())
if not ans:
return
if ans == 1:
p[i] = 1
else:
p[i] = 0
l, r = 1, m
i = 0
while True:
mid = (l + r) // 2
print(mid)
stdout.flush()
ans = int(input())
if not ans:
return
if p[i]:
if ans == 1:
l = mid
else:
r = mid
else:
if ans == 1:
r = mid
else:
l = mid
i += 1
i %= n
_main()
``` | -1 |
|
1,009 | E | Intercity Travelling | PROGRAMMING | 2,000 | [
"combinatorics",
"math",
"probabilities"
] | null | null | Leha is planning his journey from Moscow to Saratov. He hates trains, so he has decided to get from one city to another by car.
The path from Moscow to Saratov can be represented as a straight line (well, it's not that straight in reality, but in this problem we will consider it to be straight), and the distance between Moscow and Saratov is $n$ km. Let's say that Moscow is situated at the point with coordinate $0$ km, and Saratov — at coordinate $n$ km.
Driving for a long time may be really difficult. Formally, if Leha has already covered $i$ kilometers since he stopped to have a rest, he considers the difficulty of covering $(i + 1)$-th kilometer as $a_{i + 1}$. It is guaranteed that for every $i \in [1, n - 1]$ $a_i \le a_{i + 1}$. The difficulty of the journey is denoted as the sum of difficulties of each kilometer in the journey.
Fortunately, there may be some rest sites between Moscow and Saratov. Every integer point from $1$ to $n - 1$ may contain a rest site. When Leha enters a rest site, he may have a rest, and the next kilometer will have difficulty $a_1$, the kilometer after it — difficulty $a_2$, and so on.
For example, if $n = 5$ and there is a rest site in coordinate $2$, the difficulty of journey will be $2a_1 + 2a_2 + a_3$: the first kilometer will have difficulty $a_1$, the second one — $a_2$, then Leha will have a rest, and the third kilometer will have difficulty $a_1$, the fourth — $a_2$, and the last one — $a_3$. Another example: if $n = 7$ and there are rest sites in coordinates $1$ and $5$, the difficulty of Leha's journey is $3a_1 + 2a_2 + a_3 + a_4$.
Leha doesn't know which integer points contain rest sites. So he has to consider every possible situation. Obviously, there are $2^{n - 1}$ different distributions of rest sites (two distributions are different if there exists some point $x$ such that it contains a rest site in exactly one of these distributions). Leha considers all these distributions to be equiprobable. He wants to calculate $p$ — the expected value of difficulty of his journey.
Obviously, $p \cdot 2^{n - 1}$ is an integer number. You have to calculate it modulo $998244353$. | The first line contains one number $n$ ($1 \le n \le 10^6$) — the distance from Moscow to Saratov.
The second line contains $n$ integer numbers $a_1$, $a_2$, ..., $a_n$ ($1 \le a_1 \le a_2 \le \dots \le a_n \le 10^6$), where $a_i$ is the difficulty of $i$-th kilometer after Leha has rested. | Print one number — $p \cdot 2^{n - 1}$, taken modulo $998244353$. | [
"2\n1 2\n",
"4\n1 3 3 7\n"
] | [
"5\n",
"60\n"
] | none | 0 | [
{
"input": "2\n1 2",
"output": "5"
},
{
"input": "4\n1 3 3 7",
"output": "60"
},
{
"input": "100\n3 3 3 4 7 8 8 8 9 9 10 12 12 13 14 14 15 15 16 17 17 20 21 21 22 22 23 25 29 31 36 37 37 38 39 40 41 41 41 42 43 44 45 46 46 47 47 49 49 49 51 52 52 53 54 55 59 59 59 60 62 63 63 64 66 69 70 71 71 72 74 76 76 77 77 78 78 79 80 81 81 82 82 84 85 86 87 87 87 89 91 92 92 92 92 97 98 99 100 100",
"output": "758086002"
},
{
"input": "1\n12",
"output": "12"
}
] | 1,658,429,968 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 1,500 | 20,480,000 | n = int(input())
a = list(map(int, input().split()))
res = 0
for t in range(1, n + 1):
dp = [[0 for i in range(n + 1)] for j in range(n + 1)]
dp[0][0] = 1
for i in range(1, n + 1):
for j in range(n + 1):
for x in range(1, t):
if i - x < 0:
break
dp[i][j] += dp[i - x][j]
for x in range(t, i + 1):
if j >= 1:
dp[i][j] += dp[i - x][j - 1]
temp = 0
for k in range(1, n + 1):
temp += dp[n][k] * k
res += a[t - 1] * temp
print(res % 998244353) | Title: Intercity Travelling
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leha is planning his journey from Moscow to Saratov. He hates trains, so he has decided to get from one city to another by car.
The path from Moscow to Saratov can be represented as a straight line (well, it's not that straight in reality, but in this problem we will consider it to be straight), and the distance between Moscow and Saratov is $n$ km. Let's say that Moscow is situated at the point with coordinate $0$ km, and Saratov — at coordinate $n$ km.
Driving for a long time may be really difficult. Formally, if Leha has already covered $i$ kilometers since he stopped to have a rest, he considers the difficulty of covering $(i + 1)$-th kilometer as $a_{i + 1}$. It is guaranteed that for every $i \in [1, n - 1]$ $a_i \le a_{i + 1}$. The difficulty of the journey is denoted as the sum of difficulties of each kilometer in the journey.
Fortunately, there may be some rest sites between Moscow and Saratov. Every integer point from $1$ to $n - 1$ may contain a rest site. When Leha enters a rest site, he may have a rest, and the next kilometer will have difficulty $a_1$, the kilometer after it — difficulty $a_2$, and so on.
For example, if $n = 5$ and there is a rest site in coordinate $2$, the difficulty of journey will be $2a_1 + 2a_2 + a_3$: the first kilometer will have difficulty $a_1$, the second one — $a_2$, then Leha will have a rest, and the third kilometer will have difficulty $a_1$, the fourth — $a_2$, and the last one — $a_3$. Another example: if $n = 7$ and there are rest sites in coordinates $1$ and $5$, the difficulty of Leha's journey is $3a_1 + 2a_2 + a_3 + a_4$.
Leha doesn't know which integer points contain rest sites. So he has to consider every possible situation. Obviously, there are $2^{n - 1}$ different distributions of rest sites (two distributions are different if there exists some point $x$ such that it contains a rest site in exactly one of these distributions). Leha considers all these distributions to be equiprobable. He wants to calculate $p$ — the expected value of difficulty of his journey.
Obviously, $p \cdot 2^{n - 1}$ is an integer number. You have to calculate it modulo $998244353$.
Input Specification:
The first line contains one number $n$ ($1 \le n \le 10^6$) — the distance from Moscow to Saratov.
The second line contains $n$ integer numbers $a_1$, $a_2$, ..., $a_n$ ($1 \le a_1 \le a_2 \le \dots \le a_n \le 10^6$), where $a_i$ is the difficulty of $i$-th kilometer after Leha has rested.
Output Specification:
Print one number — $p \cdot 2^{n - 1}$, taken modulo $998244353$.
Demo Input:
['2\n1 2\n', '4\n1 3 3 7\n']
Demo Output:
['5\n', '60\n']
Note:
none | ```python
n = int(input())
a = list(map(int, input().split()))
res = 0
for t in range(1, n + 1):
dp = [[0 for i in range(n + 1)] for j in range(n + 1)]
dp[0][0] = 1
for i in range(1, n + 1):
for j in range(n + 1):
for x in range(1, t):
if i - x < 0:
break
dp[i][j] += dp[i - x][j]
for x in range(t, i + 1):
if j >= 1:
dp[i][j] += dp[i - x][j - 1]
temp = 0
for k in range(1, n + 1):
temp += dp[n][k] * k
res += a[t - 1] * temp
print(res % 998244353)
``` | 0 |
|
465 | A | inc ARG | PROGRAMMING | 900 | [
"implementation"
] | null | null | Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit.
Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded.
Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell.
The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit. | Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell. | [
"4\n1100\n",
"4\n1111\n"
] | [
"3\n",
"4\n"
] | In the first sample the cell ends up with value 0010, in the second sample — with 0000. | 500 | [
{
"input": "4\n1100",
"output": "3"
},
{
"input": "4\n1111",
"output": "4"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "1"
},
{
"input": "2\n01",
"output": "1"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "10\n0000000000",
"output": "1"
},
{
"input": "20\n11111111110110001100",
"output": "11"
},
{
"input": "50\n01011110100010000001010000100001001101001101101011",
"output": "1"
},
{
"input": "60\n111111111101111111111111111111111111111111111111111111111111",
"output": "11"
},
{
"input": "60\n111111111111111111111111111111111111111111111111111111111111",
"output": "60"
},
{
"input": "66\n111111010010011001110011000111000100011110011001111110011111111101",
"output": "7"
},
{
"input": "90\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "91\n1011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "2"
},
{
"input": "92\n11111111101011111011111111111111111111111011101111111111011111111111111101111111101011111101",
"output": "10"
},
{
"input": "100\n0001011110100011001100100010111001000001111101101001001001001011110100101101010000000110100101110010",
"output": "1"
},
{
"input": "100\n0111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "1"
},
{
"input": "100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "100"
},
{
"input": "100\n1011001110001000011111110011000100001110010110111101110110011011011000010100110001111100000010110010",
"output": "2"
},
{
"input": "100\n1101111011001111111111110011110111101110111111111111111111111111111111011111111111110111111111111111",
"output": "3"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "100"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110",
"output": "100"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111101",
"output": "99"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111100000",
"output": "96"
},
{
"input": "100\n1111111111111111111111111111111110011000110110001010101100100100001000010010010010000100110000010101",
"output": "34"
}
] | 1,410,104,004 | 204 | Python 3 | WRONG_ANSWER | PRETESTS | 0 | 46 | 0 | n=input()
s=input()
i=1
while s[i]=='1':
i+=1
if i==len(s):
break
print(i) | Title: inc ARG
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit.
Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded.
Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell.
The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit.
Output Specification:
Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell.
Demo Input:
['4\n1100\n', '4\n1111\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample the cell ends up with value 0010, in the second sample — with 0000. | ```python
n=input()
s=input()
i=1
while s[i]=='1':
i+=1
if i==len(s):
break
print(i)
``` | 0 |
|
228 | A | Is your horseshoe on the other hoof? | PROGRAMMING | 800 | [
"implementation"
] | null | null | Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. | The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers. | Print a single integer — the minimum number of horseshoes Valera needs to buy. | [
"1 7 3 3\n",
"7 7 7 7\n"
] | [
"1\n",
"3\n"
] | none | 500 | [
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 586975278",
"output": "1"
},
{
"input": "259504825 377489979 588153796 377489979",
"output": "1"
},
{
"input": "652588203 931100304 931100304 652588203",
"output": "2"
},
{
"input": "391958720 651507265 391958720 651507265",
"output": "2"
},
{
"input": "90793237 90793237 90793237 90793237",
"output": "3"
},
{
"input": "551651653 551651653 551651653 551651653",
"output": "3"
},
{
"input": "156630260 609654355 668943582 973622757",
"output": "0"
},
{
"input": "17061017 110313588 434481173 796661222",
"output": "0"
},
{
"input": "24975422 256716298 337790533 690960249",
"output": "0"
},
{
"input": "255635360 732742923 798648949 883146723",
"output": "0"
},
{
"input": "133315691 265159773 734556507 265159773",
"output": "1"
},
{
"input": "28442865 741657755 978106882 978106882",
"output": "1"
},
{
"input": "131245479 174845575 497483467 131245479",
"output": "1"
},
{
"input": "139159884 616215581 958341883 616215581",
"output": "1"
},
{
"input": "147784432 947653080 947653080 947653080",
"output": "2"
},
{
"input": "94055790 756126496 756126496 94055790",
"output": "2"
},
{
"input": "240458500 511952208 240458500 511952208",
"output": "2"
},
{
"input": "681828506 972810624 972810624 681828506",
"output": "2"
},
{
"input": "454961014 454961014 454961014 454961014",
"output": "3"
},
{
"input": "915819430 915819430 915819430 915819430",
"output": "3"
},
{
"input": "671645142 671645142 671645142 671645142",
"output": "3"
},
{
"input": "132503558 132503558 132503558 132503558",
"output": "3"
},
{
"input": "5 5 999999 6",
"output": "1"
},
{
"input": "1 1 2 5",
"output": "1"
},
{
"input": "2 1 2 3",
"output": "1"
},
{
"input": "1 1 3 5",
"output": "1"
},
{
"input": "1 1 3 3",
"output": "2"
},
{
"input": "2 2 2 1",
"output": "2"
},
{
"input": "3 1 1 1",
"output": "2"
},
{
"input": "1 2 2 2",
"output": "2"
}
] | 1,689,740,765 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 62 | 0 | n=list(map(int, input().split()))
st=set(n)
print(len(n)-len(st)) | Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none | ```python
n=list(map(int, input().split()))
st=set(n)
print(len(n)-len(st))
``` | 3 |
|
697 | A | Pineapple Incident | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | Ted has a pineapple. This pineapple is able to bark like a bulldog! At time *t* (in seconds) it barks for the first time. Then every *s* seconds after it, it barks twice with 1 second interval. Thus it barks at times *t*, *t*<=+<=*s*, *t*<=+<=*s*<=+<=1, *t*<=+<=2*s*, *t*<=+<=2*s*<=+<=1, etc.
Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time *x* (in seconds), so he asked you to tell him if it's gonna bark at that time. | The first and only line of input contains three integers *t*, *s* and *x* (0<=≤<=*t*,<=*x*<=≤<=109, 2<=≤<=*s*<=≤<=109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively. | Print a single "YES" (without quotes) if the pineapple will bark at time *x* or a single "NO" (without quotes) otherwise in the only line of output. | [
"3 10 4\n",
"3 10 3\n",
"3 8 51\n",
"3 8 52\n"
] | [
"NO\n",
"YES\n",
"YES\n",
"YES\n"
] | In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3.
In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52. | 500 | [
{
"input": "3 10 4",
"output": "NO"
},
{
"input": "3 10 3",
"output": "YES"
},
{
"input": "3 8 51",
"output": "YES"
},
{
"input": "3 8 52",
"output": "YES"
},
{
"input": "456947336 740144 45",
"output": "NO"
},
{
"input": "33 232603 599417964",
"output": "YES"
},
{
"input": "4363010 696782227 701145238",
"output": "YES"
},
{
"input": "9295078 2 6",
"output": "NO"
},
{
"input": "76079 281367 119938421",
"output": "YES"
},
{
"input": "93647 7 451664565",
"output": "YES"
},
{
"input": "5 18553 10908",
"output": "NO"
},
{
"input": "6 52 30",
"output": "NO"
},
{
"input": "6431 855039 352662",
"output": "NO"
},
{
"input": "749399100 103031711 761562532",
"output": "NO"
},
{
"input": "21 65767 55245",
"output": "NO"
},
{
"input": "4796601 66897 4860613",
"output": "NO"
},
{
"input": "8 6728951 860676",
"output": "NO"
},
{
"input": "914016 6 914019",
"output": "NO"
},
{
"input": "60686899 78474 60704617",
"output": "NO"
},
{
"input": "3 743604 201724",
"output": "NO"
},
{
"input": "571128 973448796 10",
"output": "NO"
},
{
"input": "688051712 67 51",
"output": "NO"
},
{
"input": "74619 213344 6432326",
"output": "NO"
},
{
"input": "6947541 698167 6",
"output": "NO"
},
{
"input": "83 6 6772861",
"output": "NO"
},
{
"input": "251132 67561 135026988",
"output": "NO"
},
{
"input": "8897216 734348516 743245732",
"output": "YES"
},
{
"input": "50 64536 153660266",
"output": "YES"
},
{
"input": "876884 55420 971613604",
"output": "YES"
},
{
"input": "0 6906451 366041903",
"output": "YES"
},
{
"input": "11750 8 446010134",
"output": "YES"
},
{
"input": "582692707 66997 925047377",
"output": "YES"
},
{
"input": "11 957526890 957526901",
"output": "YES"
},
{
"input": "556888 514614196 515171084",
"output": "YES"
},
{
"input": "6 328006 584834704",
"output": "YES"
},
{
"input": "4567998 4 204966403",
"output": "YES"
},
{
"input": "60 317278 109460971",
"output": "YES"
},
{
"input": "906385 342131991 685170368",
"output": "YES"
},
{
"input": "1 38 902410512",
"output": "YES"
},
{
"input": "29318 787017 587931018",
"output": "YES"
},
{
"input": "351416375 243431 368213115",
"output": "YES"
},
{
"input": "54 197366062 197366117",
"output": "YES"
},
{
"input": "586389 79039 850729874",
"output": "YES"
},
{
"input": "723634470 2814619 940360134",
"output": "YES"
},
{
"input": "0 2 0",
"output": "YES"
},
{
"input": "0 2 1",
"output": "NO"
},
{
"input": "0 2 2",
"output": "YES"
},
{
"input": "0 2 3",
"output": "YES"
},
{
"input": "0 2 1000000000",
"output": "YES"
},
{
"input": "0 10 23",
"output": "NO"
},
{
"input": "0 2 999999999",
"output": "YES"
},
{
"input": "10 5 11",
"output": "NO"
},
{
"input": "1 2 1000000000",
"output": "YES"
},
{
"input": "1 10 20",
"output": "NO"
},
{
"input": "1 2 999999937",
"output": "YES"
},
{
"input": "10 3 5",
"output": "NO"
},
{
"input": "3 2 5",
"output": "YES"
},
{
"input": "0 4 0",
"output": "YES"
},
{
"input": "0 215 403",
"output": "NO"
},
{
"input": "5 2 10",
"output": "YES"
},
{
"input": "0 2 900000000",
"output": "YES"
},
{
"input": "0 79 4000",
"output": "NO"
},
{
"input": "5 1000 1000",
"output": "NO"
},
{
"input": "1 5 103",
"output": "NO"
},
{
"input": "5 2 6",
"output": "NO"
},
{
"input": "120 2 1000000000",
"output": "YES"
},
{
"input": "2 2 1000000000",
"output": "YES"
},
{
"input": "5 5 13",
"output": "NO"
},
{
"input": "10 5 15",
"output": "YES"
},
{
"input": "11 2 0",
"output": "NO"
},
{
"input": "3 8 53",
"output": "NO"
},
{
"input": "2 2 4",
"output": "YES"
},
{
"input": "4 4 0",
"output": "NO"
},
{
"input": "1 2 3",
"output": "YES"
},
{
"input": "5 3 9",
"output": "YES"
},
{
"input": "5 6 19",
"output": "NO"
},
{
"input": "3 10 125",
"output": "NO"
},
{
"input": "5 3 8",
"output": "YES"
},
{
"input": "6 3 9",
"output": "YES"
},
{
"input": "0 3 5",
"output": "NO"
},
{
"input": "5 3 300000035",
"output": "YES"
},
{
"input": "5 2 7",
"output": "YES"
},
{
"input": "1 5 6",
"output": "YES"
},
{
"input": "4 2 6",
"output": "YES"
},
{
"input": "0 3 999999998",
"output": "NO"
},
{
"input": "0 10001 0",
"output": "YES"
},
{
"input": "6 5 3",
"output": "NO"
},
{
"input": "1 5 1000000000",
"output": "NO"
},
{
"input": "1 3 6",
"output": "NO"
},
{
"input": "3 3 1000000000",
"output": "YES"
},
{
"input": "3 3 4",
"output": "NO"
},
{
"input": "3 3 5",
"output": "NO"
},
{
"input": "3 3 0",
"output": "NO"
},
{
"input": "1 2 4",
"output": "YES"
},
{
"input": "5 5 10",
"output": "YES"
}
] | 1,697,770,612 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | t, s, x = map(int, input().split())
c = 0
ct = t
while ct <= x:
#print(abs(ct-t))
if abs(ct-t) < 2:
print("YES")
exit()
ct += s
print("NO") | Title: Pineapple Incident
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ted has a pineapple. This pineapple is able to bark like a bulldog! At time *t* (in seconds) it barks for the first time. Then every *s* seconds after it, it barks twice with 1 second interval. Thus it barks at times *t*, *t*<=+<=*s*, *t*<=+<=*s*<=+<=1, *t*<=+<=2*s*, *t*<=+<=2*s*<=+<=1, etc.
Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time *x* (in seconds), so he asked you to tell him if it's gonna bark at that time.
Input Specification:
The first and only line of input contains three integers *t*, *s* and *x* (0<=≤<=*t*,<=*x*<=≤<=109, 2<=≤<=*s*<=≤<=109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively.
Output Specification:
Print a single "YES" (without quotes) if the pineapple will bark at time *x* or a single "NO" (without quotes) otherwise in the only line of output.
Demo Input:
['3 10 4\n', '3 10 3\n', '3 8 51\n', '3 8 52\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n', 'YES\n']
Note:
In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3.
In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52. | ```python
t, s, x = map(int, input().split())
c = 0
ct = t
while ct <= x:
#print(abs(ct-t))
if abs(ct-t) < 2:
print("YES")
exit()
ct += s
print("NO")
``` | 0 |
|
559 | A | Gerald's Hexagon | PROGRAMMING | 1,600 | [
"brute force",
"geometry",
"math"
] | null | null | Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to . Then he measured the length of its sides, and found that each of them is equal to an integer number of centimeters. There the properties of the hexagon ended and Gerald decided to draw on it.
He painted a few lines, parallel to the sides of the hexagon. The lines split the hexagon into regular triangles with sides of 1 centimeter. Now Gerald wonders how many triangles he has got. But there were so many of them that Gerald lost the track of his counting. Help the boy count the triangles. | The first and the single line of the input contains 6 space-separated integers *a*1,<=*a*2,<=*a*3,<=*a*4,<=*a*5 and *a*6 (1<=≤<=*a**i*<=≤<=1000) — the lengths of the sides of the hexagons in centimeters in the clockwise order. It is guaranteed that the hexagon with the indicated properties and the exactly such sides exists. | Print a single integer — the number of triangles with the sides of one 1 centimeter, into which the hexagon is split. | [
"1 1 1 1 1 1\n",
"1 2 1 2 1 2\n"
] | [
"6\n",
"13\n"
] | This is what Gerald's hexagon looks like in the first sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/84d193e27b02c38eb1eadc536602a2ec0b9f9519.png" style="max-width: 100.0%;max-height: 100.0%;"/>
And that's what it looks like in the second sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/e29076a96da8ca864654cc6195654d9bf07d31ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 500 | [
{
"input": "1 1 1 1 1 1",
"output": "6"
},
{
"input": "1 2 1 2 1 2",
"output": "13"
},
{
"input": "2 4 5 3 3 6",
"output": "83"
},
{
"input": "45 19 48 18 46 21",
"output": "6099"
},
{
"input": "66 6 65 6 66 5",
"output": "5832"
},
{
"input": "7 5 4 8 4 5",
"output": "175"
},
{
"input": "3 2 1 4 1 2",
"output": "25"
},
{
"input": "7 1 7 3 5 3",
"output": "102"
},
{
"input": "9 2 9 3 8 3",
"output": "174"
},
{
"input": "1 6 1 5 2 5",
"output": "58"
},
{
"input": "41 64 48 61 44 68",
"output": "17488"
},
{
"input": "1 59 2 59 1 60",
"output": "3838"
},
{
"input": "30 36 36 32 34 38",
"output": "7052"
},
{
"input": "50 40 46 38 52 34",
"output": "11176"
},
{
"input": "4 60 4 60 4 60",
"output": "4576"
},
{
"input": "718 466 729 470 714 481",
"output": "2102808"
},
{
"input": "131 425 143 461 95 473",
"output": "441966"
},
{
"input": "125 7 128 8 124 11",
"output": "20215"
},
{
"input": "677 303 685 288 692 296",
"output": "1365807"
},
{
"input": "1 577 7 576 2 582",
"output": "342171"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "6000000"
},
{
"input": "1 1 1000 1 1 1000",
"output": "4002"
},
{
"input": "1000 1000 1 1000 1000 1",
"output": "2004000"
},
{
"input": "1000 1 1000 999 2 999",
"output": "2003997"
},
{
"input": "1 1000 1 1 1000 1",
"output": "4002"
},
{
"input": "888 888 888 887 889 887",
"output": "4729487"
}
] | 1,672,988,161 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 46 | 0 | import sys
input = sys.stdin.readline
w = list(map(int, input().split()))
for i in range(6):
if w[1] == w[5] and w[2] == w[4]:
c = 0
a, b = w[0], w[3]
for j in range(w[1]):
c += 2*a + 1
a += 1
for j in range(w[2]):
c += 2*b + 1
b += 1
print(c)
break
else:
w = w[1:] + [w[0]] | Title: Gerald's Hexagon
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to . Then he measured the length of its sides, and found that each of them is equal to an integer number of centimeters. There the properties of the hexagon ended and Gerald decided to draw on it.
He painted a few lines, parallel to the sides of the hexagon. The lines split the hexagon into regular triangles with sides of 1 centimeter. Now Gerald wonders how many triangles he has got. But there were so many of them that Gerald lost the track of his counting. Help the boy count the triangles.
Input Specification:
The first and the single line of the input contains 6 space-separated integers *a*1,<=*a*2,<=*a*3,<=*a*4,<=*a*5 and *a*6 (1<=≤<=*a**i*<=≤<=1000) — the lengths of the sides of the hexagons in centimeters in the clockwise order. It is guaranteed that the hexagon with the indicated properties and the exactly such sides exists.
Output Specification:
Print a single integer — the number of triangles with the sides of one 1 centimeter, into which the hexagon is split.
Demo Input:
['1 1 1 1 1 1\n', '1 2 1 2 1 2\n']
Demo Output:
['6\n', '13\n']
Note:
This is what Gerald's hexagon looks like in the first sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/84d193e27b02c38eb1eadc536602a2ec0b9f9519.png" style="max-width: 100.0%;max-height: 100.0%;"/>
And that's what it looks like in the second sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/e29076a96da8ca864654cc6195654d9bf07d31ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
import sys
input = sys.stdin.readline
w = list(map(int, input().split()))
for i in range(6):
if w[1] == w[5] and w[2] == w[4]:
c = 0
a, b = w[0], w[3]
for j in range(w[1]):
c += 2*a + 1
a += 1
for j in range(w[2]):
c += 2*b + 1
b += 1
print(c)
break
else:
w = w[1:] + [w[0]]
``` | 0 |
|
432 | A | Choosing Teams | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times? | The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship. | Print a single number — the answer to the problem. | [
"5 2\n0 4 5 1 0\n",
"6 4\n0 1 2 3 4 5\n",
"6 5\n0 0 0 0 0 0\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | 500 | [
{
"input": "5 2\n0 4 5 1 0",
"output": "1"
},
{
"input": "6 4\n0 1 2 3 4 5",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0",
"output": "1"
},
{
"input": "3 4\n0 2 0",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "12 2\n0 1 2 3 4 0 1 2 3 4 0 1",
"output": "3"
},
{
"input": "15 2\n0 1 2 3 4 0 1 2 3 4 0 1 2 3 4",
"output": "4"
},
{
"input": "13 1\n5 0 5 0 1 5 0 4 1 1 1 4 1",
"output": "3"
},
{
"input": "20 1\n5 0 4 2 2 3 2 1 2 4 1 3 5 5 5 4 4 1 3 0",
"output": "5"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "6 3\n4 4 4 4 4 4",
"output": "0"
},
{
"input": "1 4\n1",
"output": "0"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "2 3\n0 0",
"output": "0"
}
] | 1,690,557,152 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 2 | 30 | 0 | def f(k, members):
team=0
members.sort()
while(members[0]<=(5-k) and members[1]<=(5-k) and members[2]<=(5-k)):
members = members[3:]
team+=1
return team
nk = input().split(" ")
mem = input().split(" ")
mem = [int(i) for i in mem]
print(f(int(nk[1]),mem)) | Title: Choosing Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
Input Specification:
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | ```python
def f(k, members):
team=0
members.sort()
while(members[0]<=(5-k) and members[1]<=(5-k) and members[2]<=(5-k)):
members = members[3:]
team+=1
return team
nk = input().split(" ")
mem = input().split(" ")
mem = [int(i) for i in mem]
print(f(int(nk[1]),mem))
``` | -1 |
|
711 | A | Bus to Udayland | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit? | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details. | If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them. | [
"6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n",
"5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n"
] | [
"YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n",
"NO\n",
"YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n"
] | Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX | 500 | [
{
"input": "6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX",
"output": "YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX"
},
{
"input": "4\nXO|OX\nXO|XX\nOX|OX\nXX|OX",
"output": "NO"
},
{
"input": "5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO",
"output": "YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO"
},
{
"input": "1\nXO|OX",
"output": "NO"
},
{
"input": "1\nOO|OO",
"output": "YES\n++|OO"
},
{
"input": "4\nXO|XX\nXX|XO\nOX|XX\nXO|XO",
"output": "NO"
},
{
"input": "9\nOX|XO\nOX|XO\nXO|OX\nOX|OX\nXO|OX\nXX|OO\nOX|OX\nOX|XO\nOX|OX",
"output": "YES\nOX|XO\nOX|XO\nXO|OX\nOX|OX\nXO|OX\nXX|++\nOX|OX\nOX|XO\nOX|OX"
},
{
"input": "61\nOX|XX\nOX|XX\nOX|XX\nXO|XO\nXX|XO\nXX|XX\nXX|XX\nOX|XX\nXO|XO\nOX|XO\nXO|OX\nXX|XX\nXX|XX\nOX|OX\nXX|OX\nOX|XO\nOX|XO\nXO|OX\nXO|XX\nOX|XX\nOX|XX\nXO|OX\nXO|XX\nXO|XX\nOX|XX\nXX|XX\nXX|XO\nXO|XX\nXX|XX\nXO|OX\nXX|XO\nXO|XX\nXO|XO\nXO|OX\nXX|OX\nXO|OX\nOX|XX\nXX|OX\nXX|XX\nOX|XO\nOX|XX\nXO|OX\nOX|XX\nOX|XX\nXO|XO\nXO|XX\nOX|XX\nXO|XO\nOX|XX\nXX|XX\nOX|XO\nXO|XO\nXO|XO\nOX|OX\nXX|OX\nXX|OX\nOX|XO\nOX|XX\nOX|OX\nXO|XX\nOX|XX",
"output": "NO"
},
{
"input": "1\nOO|XX",
"output": "YES\n++|XX"
},
{
"input": "10\nOO|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|XX",
"output": "YES\n++|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|XX"
},
{
"input": "10\nXX|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|OO",
"output": "YES\nXX|XX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXO|OX\nXX|++"
},
{
"input": "5\nXX|XX\nXX|XX\nXO|OX\nOO|OX\nOX|XO",
"output": "YES\nXX|XX\nXX|XX\nXO|OX\n++|OX\nOX|XO"
},
{
"input": "6\nOO|XX\nXO|XX\nOX|OO\nXX|OX\nOO|XX\nOX|XX",
"output": "YES\n++|XX\nXO|XX\nOX|OO\nXX|OX\nOO|XX\nOX|XX"
}
] | 1,631,048,321 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 61 | 6,963,200 | w = True
for i in range(int(input())):
t = input()
if t[0] == t[1] == 'O' and w == True:
w = False
print('++|'+ t[3] + t[4])
elif t[3] == t[4] == 'O' and w == True:
w = False
print(t[3] + t[4] + '|++')
else:
print(t) | Title: Bus to Udayland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has *n* rows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.
ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows of seats in the bus.
Then, *n* lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.
Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details.
Output Specification:
If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next *n* lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).
If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.
If there are multiple solutions, you may print any of them.
Demo Input:
['6\nOO|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', '4\nXO|OX\nXO|XX\nOX|OX\nXX|OX\n', '5\nXX|XX\nXX|XX\nXO|OX\nXO|OO\nOX|XO\n']
Demo Output:
['YES\n++|OX\nXO|XX\nOX|OO\nXX|OX\nOO|OO\nOO|XX\n', 'NO\n', 'YES\nXX|XX\nXX|XX\nXO|OX\nXO|++\nOX|XO\n']
Note:
Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.
O+|+X
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX | ```python
w = True
for i in range(int(input())):
t = input()
if t[0] == t[1] == 'O' and w == True:
w = False
print('++|'+ t[3] + t[4])
elif t[3] == t[4] == 'O' and w == True:
w = False
print(t[3] + t[4] + '|++')
else:
print(t)
``` | 0 |
|
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,682,605,273 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 122 | 0 | word = input()
big_letter = 0
small_letter = 0
for i in word:
if ord(i) < 95:
big_letter += 1
else:
small_letter += 1
print(word.lower() if small_letter >= big_letter else word.upper()) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
word = input()
big_letter = 0
small_letter = 0
for i in word:
if ord(i) < 95:
big_letter += 1
else:
small_letter += 1
print(word.lower() if small_letter >= big_letter else word.upper())
``` | 3.9695 |
977 | A | Wrong Subtraction | PROGRAMMING | 800 | [
"implementation"
] | null | null | Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number. | The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly. | Print one integer number — the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number. | [
"512 4\n",
"1000000000 9\n"
] | [
"50\n",
"1\n"
] | The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$. | 0 | [
{
"input": "512 4",
"output": "50"
},
{
"input": "1000000000 9",
"output": "1"
},
{
"input": "131203 11",
"output": "12"
},
{
"input": "999999999 50",
"output": "9999"
},
{
"input": "999999999 49",
"output": "99990"
},
{
"input": "131203 9",
"output": "130"
},
{
"input": "900000000 16",
"output": "1"
},
{
"input": "909090909 50",
"output": "3"
},
{
"input": "1001 2",
"output": "100"
},
{
"input": "5 2",
"output": "3"
},
{
"input": "2 1",
"output": "1"
}
] | 1,696,433,373 | 2,147,483,647 | Python 3 | OK | TESTS | 11 | 46 | 0 | s = input()
l = list(map(int, s.split()))
for i in range(0,l[1]):
if(l[0]%10==0):
l[0]=l[0]//10
else:
l[0]=l[0]-1
print(l[0])
| Title: Wrong Subtraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number.
Input Specification:
The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly.
Output Specification:
Print one integer number — the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number.
Demo Input:
['512 4\n', '1000000000 9\n']
Demo Output:
['50\n', '1\n']
Note:
The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$. | ```python
s = input()
l = list(map(int, s.split()))
for i in range(0,l[1]):
if(l[0]%10==0):
l[0]=l[0]//10
else:
l[0]=l[0]-1
print(l[0])
``` | 3 |
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