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279 | B | Books | PROGRAMMING | 1,400 | [
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null | When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read. | The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book. | Print a single integer — the maximum number of books Valera can read. | [
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] | [
"3\n",
"1\n"
] | none | 1,000 | [
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4"
},
{
"input": "7 13\n6 8 14 9 4 11 10",
"output": "2"
},
{
"input": "10 15\n10 9 1 1 5 10 5 3 7 2",
"output": "3"
},
{
"input": "20 30\n8 1 2 6 9 4 1 9 9 10 4 7 8 9 5 7 1 8 7 4",
"output": "6"
},
{
"input": "30 60\n16 13 22 38 13 35 17 17 20 38 12 19 9 22 20 3 35 34 34 21 35 40 22 3 27 19 12 4 8 19",
"output": "4"
},
{
"input": "100 100\n75 92 18 6 81 67 7 92 100 65 82 32 50 67 85 31 80 91 84 63 39 52 92 81 1 98 24 12 43 48 17 86 51 72 48 95 45 50 12 66 19 79 49 89 34 1 97 75 20 33 96 27 42 23 73 71 93 1 85 19 66 14 17 61 20 39 36 33 42 61 56 64 23 91 80 99 40 74 13 18 98 85 74 39 62 84 46 74 50 23 38 11 79 14 9 25 66 100 25 52",
"output": "3"
},
{
"input": "10 1\n4418 7528 8170 1736 1317 3205 8183 4995 8039 4708",
"output": "0"
},
{
"input": "50 2\n124 214 63 73 996 760 38 571 451 300 970 1 706 937 837 494 619 88 851 411 957 990 842 613 821 649 627 34 693 678 734 116 816 985 705 940 499 493 922 967 854 439 112 644 961 438 189 572 655 550",
"output": "1"
}
] | 1,672,980,444 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 5 | 186 | 2,150,400 | from math import gcd, lcm, floor, sqrt, isqrt
import sys
from collections import defaultdict, deque
from functools import reduce
from heapq import heapify, heappop, heappush
from sys import maxsize as INF
from time import perf_counter
from random import randint
# === i/o ===
def reads(): return sys.stdin.readline().strip("\n")
def readi(): return int(sys.stdin.readline().strip("\n"))
def readf(): return float(sys.stdin.readline().strip("\n"))
def readp(): return sys.stdin.readline().strip("\n").split(" ")
def readpi(): return map(int, sys.stdin.readline().strip("\n").split(" "))
def readli(): return list(map(int, sys.stdin.readline().strip("\n").split(" ")))
def readpt(*args):
s = sys.stdin.readline().strip("\n").split(" ")
return (args[i](s[i]) for i in range(len(s)))
def yes(): sys.stdout.write("YES")
def no(): sys.stdout.write("NO")
def out(s): sys.stdout.write(str(s) + "\n")
def outpl(a):
for n in a: sys.stdout.write(str(n) + " ")
sys.stdout.write("\n")
def outpa(*args):
for n in args: sys.stdout.write(str(n) + " ")
sys.stdout.write("\n")
# === CODE HERE ===
def solve(n, t, a):
a.sort()
ans = 0
for i in a:
if t >= i:
t -= i
ans += 1
else:
break
return ans
n, t = readpi()
a = readli()
ans = solve(n, t, a)
out(ans) | Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer — the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
from math import gcd, lcm, floor, sqrt, isqrt
import sys
from collections import defaultdict, deque
from functools import reduce
from heapq import heapify, heappop, heappush
from sys import maxsize as INF
from time import perf_counter
from random import randint
# === i/o ===
def reads(): return sys.stdin.readline().strip("\n")
def readi(): return int(sys.stdin.readline().strip("\n"))
def readf(): return float(sys.stdin.readline().strip("\n"))
def readp(): return sys.stdin.readline().strip("\n").split(" ")
def readpi(): return map(int, sys.stdin.readline().strip("\n").split(" "))
def readli(): return list(map(int, sys.stdin.readline().strip("\n").split(" ")))
def readpt(*args):
s = sys.stdin.readline().strip("\n").split(" ")
return (args[i](s[i]) for i in range(len(s)))
def yes(): sys.stdout.write("YES")
def no(): sys.stdout.write("NO")
def out(s): sys.stdout.write(str(s) + "\n")
def outpl(a):
for n in a: sys.stdout.write(str(n) + " ")
sys.stdout.write("\n")
def outpa(*args):
for n in args: sys.stdout.write(str(n) + " ")
sys.stdout.write("\n")
# === CODE HERE ===
def solve(n, t, a):
a.sort()
ans = 0
for i in a:
if t >= i:
t -= i
ans += 1
else:
break
return ans
n, t = readpi()
a = readli()
ans = solve(n, t, a)
out(ans)
``` | 0 |
|
32 | A | Reconnaissance | PROGRAMMING | 800 | [
"brute force"
] | A. Reconnaissance | 2 | 256 | According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most *d* centimeters. Captain Bob has *n* soldiers in his detachment. Their heights are *a*1,<=*a*2,<=...,<=*a**n* centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1,<=2) and (2,<=1) should be regarded as different. | The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*d*<=≤<=109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains *n* space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109. | Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed *d*. | [
"5 10\n10 20 50 60 65\n",
"5 1\n55 30 29 31 55\n"
] | [
"6\n",
"6\n"
] | none | 500 | [
{
"input": "5 10\n10 20 50 60 65",
"output": "6"
},
{
"input": "5 1\n55 30 29 31 55",
"output": "6"
},
{
"input": "6 10\n4 6 4 1 9 3",
"output": "30"
},
{
"input": "7 100\n19 1694 261 162 1 234 513",
"output": "8"
},
{
"input": "8 42\n37 53 74 187 568 22 5 65",
"output": "20"
},
{
"input": "10 4\n11 6 76 49 28 20 57 152 5 32",
"output": "4"
},
{
"input": "100 100\n51 93 101 960 2 477 213 129 663 925 254 78 1486 274 160 481 132 156 412 372 5 57 152 298 1771 7 359 468 254 406 202 929 221 366 552 97 555 29 822 118 539 140 992 854 7 163 134 103 940 30 409 1003 398 43 555 79 107 40 23 103 643 171 310 382 770 337 18 189 570 177 29 54 855 171 205 291 299 935 620 180 114 358 88 292 118 400 218 537 369 60 683 192 13 537 59 824 264 191 3 300",
"output": "2404"
}
] | 1,544,364,532 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 624 | 0 | n,d = map(int, input().split())
h = [int(x) for x in input().split()]
cnt = 0
for i in range(0,n-1):
for j in range(i+1,n):
if abs(h[i]-h[j]) <= d:
cnt += 2
print(cnt) | Title: Reconnaissance
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most *d* centimeters. Captain Bob has *n* soldiers in his detachment. Their heights are *a*1,<=*a*2,<=...,<=*a**n* centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1,<=2) and (2,<=1) should be regarded as different.
Input Specification:
The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*d*<=≤<=109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains *n* space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109.
Output Specification:
Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed *d*.
Demo Input:
['5 10\n10 20 50 60 65\n', '5 1\n55 30 29 31 55\n']
Demo Output:
['6\n', '6\n']
Note:
none | ```python
n,d = map(int, input().split())
h = [int(x) for x in input().split()]
cnt = 0
for i in range(0,n-1):
for j in range(i+1,n):
if abs(h[i]-h[j]) <= d:
cnt += 2
print(cnt)
``` | 3.844 |
893 | C | Rumor | PROGRAMMING | 1,300 | [
"dfs and similar",
"graphs",
"greedy"
] | null | null | Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely. | The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once. | Print one number — the minimum amount of gold Vova has to spend in order to finish the quest. | [
"5 2\n2 5 3 4 8\n1 4\n4 5\n",
"10 0\n1 2 3 4 5 6 7 8 9 10\n",
"10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n"
] | [
"10\n",
"55\n",
"15\n"
] | In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters. | 0 | [
{
"input": "5 2\n2 5 3 4 8\n1 4\n4 5",
"output": "10"
},
{
"input": "10 0\n1 2 3 4 5 6 7 8 9 10",
"output": "55"
},
{
"input": "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10",
"output": "15"
},
{
"input": "1 0\n0",
"output": "0"
},
{
"input": "1 0\n1000000000",
"output": "1000000000"
},
{
"input": "2 0\n0 0",
"output": "0"
},
{
"input": "2 0\n1000000000 0",
"output": "1000000000"
},
{
"input": "2 0\n0 1000000000",
"output": "1000000000"
},
{
"input": "2 0\n1000000000 1000000000",
"output": "2000000000"
},
{
"input": "2 1\n0 0\n1 2",
"output": "0"
}
] | 1,656,502,595 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 93 | 2,150,400 | import sys
input = sys.stdin.readline
n, m = map(int, input().split())
w = list(map(int, input().split()))
d = list(range(n+1))
def F(x):
while d[x] != x:
x = d[x]
return x
for i in range(m):
a, b = map(F, map(int, input().split()))
print(a,b)
if w[a] > w[b]:
d[a] = b
else:
d[b] = a
print(sum(w[i] for i in range(1, n+1) if d[i] == i)) | Title: Rumor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely.
Input Specification:
The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.
Output Specification:
Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.
Demo Input:
['5 2\n2 5 3 4 8\n1 4\n4 5\n', '10 0\n1 2 3 4 5 6 7 8 9 10\n', '10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n']
Demo Output:
['10\n', '55\n', '15\n']
Note:
In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters. | ```python
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
w = list(map(int, input().split()))
d = list(range(n+1))
def F(x):
while d[x] != x:
x = d[x]
return x
for i in range(m):
a, b = map(F, map(int, input().split()))
print(a,b)
if w[a] > w[b]:
d[a] = b
else:
d[b] = a
print(sum(w[i] for i in range(1, n+1) if d[i] == i))
``` | -1 |
|
604 | B | More Cowbell | PROGRAMMING | 1,400 | [
"binary search",
"greedy"
] | null | null | Kevin Sun wants to move his precious collection of *n* cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into *k* boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his *i*-th (1<=≤<=*i*<=≤<=*n*) cowbell is an integer *s**i*. In fact, he keeps his cowbells sorted by size, so *s**i*<=-<=1<=≤<=*s**i* for any *i*<=><=1. Also an expert packer, Kevin can fit one or two cowbells into a box of size *s* if and only if the sum of their sizes does not exceed *s*. Given this information, help Kevin determine the smallest *s* for which it is possible to put all of his cowbells into *k* boxes of size *s*. | The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=2·*k*<=≤<=100<=000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains *n* space-separated integers *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*s*1<=≤<=*s*2<=≤<=...<=≤<=*s**n*<=≤<=1<=000<=000), the sizes of Kevin's cowbells. It is guaranteed that the sizes *s**i* are given in non-decreasing order. | Print a single integer, the smallest *s* for which it is possible for Kevin to put all of his cowbells into *k* boxes of size *s*. | [
"2 1\n2 5\n",
"4 3\n2 3 5 9\n",
"3 2\n3 5 7\n"
] | [
"7\n",
"9\n",
"8\n"
] | In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | 1,000 | [
{
"input": "2 1\n2 5",
"output": "7"
},
{
"input": "4 3\n2 3 5 9",
"output": "9"
},
{
"input": "3 2\n3 5 7",
"output": "8"
},
{
"input": "20 11\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10 10\n3 15 31 61 63 63 68 94 98 100",
"output": "100"
},
{
"input": "100 97\n340 402 415 466 559 565 649 689 727 771 774 776 789 795 973 1088 1212 1293 1429 1514 1587 1599 1929 1997 2278 2529 2656 2677 2839 2894 2951 3079 3237 3250 3556 3568 3569 3578 3615 3641 3673 3892 4142 4418 4515 4766 4846 4916 5225 5269 5352 5460 5472 5635 5732 5886 5941 5976 5984 6104 6113 6402 6409 6460 6550 6563 6925 7006 7289 7401 7441 7451 7709 7731 7742 7750 7752 7827 8101 8154 8376 8379 8432 8534 8578 8630 8706 8814 8882 8972 9041 9053 9109 9173 9473 9524 9547 9775 9791 9983",
"output": "9983"
},
{
"input": "10 9\n7 29 35 38 41 47 54 56 73 74",
"output": "74"
},
{
"input": "1 2342\n12345",
"output": "12345"
},
{
"input": "10 5\n15 15 20 28 38 44 46 52 69 94",
"output": "109"
},
{
"input": "10 9\n6 10 10 32 36 38 69 80 82 93",
"output": "93"
},
{
"input": "10 10\n4 19 22 24 25 43 49 56 78 88",
"output": "88"
},
{
"input": "100 89\n474 532 759 772 803 965 1043 1325 1342 1401 1411 1452 1531 1707 1906 1928 2034 2222 2335 2606 2757 2968 2978 3211 3513 3734 3772 3778 3842 3948 3976 4038 4055 4113 4182 4267 4390 4408 4478 4595 4668 4792 4919 5133 5184 5255 5312 5341 5476 5628 5683 5738 5767 5806 5973 6051 6134 6254 6266 6279 6314 6342 6599 6676 6747 6777 6827 6842 7057 7097 7259 7340 7378 7405 7510 7520 7698 7796 8148 8351 8507 8601 8805 8814 8826 8978 9116 9140 9174 9338 9394 9403 9407 9423 9429 9519 9764 9784 9838 9946",
"output": "9946"
},
{
"input": "100 74\n10 211 323 458 490 592 979 981 1143 1376 1443 1499 1539 1612 1657 1874 2001 2064 2123 2274 2346 2471 2522 2589 2879 2918 2933 2952 3160 3164 3167 3270 3382 3404 3501 3522 3616 3802 3868 3985 4007 4036 4101 4580 4687 4713 4714 4817 4955 5257 5280 5343 5428 5461 5566 5633 5727 5874 5925 6233 6309 6389 6500 6701 6731 6847 6916 7088 7088 7278 7296 7328 7564 7611 7646 7887 7887 8065 8075 8160 8300 8304 8316 8355 8404 8587 8758 8794 8890 9038 9163 9235 9243 9339 9410 9587 9868 9916 9923 9986",
"output": "9986"
},
{
"input": "100 61\n82 167 233 425 432 456 494 507 562 681 683 921 1218 1323 1395 1531 1586 1591 1675 1766 1802 1842 2116 2625 2697 2735 2739 3337 3349 3395 3406 3596 3610 3721 4059 4078 4305 4330 4357 4379 4558 4648 4651 4784 4819 4920 5049 5312 5361 5418 5440 5463 5547 5594 5821 5951 5972 6141 6193 6230 6797 6842 6853 6854 7017 7026 7145 7322 7391 7460 7599 7697 7756 7768 7872 7889 8094 8215 8408 8440 8462 8714 8756 8760 8881 9063 9111 9184 9281 9373 9406 9417 9430 9511 9563 9634 9660 9788 9883 9927",
"output": "9927"
},
{
"input": "100 84\n53 139 150 233 423 570 786 861 995 1017 1072 1196 1276 1331 1680 1692 1739 1748 1826 2067 2280 2324 2368 2389 2607 2633 2760 2782 2855 2996 3030 3093 3513 3536 3557 3594 3692 3707 3823 3832 4009 4047 4088 4095 4408 4537 4565 4601 4784 4878 4935 5029 5252 5322 5389 5407 5511 5567 5857 6182 6186 6198 6280 6290 6353 6454 6458 6567 6843 7166 7216 7257 7261 7375 7378 7539 7542 7762 7771 7797 7980 8363 8606 8612 8663 8801 8808 8823 8918 8975 8997 9240 9245 9259 9356 9755 9759 9760 9927 9970",
"output": "9970"
},
{
"input": "100 50\n130 248 312 312 334 589 702 916 921 1034 1047 1346 1445 1500 1585 1744 1951 2123 2273 2362 2400 2455 2496 2530 2532 2944 3074 3093 3094 3134 3698 3967 4047 4102 4109 4260 4355 4466 4617 4701 4852 4892 4915 4917 4936 4981 4999 5106 5152 5203 5214 5282 5412 5486 5525 5648 5897 5933 5969 6251 6400 6421 6422 6558 6805 6832 6908 6924 6943 6980 7092 7206 7374 7417 7479 7546 7672 7756 7973 8020 8028 8079 8084 8085 8137 8153 8178 8239 8639 8667 8829 9263 9333 9370 9420 9579 9723 9784 9841 9993",
"output": "11103"
},
{
"input": "100 50\n156 182 208 409 496 515 659 761 772 794 827 912 1003 1236 1305 1388 1412 1422 1428 1465 1613 2160 2411 2440 2495 2684 2724 2925 3033 3035 3155 3260 3378 3442 3483 3921 4031 4037 4091 4113 4119 4254 4257 4442 4559 4614 4687 4839 4896 5054 5246 5316 5346 5859 5928 5981 6148 6250 6422 6433 6448 6471 6473 6485 6503 6779 6812 7050 7064 7074 7141 7378 7424 7511 7574 7651 7808 7858 8286 8291 8446 8536 8599 8628 8636 8768 8900 8981 9042 9055 9114 9146 9186 9411 9480 9590 9681 9749 9757 9983",
"output": "10676"
},
{
"input": "100 50\n145 195 228 411 577 606 629 775 1040 1040 1058 1187 1307 1514 1784 1867 1891 2042 2042 2236 2549 2555 2560 2617 2766 2807 2829 2917 3070 3072 3078 3095 3138 3147 3149 3196 3285 3287 3309 3435 3531 3560 3563 3769 3830 3967 4081 4158 4315 4387 4590 4632 4897 4914 5128 5190 5224 5302 5402 5416 5420 5467 5517 5653 5820 5862 5941 6053 6082 6275 6292 6316 6490 6530 6619 6632 6895 7071 7234 7323 7334 7412 7626 7743 8098 8098 8136 8158 8264 8616 8701 8718 8770 8803 8809 8983 9422 9530 9811 9866",
"output": "10011"
},
{
"input": "100 50\n56 298 387 456 518 532 589 792 870 1041 1055 1122 1141 1166 1310 1329 1523 1548 1626 1730 1780 1833 1850 1911 2006 2157 2303 2377 2403 2442 2450 2522 2573 2822 2994 3200 3238 3252 3280 3311 3345 3422 3429 3506 3526 3617 3686 3791 4134 4467 4525 4614 4633 4792 5017 5220 5243 5338 5445 5536 5639 5675 5763 5875 6129 6220 6228 6287 6385 6616 6789 6822 6940 6959 6985 7297 7304 7391 7443 7580 7824 7884 7981 8055 8063 8223 8280 8322 8346 8473 8688 8986 9087 9323 9477 9489 9518 9621 9699 9804",
"output": "10009"
},
{
"input": "100 49990\n221 257 543 613 818 855 884 1195 1233 1293 1327 1474 1567 1568 1728 1789 1809 1957 1994 2198 2217 2289 2298 2360 2441 2454 2517 2519 2597 2652 2909 2958 2970 3271 3310 3409 3450 3646 3950 3965 4016 4116 4134 4159 4462 4624 4631 4783 4804 5031 5276 5538 5729 5845 5880 5894 5959 5961 6028 6105 6184 6335 6370 6523 6532 6677 6762 6868 6998 7066 7245 7360 7490 7492 7645 7677 7786 7804 8038 8131 8207 8400 8507 8556 8575 8599 8673 8676 8712 8816 9123 9158 9216 9369 9628 9707 9768 9800 9876 9934",
"output": "9934"
},
{
"input": "1 50000\n2",
"output": "2"
},
{
"input": "1 40000\n23456",
"output": "23456"
},
{
"input": "1 2\n10",
"output": "10"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 2\n2 3",
"output": "3"
},
{
"input": "4 2\n1 2 8 9",
"output": "10"
}
] | 1,549,467,395 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 171 | 7,680,000 | n,k = map(int,input().split())
a = list(map(int,input().split()))
if k >= n :
print(max(a))
else:
sum = max(a)
a = a[:2*(n-k)]
#print(a)
for i in range(len(a)):
sum = max(a[i]+a[- (i + 1)],sum)
print(sum) | Title: More Cowbell
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin Sun wants to move his precious collection of *n* cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into *k* boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his *i*-th (1<=≤<=*i*<=≤<=*n*) cowbell is an integer *s**i*. In fact, he keeps his cowbells sorted by size, so *s**i*<=-<=1<=≤<=*s**i* for any *i*<=><=1. Also an expert packer, Kevin can fit one or two cowbells into a box of size *s* if and only if the sum of their sizes does not exceed *s*. Given this information, help Kevin determine the smallest *s* for which it is possible to put all of his cowbells into *k* boxes of size *s*.
Input Specification:
The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=2·*k*<=≤<=100<=000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains *n* space-separated integers *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*s*1<=≤<=*s*2<=≤<=...<=≤<=*s**n*<=≤<=1<=000<=000), the sizes of Kevin's cowbells. It is guaranteed that the sizes *s**i* are given in non-decreasing order.
Output Specification:
Print a single integer, the smallest *s* for which it is possible for Kevin to put all of his cowbells into *k* boxes of size *s*.
Demo Input:
['2 1\n2 5\n', '4 3\n2 3 5 9\n', '3 2\n3 5 7\n']
Demo Output:
['7\n', '9\n', '8\n']
Note:
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | ```python
n,k = map(int,input().split())
a = list(map(int,input().split()))
if k >= n :
print(max(a))
else:
sum = max(a)
a = a[:2*(n-k)]
#print(a)
for i in range(len(a)):
sum = max(a[i]+a[- (i + 1)],sum)
print(sum)
``` | 3 |
|
385 | A | Bear and Raspberry | PROGRAMMING | 1,000 | [
"brute force",
"greedy",
"implementation"
] | null | null | The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following *n* days. According to the bear's data, on the *i*-th (1<=≤<=*i*<=≤<=*n*) day, the price for one barrel of honey is going to is *x**i* kilos of raspberry.
Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for *c* kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day *d* (1<=≤<=*d*<=<<=*n*), lent a barrel of honey and immediately (on day *d*) sell it according to a daily exchange rate. The next day (*d*<=+<=1) the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day *d*<=+<=1) give his friend the borrowed barrel of honey as well as *c* kilograms of raspberry for renting the barrel.
The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan. | The first line contains two space-separated integers, *n* and *c* (2<=≤<=*n*<=≤<=100,<=0<=≤<=*c*<=≤<=100), — the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100), the price of a honey barrel on day *i*. | Print a single integer — the answer to the problem. | [
"5 1\n5 10 7 3 20\n",
"6 2\n100 1 10 40 10 40\n",
"3 0\n1 2 3\n"
] | [
"3\n",
"97\n",
"0\n"
] | In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3.
In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97. | 500 | [
{
"input": "5 1\n5 10 7 3 20",
"output": "3"
},
{
"input": "6 2\n100 1 10 40 10 40",
"output": "97"
},
{
"input": "3 0\n1 2 3",
"output": "0"
},
{
"input": "2 0\n2 1",
"output": "1"
},
{
"input": "10 5\n10 1 11 2 12 3 13 4 14 5",
"output": "4"
},
{
"input": "100 4\n2 57 70 8 44 10 88 67 50 44 93 79 72 50 69 19 21 9 71 47 95 13 46 10 68 72 54 40 15 83 57 92 58 25 4 22 84 9 8 55 87 0 16 46 86 58 5 21 32 28 10 46 11 29 13 33 37 34 78 33 33 21 46 70 77 51 45 97 6 21 68 61 87 54 8 91 37 12 76 61 57 9 100 45 44 88 5 71 98 98 26 45 37 87 34 50 33 60 64 77",
"output": "87"
},
{
"input": "100 5\n15 91 86 53 18 52 26 89 8 4 5 100 11 64 88 91 35 57 67 72 71 71 69 73 97 23 11 1 59 86 37 82 6 67 71 11 7 31 11 68 21 43 89 54 27 10 3 33 8 57 79 26 90 81 6 28 24 7 33 50 24 13 27 85 4 93 14 62 37 67 33 40 7 48 41 4 14 9 95 10 64 62 7 93 23 6 28 27 97 64 26 83 70 0 97 74 11 82 70 93",
"output": "84"
},
{
"input": "6 100\n10 9 8 7 6 5",
"output": "0"
},
{
"input": "100 9\n66 71 37 41 23 38 77 11 74 13 51 26 93 56 81 17 12 70 85 37 54 100 14 99 12 83 44 16 99 65 13 48 92 32 69 33 100 57 58 88 25 45 44 85 5 41 82 15 37 18 21 45 3 68 33 9 52 64 8 73 32 41 87 99 26 26 47 24 79 93 9 44 11 34 85 26 14 61 49 38 25 65 49 81 29 82 28 23 2 64 38 13 77 68 67 23 58 57 83 46",
"output": "78"
},
{
"input": "100 100\n9 72 46 37 26 94 80 1 43 85 26 53 58 18 24 19 67 2 100 52 61 81 48 15 73 41 97 93 45 1 73 54 75 51 28 79 0 14 41 42 24 50 70 18 96 100 67 1 68 48 44 39 63 77 78 18 10 51 32 53 26 60 1 13 66 39 55 27 23 71 75 0 27 88 73 31 16 95 87 84 86 71 37 40 66 70 65 83 19 4 81 99 26 51 67 63 80 54 23 44",
"output": "0"
},
{
"input": "43 65\n32 58 59 75 85 18 57 100 69 0 36 38 79 95 82 47 7 55 28 88 27 88 63 71 80 86 67 53 69 37 99 54 81 19 55 12 2 17 84 77 25 26 62",
"output": "4"
},
{
"input": "12 64\n14 87 40 24 32 36 4 41 38 77 68 71",
"output": "0"
},
{
"input": "75 94\n80 92 25 48 78 17 69 52 79 73 12 15 59 55 25 61 96 27 98 43 30 43 36 94 67 54 86 99 100 61 65 8 65 19 18 21 75 31 2 98 55 87 14 1 17 97 94 11 57 29 34 71 76 67 45 0 78 29 86 82 29 23 77 100 48 43 65 62 88 34 7 28 13 1 1",
"output": "0"
},
{
"input": "59 27\n76 61 24 66 48 18 69 84 21 8 64 90 19 71 36 90 9 36 30 37 99 37 100 56 9 79 55 37 54 63 11 11 49 71 91 70 14 100 10 44 52 23 21 19 96 13 93 66 52 79 76 5 62 6 90 35 94 7 27",
"output": "63"
},
{
"input": "86 54\n41 84 16 5 20 79 73 13 23 24 42 73 70 80 69 71 33 44 62 29 86 88 40 64 61 55 58 19 16 23 84 100 38 91 89 98 47 50 55 87 12 94 2 12 0 1 4 26 50 96 68 34 94 80 8 22 60 3 72 84 65 89 44 52 50 9 24 34 81 28 56 17 38 85 78 90 62 60 1 40 91 2 7 41 84 22",
"output": "38"
},
{
"input": "37 2\n65 36 92 92 92 76 63 56 15 95 75 26 15 4 73 50 41 92 26 20 19 100 63 55 25 75 61 96 35 0 14 6 96 3 28 41 83",
"output": "91"
},
{
"input": "19 4\n85 2 56 70 33 75 89 60 100 81 42 28 18 92 29 96 49 23 14",
"output": "79"
},
{
"input": "89 1\n50 53 97 41 68 27 53 66 93 19 11 78 46 49 38 69 96 9 43 16 1 63 95 64 96 6 34 34 45 40 19 4 53 8 11 18 95 25 50 16 64 33 97 49 23 81 63 10 30 73 76 55 7 70 9 98 6 36 75 78 3 92 85 75 40 75 55 71 9 91 15 17 47 55 44 35 55 88 53 87 61 22 100 56 14 87 36 84 24",
"output": "91"
},
{
"input": "67 0\n40 48 15 46 90 7 65 52 24 15 42 81 2 6 71 94 32 18 97 67 83 98 48 51 10 47 8 68 36 46 65 75 90 30 62 9 5 35 80 60 69 58 62 68 58 73 80 9 22 46 56 64 44 11 93 73 62 54 15 20 17 69 16 33 85 62 49",
"output": "83"
},
{
"input": "96 0\n38 97 82 43 80 40 1 99 50 94 81 63 92 13 57 24 4 10 25 32 79 56 96 19 25 14 69 56 66 22 23 78 87 76 37 30 75 77 61 64 35 64 62 32 44 62 6 84 91 44 99 5 71 19 17 12 35 52 1 14 35 18 8 36 54 42 4 67 80 11 88 44 34 35 12 38 66 42 4 90 45 10 1 44 37 96 23 28 100 90 75 17 27 67 51 70",
"output": "94"
},
{
"input": "14 14\n87 63 62 31 59 47 40 89 92 43 80 30 99 42",
"output": "43"
},
{
"input": "12 0\n100 1 100 2 100 3 100 4 100 5 100 0",
"output": "100"
},
{
"input": "3 1\n1 2 3",
"output": "0"
},
{
"input": "3 2\n3 3 3",
"output": "0"
},
{
"input": "3 3\n3 2 1",
"output": "0"
},
{
"input": "3 100\n1 2 3",
"output": "0"
},
{
"input": "2 100\n0 0",
"output": "0"
},
{
"input": "2 90\n10 5",
"output": "0"
},
{
"input": "2 5\n5 4",
"output": "0"
},
{
"input": "3 1\n19 20 1",
"output": "18"
},
{
"input": "5 1\n5 10 7 4 20",
"output": "2"
},
{
"input": "5 1\n1 2 3 4 5",
"output": "0"
}
] | 1,596,597,090 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 109 | 6,963,200 | x,y =[int(x) for x in input("").split()]
ar= list(map(int, input("").split()))
diff=[]
for i in range(len(ar)-1):
di=ar[i]-ar[i+1]
diff.append(di)
d =diff.index(max(diff))
j=ar[d]-ar[d+1]-y
if(j>0):
print(j)
else:
print(0)
| Title: Bear and Raspberry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following *n* days. According to the bear's data, on the *i*-th (1<=≤<=*i*<=≤<=*n*) day, the price for one barrel of honey is going to is *x**i* kilos of raspberry.
Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for *c* kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day *d* (1<=≤<=*d*<=<<=*n*), lent a barrel of honey and immediately (on day *d*) sell it according to a daily exchange rate. The next day (*d*<=+<=1) the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day *d*<=+<=1) give his friend the borrowed barrel of honey as well as *c* kilograms of raspberry for renting the barrel.
The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan.
Input Specification:
The first line contains two space-separated integers, *n* and *c* (2<=≤<=*n*<=≤<=100,<=0<=≤<=*c*<=≤<=100), — the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100), the price of a honey barrel on day *i*.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['5 1\n5 10 7 3 20\n', '6 2\n100 1 10 40 10 40\n', '3 0\n1 2 3\n']
Demo Output:
['3\n', '97\n', '0\n']
Note:
In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3.
In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97. | ```python
x,y =[int(x) for x in input("").split()]
ar= list(map(int, input("").split()))
diff=[]
for i in range(len(ar)-1):
di=ar[i]-ar[i+1]
diff.append(di)
d =diff.index(max(diff))
j=ar[d]-ar[d+1]-y
if(j>0):
print(j)
else:
print(0)
``` | 3 |
|
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"
}
] | 1,622,300,950 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 154 | 512,000 | #!/usr/bin/env python
# Header
import re
# Input
get_INPUT = input()
# Logic for Word
lower,upper = 0,0
for i in get_INPUT:
if i.isupper():
upper = upper+1
elif i.islower():
lower = lower+1
if lower > upper or lower == upper:
print(get_INPUT.lower())
elif upper > lower:
print(get_INPUT.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
#!/usr/bin/env python
# Header
import re
# Input
get_INPUT = input()
# Logic for Word
lower,upper = 0,0
for i in get_INPUT:
if i.isupper():
upper = upper+1
elif i.islower():
lower = lower+1
if lower > upper or lower == upper:
print(get_INPUT.lower())
elif upper > lower:
print(get_INPUT.upper())
``` | 3.960546 |
998 | B | Cutting | PROGRAMMING | 1,200 | [
"dp",
"greedy",
"sortings"
] | null | null | There are a lot of things which could be cut — trees, paper, "the rope". In this problem you are going to cut a sequence of integers.
There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.
Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$ $\to$ two cuts $\to$ $[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$. On each segment the number of even elements should be equal to the number of odd elements.
The cost of the cut between $x$ and $y$ numbers is $|x - y|$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $B$ bitcoins. | First line of the input contains an integer $n$ ($2 \le n \le 100$) and an integer $B$ ($1 \le B \le 100$) — the number of elements in the sequence and the number of bitcoins you have.
Second line contains $n$ integers: $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 100$) — elements of the sequence, which contains the equal number of even and odd numbers | Print the maximum possible number of cuts which can be made while spending no more than $B$ bitcoins. | [
"6 4\n1 2 5 10 15 20\n",
"4 10\n1 3 2 4\n",
"6 100\n1 2 3 4 5 6\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first sample the optimal answer is to split sequence between $2$ and $5$. Price of this cut is equal to $3$ bitcoins.
In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.
In the third sample the sequence should be cut between $2$ and $3$, and between $4$ and $5$. The total price of the cuts is $1 + 1 = 2$ bitcoins. | 1,000 | [
{
"input": "6 4\n1 2 5 10 15 20",
"output": "1"
},
{
"input": "4 10\n1 3 2 4",
"output": "0"
},
{
"input": "6 100\n1 2 3 4 5 6",
"output": "2"
},
{
"input": "2 100\n13 78",
"output": "0"
},
{
"input": "10 1\n56 56 98 2 11 64 97 41 95 53",
"output": "0"
},
{
"input": "10 100\n94 65 24 47 29 98 20 65 6 17",
"output": "2"
},
{
"input": "100 1\n35 6 19 84 49 64 36 91 50 65 21 86 20 89 10 52 50 24 98 74 11 48 58 98 51 85 1 29 44 83 9 97 68 41 83 57 1 57 46 42 87 2 32 50 3 57 17 77 22 100 36 27 3 34 55 8 90 61 34 20 15 39 43 46 60 60 14 23 4 22 75 51 98 23 69 22 99 57 63 30 79 7 16 8 34 84 13 47 93 40 48 25 93 1 80 6 82 93 6 21",
"output": "0"
},
{
"input": "100 10\n3 20 3 29 90 69 2 30 70 28 71 99 22 99 34 70 87 48 3 92 71 61 26 90 14 38 51 81 16 33 49 71 14 52 50 95 65 16 80 57 87 47 29 14 40 31 74 15 87 76 71 61 30 91 44 10 87 48 84 12 77 51 25 68 49 38 79 8 7 9 39 19 48 40 15 53 29 4 60 86 76 84 6 37 45 71 46 38 80 68 94 71 64 72 41 51 71 60 79 7",
"output": "2"
},
{
"input": "100 100\n60 83 82 16 17 7 89 6 83 100 85 41 72 44 23 28 64 84 3 23 33 52 93 30 81 38 67 25 26 97 94 78 41 74 74 17 53 51 54 17 20 81 95 76 42 16 16 56 74 69 30 9 82 91 32 13 47 45 97 40 56 57 27 28 84 98 91 5 61 20 3 43 42 26 83 40 34 100 5 63 62 61 72 5 32 58 93 79 7 18 50 43 17 24 77 73 87 74 98 2",
"output": "11"
},
{
"input": "100 100\n70 54 10 72 81 84 56 15 27 19 43 100 49 44 52 33 63 40 95 17 58 2 51 39 22 18 82 1 16 99 32 29 24 94 9 98 5 37 47 14 42 73 41 31 79 64 12 6 53 26 68 67 89 13 90 4 21 93 46 74 75 88 66 57 23 7 25 48 92 62 30 8 50 61 38 87 71 34 97 28 80 11 60 91 3 35 86 96 36 20 59 65 83 45 76 77 78 69 85 55",
"output": "3"
},
{
"input": "100 100\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": "49"
},
{
"input": "10 10\n94 32 87 13 4 22 85 81 18 95",
"output": "1"
},
{
"input": "10 50\n40 40 9 3 64 96 67 19 21 30",
"output": "1"
},
{
"input": "100 50\n13 31 29 86 46 10 2 87 94 2 28 31 29 15 64 3 94 71 37 76 9 91 89 38 12 46 53 33 58 11 98 4 37 72 30 52 6 86 40 98 28 6 34 80 61 47 45 69 100 47 91 64 87 41 67 58 88 75 13 81 36 58 66 29 10 27 54 83 44 15 11 33 49 36 61 18 89 26 87 1 99 19 57 21 55 84 20 74 14 43 15 51 2 76 22 92 43 14 72 77",
"output": "3"
},
{
"input": "100 1\n78 52 95 76 96 49 53 59 77 100 64 11 9 48 15 17 44 46 21 54 39 68 43 4 32 28 73 6 16 62 72 84 65 86 98 75 33 45 25 3 91 82 2 92 63 88 7 50 97 93 14 22 20 42 60 55 80 85 29 34 56 71 83 38 26 47 90 70 51 41 40 31 37 12 35 99 67 94 1 87 57 8 61 19 23 79 36 18 66 74 5 27 81 69 24 58 13 10 89 30",
"output": "0"
},
{
"input": "100 10\n19 55 91 50 31 23 60 84 38 1 22 51 27 76 28 98 11 44 61 63 15 93 52 3 66 16 53 36 18 62 35 85 78 37 73 64 87 74 46 26 82 69 49 33 83 89 56 67 71 25 39 94 96 17 21 6 47 68 34 42 57 81 13 10 54 2 48 80 20 77 4 5 59 30 90 95 45 75 8 88 24 41 40 14 97 32 7 9 65 70 100 99 72 58 92 29 79 12 86 43",
"output": "0"
},
{
"input": "100 50\n2 4 82 12 47 63 52 91 87 45 53 1 17 25 64 50 9 13 22 54 21 30 43 24 38 33 68 11 41 78 99 23 28 18 58 67 79 10 71 56 49 61 26 29 59 20 90 74 5 75 89 8 39 95 72 42 66 98 44 32 88 35 92 3 97 55 65 51 77 27 81 76 84 69 73 85 19 46 62 100 60 37 7 36 57 6 14 83 40 48 16 70 96 15 31 93 80 86 94 34",
"output": "1"
},
{
"input": "100 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 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": "1"
},
{
"input": "100 10\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": "10"
},
{
"input": "100 50\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": "49"
},
{
"input": "100 30\n2 1 2 2 2 2 1 1 1 2 1 1 2 2 1 2 1 2 2 2 2 1 2 1 2 1 1 2 1 1 2 2 2 1 1 2 1 2 2 2 1 1 1 1 1 2 1 1 1 1 1 2 2 2 2 1 2 1 1 1 2 2 2 2 1 2 2 1 1 1 1 2 2 2 1 2 2 1 2 1 1 2 2 2 1 2 2 1 2 1 1 2 1 1 1 1 2 1 1 2",
"output": "11"
},
{
"input": "100 80\n1 1 1 2 2 1 1 2 1 1 1 1 2 2 2 1 2 2 2 2 1 1 2 2 1 1 1 1 2 2 2 1 1 1 1 1 1 1 2 2 2 2 1 2 2 1 2 1 1 1 1 2 2 1 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 1 1 2 1 1 1 2 1 1 2 1 2 1 2 2 1 1 2 1 1 1 1 2 2 2 1 2 2 1 2",
"output": "12"
},
{
"input": "100 30\n100 99 100 99 99 100 100 99 100 99 99 100 100 100 99 99 99 100 99 99 99 99 100 99 99 100 100 99 100 99 99 99 100 99 100 100 99 100 100 100 100 100 99 99 100 99 99 100 99 100 99 99 100 100 99 100 99 99 100 99 100 100 100 100 99 99 99 100 99 100 99 100 100 100 99 100 100 100 99 100 99 99 100 100 100 100 99 99 99 100 99 100 100 99 99 99 100 100 99 99",
"output": "14"
},
{
"input": "100 80\n99 100 100 100 99 99 99 99 100 99 99 99 99 99 99 99 99 100 100 99 99 99 99 99 100 99 100 99 100 100 100 100 100 99 100 100 99 99 100 100 100 100 100 99 100 99 100 99 99 99 100 99 99 99 99 99 99 99 99 100 99 100 100 99 99 99 99 100 100 100 99 100 100 100 100 100 99 100 100 100 100 100 100 100 100 99 99 99 99 100 99 100 100 100 100 100 99 100 99 100",
"output": "4"
},
{
"input": "100 30\n100 100 39 39 39 100 100 39 39 100 39 39 100 39 100 39 100 100 100 100 100 39 100 100 100 39 39 39 100 39 100 100 39 39 100 39 39 39 100 100 39 100 39 100 39 39 100 100 39 100 39 100 39 39 39 100 39 100 39 39 39 100 39 39 100 100 39 39 39 100 100 39 39 39 100 100 100 100 39 100 100 100 39 39 100 39 100 100 39 100 39 100 39 39 100 39 39 100 100 100",
"output": "5"
},
{
"input": "100 80\n39 100 39 100 100 100 100 39 39 100 100 39 39 100 39 39 39 39 100 39 39 39 39 100 100 100 100 39 100 39 39 100 100 39 39 100 39 100 39 100 100 39 39 100 39 39 39 100 39 100 39 100 100 100 100 100 100 100 39 100 39 100 100 100 39 39 39 39 39 100 100 100 39 100 100 100 100 39 100 100 39 39 100 39 39 39 100 39 100 39 39 100 100 39 100 39 39 39 100 39",
"output": "6"
},
{
"input": "4 1\n1 2 3 4",
"output": "1"
},
{
"input": "4 1\n1 2 1 2",
"output": "1"
},
{
"input": "4 4\n1 2 6 7",
"output": "1"
},
{
"input": "4 8\n1 2 10 11",
"output": "1"
},
{
"input": "6 2\n1 2 3 4 5 6",
"output": "2"
},
{
"input": "6 1\n1 2 1 2 1 2",
"output": "1"
},
{
"input": "6 4\n1 2 4 5 7 8",
"output": "2"
},
{
"input": "6 3\n1 2 5 10 15 20",
"output": "1"
}
] | 1,547,630,869 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 109 | 0 | n,m=map(int,input().split())
l=list(map(int,input().split()))
x,y,ans,z,p=0,0,[],0,0
for i in range(n-1):
if l[i]%2==1:
x+=1
else:
y+=1
if x==y:
ans.append(abs(l[i]-l[i+1]))
ans.sort()
for i in range(len(ans)):
if p + ans[i]<=m:
z += 1
p += ans[i]
else:
break
print(z) | Title: Cutting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are a lot of things which could be cut — trees, paper, "the rope". In this problem you are going to cut a sequence of integers.
There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.
Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$ $\to$ two cuts $\to$ $[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$. On each segment the number of even elements should be equal to the number of odd elements.
The cost of the cut between $x$ and $y$ numbers is $|x - y|$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $B$ bitcoins.
Input Specification:
First line of the input contains an integer $n$ ($2 \le n \le 100$) and an integer $B$ ($1 \le B \le 100$) — the number of elements in the sequence and the number of bitcoins you have.
Second line contains $n$ integers: $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 100$) — elements of the sequence, which contains the equal number of even and odd numbers
Output Specification:
Print the maximum possible number of cuts which can be made while spending no more than $B$ bitcoins.
Demo Input:
['6 4\n1 2 5 10 15 20\n', '4 10\n1 3 2 4\n', '6 100\n1 2 3 4 5 6\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample the optimal answer is to split sequence between $2$ and $5$. Price of this cut is equal to $3$ bitcoins.
In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.
In the third sample the sequence should be cut between $2$ and $3$, and between $4$ and $5$. The total price of the cuts is $1 + 1 = 2$ bitcoins. | ```python
n,m=map(int,input().split())
l=list(map(int,input().split()))
x,y,ans,z,p=0,0,[],0,0
for i in range(n-1):
if l[i]%2==1:
x+=1
else:
y+=1
if x==y:
ans.append(abs(l[i]-l[i+1]))
ans.sort()
for i in range(len(ans)):
if p + ans[i]<=m:
z += 1
p += ans[i]
else:
break
print(z)
``` | 3 |
|
255 | A | Greg's Workout | PROGRAMMING | 800 | [
"implementation"
] | null | null | Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training. | The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises. | Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous. | [
"2\n2 8\n",
"3\n5 1 10\n",
"7\n3 3 2 7 9 6 8\n"
] | [
"biceps\n",
"back\n",
"chest\n"
] | In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise. | 500 | [
{
"input": "2\n2 8",
"output": "biceps"
},
{
"input": "3\n5 1 10",
"output": "back"
},
{
"input": "7\n3 3 2 7 9 6 8",
"output": "chest"
},
{
"input": "4\n5 6 6 2",
"output": "chest"
},
{
"input": "5\n8 2 2 6 3",
"output": "chest"
},
{
"input": "6\n8 7 2 5 3 4",
"output": "chest"
},
{
"input": "8\n7 2 9 10 3 8 10 6",
"output": "chest"
},
{
"input": "9\n5 4 2 3 4 4 5 2 2",
"output": "chest"
},
{
"input": "10\n4 9 8 5 3 8 8 10 4 2",
"output": "biceps"
},
{
"input": "11\n10 9 7 6 1 3 9 7 1 3 5",
"output": "chest"
},
{
"input": "12\n24 22 6 16 5 21 1 7 2 19 24 5",
"output": "chest"
},
{
"input": "13\n24 10 5 7 16 17 2 7 9 20 15 2 24",
"output": "chest"
},
{
"input": "14\n13 14 19 8 5 17 9 16 15 9 5 6 3 7",
"output": "back"
},
{
"input": "15\n24 12 22 21 25 23 21 5 3 24 23 13 12 16 12",
"output": "chest"
},
{
"input": "16\n12 6 18 6 25 7 3 1 1 17 25 17 6 8 17 8",
"output": "biceps"
},
{
"input": "17\n13 8 13 4 9 21 10 10 9 22 14 23 22 7 6 14 19",
"output": "chest"
},
{
"input": "18\n1 17 13 6 11 10 25 13 24 9 21 17 3 1 17 12 25 21",
"output": "back"
},
{
"input": "19\n22 22 24 25 19 10 7 10 4 25 19 14 1 14 3 18 4 19 24",
"output": "chest"
},
{
"input": "20\n9 8 22 11 18 14 15 10 17 11 2 1 25 20 7 24 4 25 9 20",
"output": "chest"
},
{
"input": "1\n10",
"output": "chest"
},
{
"input": "2\n15 3",
"output": "chest"
},
{
"input": "3\n21 11 19",
"output": "chest"
},
{
"input": "4\n19 24 13 15",
"output": "chest"
},
{
"input": "5\n4 24 1 9 19",
"output": "biceps"
},
{
"input": "6\n6 22 24 7 15 24",
"output": "back"
},
{
"input": "7\n10 8 23 23 14 18 14",
"output": "chest"
},
{
"input": "8\n5 16 8 9 17 16 14 7",
"output": "biceps"
},
{
"input": "9\n12 3 10 23 6 4 22 13 12",
"output": "chest"
},
{
"input": "10\n1 9 20 18 20 17 7 24 23 2",
"output": "back"
},
{
"input": "11\n22 25 8 2 18 15 1 13 1 11 4",
"output": "biceps"
},
{
"input": "12\n20 12 14 2 15 6 24 3 11 8 11 14",
"output": "chest"
},
{
"input": "13\n2 18 8 8 8 20 5 22 15 2 5 19 18",
"output": "back"
},
{
"input": "14\n1 6 10 25 17 13 21 11 19 4 15 24 5 22",
"output": "biceps"
},
{
"input": "15\n13 5 25 13 17 25 19 21 23 17 12 6 14 8 6",
"output": "back"
},
{
"input": "16\n10 15 2 17 22 12 14 14 6 11 4 13 9 8 21 14",
"output": "chest"
},
{
"input": "17\n7 22 9 22 8 7 20 22 23 5 12 11 1 24 17 20 10",
"output": "biceps"
},
{
"input": "18\n18 15 4 25 5 11 21 25 12 14 25 23 19 19 13 6 9 17",
"output": "chest"
},
{
"input": "19\n3 1 3 15 15 25 10 25 23 10 9 21 13 23 19 3 24 21 14",
"output": "back"
},
{
"input": "20\n19 18 11 3 6 14 3 3 25 3 1 19 25 24 23 12 7 4 8 6",
"output": "back"
},
{
"input": "1\n19",
"output": "chest"
},
{
"input": "2\n1 7",
"output": "biceps"
},
{
"input": "3\n18 18 23",
"output": "back"
},
{
"input": "4\n12 15 1 13",
"output": "chest"
},
{
"input": "5\n11 14 25 21 21",
"output": "biceps"
},
{
"input": "6\n11 9 12 11 22 18",
"output": "biceps"
},
{
"input": "7\n11 1 16 20 21 25 20",
"output": "chest"
},
{
"input": "8\n1 2 20 9 3 22 17 4",
"output": "back"
},
{
"input": "9\n19 2 10 19 15 20 3 1 13",
"output": "back"
},
{
"input": "10\n11 2 11 8 21 16 2 3 19 9",
"output": "back"
},
{
"input": "20\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 24",
"output": "chest"
},
{
"input": "12\n4 24 21 3 13 24 22 13 12 21 1 15",
"output": "back"
},
{
"input": "13\n14 14 16 2 13 5 1 14 9 4 16 8 3",
"output": "biceps"
},
{
"input": "14\n1 9 15 4 11 8 25 3 9 14 13 2 1 11",
"output": "biceps"
},
{
"input": "15\n4 19 10 6 16 12 5 11 7 23 1 24 11 7 17",
"output": "back"
},
{
"input": "16\n2 8 2 8 13 22 20 12 22 23 18 13 18 22 11 17",
"output": "chest"
},
{
"input": "17\n24 5 5 16 10 8 22 6 4 13 10 10 5 23 8 20 8",
"output": "chest"
},
{
"input": "18\n14 8 9 12 11 18 24 1 14 24 18 5 12 17 1 10 1 22",
"output": "chest"
},
{
"input": "19\n21 2 10 6 9 1 24 5 2 19 10 13 10 7 19 2 6 13 24",
"output": "chest"
},
{
"input": "20\n7 1 14 17 6 6 18 13 12 3 25 4 3 19 22 24 16 14 1 23",
"output": "biceps"
},
{
"input": "1\n19",
"output": "chest"
},
{
"input": "20\n2 1 2 2 1 2 2 1 2 1 1 1 1 1 1 1 1 1 1 22",
"output": "biceps"
}
] | 1,579,437,636 | 2,147,483,647 | Python 3 | OK | TESTS | 61 | 218 | 512,000 | n = int(input())
chest = 0
bicep = 0
back = 0
line = input()
line = line.split()
for i in range(0,n):
line[i] = int(line[i])
if(i % 3 == 0):
chest = chest + line[i]
elif(i % 3 == 1):
bicep = bicep + line[i]
else:
back = back + line[i]
if(chest > bicep and chest > back):
print("chest")
elif(bicep > chest and bicep > back):
print("biceps")
else:
print("back") | Title: Greg's Workout
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.
Output Specification:
Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous.
Demo Input:
['2\n2 8\n', '3\n5 1 10\n', '7\n3 3 2 7 9 6 8\n']
Demo Output:
['biceps\n', 'back\n', 'chest\n']
Note:
In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise. | ```python
n = int(input())
chest = 0
bicep = 0
back = 0
line = input()
line = line.split()
for i in range(0,n):
line[i] = int(line[i])
if(i % 3 == 0):
chest = chest + line[i]
elif(i % 3 == 1):
bicep = bicep + line[i]
else:
back = back + line[i]
if(chest > bicep and chest > back):
print("chest")
elif(bicep > chest and bicep > back):
print("biceps")
else:
print("back")
``` | 3 |
|
375 | A | Divisible by Seven | PROGRAMMING | 1,600 | [
"math",
"number theory"
] | null | null | You have number *a*, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number *a* doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes. | The first line contains positive integer *a* in the decimal record. It is guaranteed that the record of number *a* contains digits: 1, 6, 8, 9. Number *a* doesn't contain any leading zeroes. The decimal representation of number *a* contains at least 4 and at most 106 characters. | Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number *a* in the required manner, print 0. | [
"1689\n",
"18906\n"
] | [
"1869\n",
"18690\n"
] | none | 500 | [
{
"input": "1689",
"output": "1869"
},
{
"input": "18906",
"output": "18690"
},
{
"input": "2419323689",
"output": "2432391689"
},
{
"input": "8589157262",
"output": "5857221986"
},
{
"input": "2717172350336955863014903670481525170997949309274087058935108848979319747543008692128164875210350026",
"output": "2771723503355630149036704815251709979493092740870589351088489793197475430086921281648752103500261986"
},
{
"input": "9825995656040286793128006047268547610068699214477842995873286607346639816314908021369221299622234988",
"output": "2599556040286793280060472685476100686992144778429958732866073466398163149080213692212996222349881968"
},
{
"input": "100000000689",
"output": "186900000000"
},
{
"input": "16891",
"output": "16198"
},
{
"input": "16892",
"output": "21896"
},
{
"input": "16893",
"output": "31689"
},
{
"input": "16894",
"output": "41986"
},
{
"input": "16895",
"output": "51968"
},
{
"input": "16896",
"output": "61698"
},
{
"input": "16897",
"output": "71869"
},
{
"input": "16898",
"output": "86198"
},
{
"input": "16899",
"output": "91896"
},
{
"input": "4048169",
"output": "4041968"
},
{
"input": "10994168",
"output": "94116890"
},
{
"input": "168903",
"output": "316890"
},
{
"input": "11689",
"output": "16198"
},
{
"input": "91111168",
"output": "11111968"
},
{
"input": "16890000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "18690000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "1689999999999",
"output": "9999999991968"
},
{
"input": "9883291673084",
"output": "8329730841698"
},
{
"input": "6198",
"output": "1869"
}
] | 1,671,640,231 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 102,400 | from collections import Counter
n = input()
d = Counter(n)
if d['1']==d['8'] and d['8']==d['6'] and d['9']==d['6']:
print('1869'*d['1']+'0'*d['0'])
else:
print(0) | Title: Divisible by Seven
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have number *a*, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number *a* doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input Specification:
The first line contains positive integer *a* in the decimal record. It is guaranteed that the record of number *a* contains digits: 1, 6, 8, 9. Number *a* doesn't contain any leading zeroes. The decimal representation of number *a* contains at least 4 and at most 106 characters.
Output Specification:
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number *a* in the required manner, print 0.
Demo Input:
['1689\n', '18906\n']
Demo Output:
['1869\n', '18690\n']
Note:
none | ```python
from collections import Counter
n = input()
d = Counter(n)
if d['1']==d['8'] and d['8']==d['6'] and d['9']==d['6']:
print('1869'*d['1']+'0'*d['0'])
else:
print(0)
``` | 0 |
|
445 | A | DZY Loves Chessboard | PROGRAMMING | 1,200 | [
"dfs and similar",
"implementation"
] | null | null | DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad. | Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists. | [
"1 1\n.\n",
"2 2\n..\n..\n",
"3 3\n.-.\n---\n--."
] | [
"B\n",
"BW\nWB\n",
"B-B\n---\n--B"
] | In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | 500 | [
{
"input": "1 1\n.",
"output": "B"
},
{
"input": "2 2\n..\n..",
"output": "BW\nWB"
},
{
"input": "3 3\n.-.\n---\n--.",
"output": "B-B\n---\n--B"
},
{
"input": "3 1\n-\n.\n.",
"output": "-\nW\nB"
},
{
"input": "11 11\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------",
"output": "-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------"
},
{
"input": "1 1\n-",
"output": "-"
},
{
"input": "4 2\n..\n..\n..\n..",
"output": "BW\nWB\nBW\nWB"
},
{
"input": "5 1\n.\n.\n.\n.\n.",
"output": "B\nW\nB\nW\nB"
}
] | 1,615,367,839 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 77 | 0 | n,m=map(int,input().split())
matrix=[]
for i in range(n):
matrix.append(list(map(str,input())))
list1=['W','B']
flag=0
for i in range(n):
for j in range(m):
if matrix[i][j]=='.':
matrix[i][j]=list1[flag]
if j!=m-1:
flag^=1
flag^=1
for row in matrix:
print(''.join(row))
| Title: DZY Loves Chessboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves chessboard, and he enjoys playing with it.
He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge.
You task is to find any suitable placement of chessmen on the given chessboard.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output Specification:
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell.
If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
Demo Input:
['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.']
Demo Output:
['B\n', 'BW\nWB\n', 'B-B\n---\n--B']
Note:
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK.
In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output.
In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are. | ```python
n,m=map(int,input().split())
matrix=[]
for i in range(n):
matrix.append(list(map(str,input())))
list1=['W','B']
flag=0
for i in range(n):
for j in range(m):
if matrix[i][j]=='.':
matrix[i][j]=list1[flag]
if j!=m-1:
flag^=1
flag^=1
for row in matrix:
print(''.join(row))
``` | 0 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | You are given two integers $a$ and $b$. Moreover, you are given a sequence $s_0, s_1, \dots, s_{n}$. All values in $s$ are integers $1$ or $-1$. It's known that sequence is $k$-periodic and $k$ divides $n+1$. In other words, for each $k \leq i \leq n$ it's satisfied that $s_{i} = s_{i - k}$.
Find out the non-negative remainder of division of $\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}$ by $10^{9} + 9$.
Note that the modulo is unusual! | The first line contains four integers $n, a, b$ and $k$ $(1 \leq n \leq 10^{9}, 1 \leq a, b \leq 10^{9}, 1 \leq k \leq 10^{5})$.
The second line contains a sequence of length $k$ consisting of characters '+' and '-'.
If the $i$-th character (0-indexed) is '+', then $s_{i} = 1$, otherwise $s_{i} = -1$.
Note that only the first $k$ members of the sequence are given, the rest can be obtained using the periodicity property. | Output a single integer — value of given expression modulo $10^{9} + 9$. | [
"2 2 3 3\n+-+\n",
"4 1 5 1\n-\n"
] | [
"7\n",
"999999228\n"
] | In the first example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i})$ = $2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2}$ = 7
In the second example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 \equiv 999999228 \pmod{10^{9} + 9}$. | 0 | [
{
"input": "2 2 3 3\n+-+",
"output": "7"
},
{
"input": "4 1 5 1\n-",
"output": "999999228"
},
{
"input": "1 1 4 2\n-+",
"output": "3"
},
{
"input": "3 1 4 4\n+--+",
"output": "45"
},
{
"input": "5 1 1 6\n++---+",
"output": "0"
},
{
"input": "5 2 2 6\n+--++-",
"output": "0"
},
{
"input": "686653196 115381398 884618610 3\n+-+",
"output": "542231211"
},
{
"input": "608663287 430477711 172252358 8\n-+--+-+-",
"output": "594681696"
},
{
"input": "904132655 827386249 118827660 334\n+++-+++++--+++----+-+-+-+-+--+-+---++--++--++--+-+-+++-+++--+-+-+----+-+-++++-----+--++++------+++-+-+-++-++++++++-+-++-+++--+--++------+--+-+++--++--+---++-++-+-+-++---++-++--+-+-++-+------+-+----+++-+++--+-+-+--+--+--+------+--+---+--+-++--+++---+-+-++--------+-++--++-+-+-+-+-+-+--+-++++-+++--+--++----+--+-++-++--+--+-+-++-+-++++-",
"output": "188208979"
},
{
"input": "234179195 430477711 115381398 12\n++++-+-+-+++",
"output": "549793323"
},
{
"input": "75952547 967294208 907708706 252\n++--++--+++-+-+--++--++++++---+++-++-+-----++++--++-+-++------+-+-+-++-+-+-++++------++---+-++++---+-+-++++--++++++--+-+++-++--+--+---++++---+-+++-+++--+-+--+++++---+--++-++++--++++-+-++-+++-++-----+-+++++----++--+++-+-+++++-+--++-++-+--+-++++--+-+-+-+",
"output": "605712499"
},
{
"input": "74709071 801809249 753674746 18\n++++++-+-+---+-+--",
"output": "13414893"
},
{
"input": "743329 973758 92942 82\n++----+-++++----+--+++---+--++++-+-+---+++++--+--+++++++--++-+++----+--+++++-+--+-",
"output": "299311566"
},
{
"input": "18111 291387 518587 2\n++",
"output": "724471355"
},
{
"input": "996144 218286 837447 1\n-",
"output": "549104837"
},
{
"input": "179358 828426 548710 67\n++++---+--++----+-+-++++----+--+---+------++-+-++++--+----+---+-+--",
"output": "759716474"
},
{
"input": "397521 174985 279760 1\n+",
"output": "25679493"
},
{
"input": "613632 812232 482342 1\n-",
"output": "891965141"
},
{
"input": "936810 183454 647048 1\n+",
"output": "523548992"
},
{
"input": "231531 250371 921383 28\n++-+------+--+--++++--+-+++-",
"output": "134450934"
},
{
"input": "947301 87242 360762 97\n--+++--+++-++--++-++--++--+++---+++--++++--+++++--+-++-++-----+-++-+--++-----+-++-+--++-++-+-----",
"output": "405016159"
},
{
"input": "425583346 814209084 570987274 1\n+",
"output": "63271171"
},
{
"input": "354062556 688076879 786825319 1\n+",
"output": "545304776"
},
{
"input": "206671954 13571766 192250278 1\n+",
"output": "717117421"
},
{
"input": "23047921 621656196 160244047 1\n-",
"output": "101533009"
},
{
"input": "806038018 740585177 987616107 293\n-+++++--++++---++-+--+-+---+-++++--+--+++--++---++++++++--+++++-+-++-+--+----+--+++-+-++-+++-+-+-+----------++-+-+++++++-+-+-+-++---+++-+-+-------+-+-++--++-++-++-++-+---+--++-++--+++--+++-+-+----++--+-+-++-+---+---+-+-+++------+-+++-+---++-+--+++----+++++---++-++--+----+++-+--+++-+------+-++",
"output": "441468166"
},
{
"input": "262060935 184120408 148332034 148\n+--+-------+-+-+--++-+++--++-+-++++++--++-+++-+++--+-------+-+--+++-+-+-+---++-++-+-++---+--+-+-+--+------+++--+--+-+-+---+---+-+-++++---+++--+++---",
"output": "700325386"
},
{
"input": "919350941 654611542 217223605 186\n++-++-+++++-+++--+---+++++++-++-+----+-++--+-++--++--+++-+++---+--+--++-+-+++-+-+++-++---+--+++-+-+++--+-+-------+-++------++---+-+---++-++-++---+-+--+-+--+++++---+--+--++++-++-++--+--++",
"output": "116291420"
},
{
"input": "289455627 906207104 512692624 154\n-------++--+++---++-++------++----------+--+++-+-+++---+---+++--++++++--+-+-+--+---+-+-++-++--+-++--++++---+-+---+-----+--+-+---------+++-++---++-+-+-----",
"output": "48198216"
},
{
"input": "258833760 515657142 791267045 1\n-",
"output": "935800888"
},
{
"input": "691617927 66917103 843055237 8\n--+++---",
"output": "147768186"
},
{
"input": "379582849 362892355 986900829 50\n++-++---+-+++++--++++--+--++--++-----+------++--+-",
"output": "927469713"
},
{
"input": "176799169 363368399 841293419 1\n+",
"output": "746494802"
},
{
"input": "144808247 203038656 166324035 4\n-+-+",
"output": "909066471"
},
{
"input": "477607531 177367565 20080950 2\n++",
"output": "928662830"
},
{
"input": "682074525 289438443 917164266 1\n+",
"output": "28048785"
},
{
"input": "938449224 59852396 219719125 1\n-",
"output": "648647459"
},
{
"input": "395171426 872478622 193568600 147\n+---++---+-+--+++++--+---+-++++-+-++---++++--+--+-+-++-+-++--------++---+++-+---++---+---+-+--+-++++-+++-+-+-++-+--+++-++-+-+-+-++++++-+---+---++--",
"output": "460881399"
},
{
"input": "403493428 317461491 556701240 1\n-",
"output": "936516261"
},
{
"input": "917751169 330191895 532837377 70\n-+-+++++++--++---++-+++++-+++-----+-+++---+--+-+-++-++-+-+-++-++-+----",
"output": "908035409"
},
{
"input": "252089413 552678586 938424519 1\n-",
"output": "627032736"
},
{
"input": "649316142 320010793 200197645 1\n-",
"output": "323650777"
},
{
"input": "116399299 784781190 299072480 5\n++++-",
"output": "754650814"
}
] | 1,691,790,651 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 77 | 2,867,200 | def binary_search(a, first, last, v):
while first <= last:
mid = (first + last) // 2
if a[mid] < v:
first = mid + 1
else:
last = mid - 1
return last
def dist(a, b):
return abs(a - b)
def go(pos, elv):
if y0 == y1:
return dist(x0, x1)
res = dist(x0, pos) + dist(pos, x1) + (dist(y0, y1) + elv - 1) // elv
return res
n, m, cl, ce, v = map(int, input().split())
l = sorted(list(map(int, input().split())))
e = sorted(list(map(int, input().split())))
q = int(input())
for _ in range(q):
y0, x0, y1, x1 = map(int, input().split())
lid = binary_search(l, 0, cl - 1, x0)
eid = binary_search(e, 0, ce - 1, x0)
res = float('inf')
if 0 <= lid < cl:
res = min(res, go(l[lid], 1))
if 0 <= lid + 1 < cl:
res = min(res, go(l[lid + 1], 1))
if 0 <= eid < ce:
res = min(res, go(e[eid], v))
if 0 <= eid + 1 < ce:
res = min(res, go(e[eid + 1], v))
print(res)# 1691790651.247151 | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two integers $a$ and $b$. Moreover, you are given a sequence $s_0, s_1, \dots, s_{n}$. All values in $s$ are integers $1$ or $-1$. It's known that sequence is $k$-periodic and $k$ divides $n+1$. In other words, for each $k \leq i \leq n$ it's satisfied that $s_{i} = s_{i - k}$.
Find out the non-negative remainder of division of $\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}$ by $10^{9} + 9$.
Note that the modulo is unusual!
Input Specification:
The first line contains four integers $n, a, b$ and $k$ $(1 \leq n \leq 10^{9}, 1 \leq a, b \leq 10^{9}, 1 \leq k \leq 10^{5})$.
The second line contains a sequence of length $k$ consisting of characters '+' and '-'.
If the $i$-th character (0-indexed) is '+', then $s_{i} = 1$, otherwise $s_{i} = -1$.
Note that only the first $k$ members of the sequence are given, the rest can be obtained using the periodicity property.
Output Specification:
Output a single integer — value of given expression modulo $10^{9} + 9$.
Demo Input:
['2 2 3 3\n+-+\n', '4 1 5 1\n-\n']
Demo Output:
['7\n', '999999228\n']
Note:
In the first example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i})$ = $2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2}$ = 7
In the second example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 \equiv 999999228 \pmod{10^{9} + 9}$. | ```python
def binary_search(a, first, last, v):
while first <= last:
mid = (first + last) // 2
if a[mid] < v:
first = mid + 1
else:
last = mid - 1
return last
def dist(a, b):
return abs(a - b)
def go(pos, elv):
if y0 == y1:
return dist(x0, x1)
res = dist(x0, pos) + dist(pos, x1) + (dist(y0, y1) + elv - 1) // elv
return res
n, m, cl, ce, v = map(int, input().split())
l = sorted(list(map(int, input().split())))
e = sorted(list(map(int, input().split())))
q = int(input())
for _ in range(q):
y0, x0, y1, x1 = map(int, input().split())
lid = binary_search(l, 0, cl - 1, x0)
eid = binary_search(e, 0, ce - 1, x0)
res = float('inf')
if 0 <= lid < cl:
res = min(res, go(l[lid], 1))
if 0 <= lid + 1 < cl:
res = min(res, go(l[lid + 1], 1))
if 0 <= eid < ce:
res = min(res, go(e[eid], v))
if 0 <= eid + 1 < ce:
res = min(res, go(e[eid + 1], v))
print(res)# 1691790651.247151
``` | -1 |
|
513 | A | Game | PROGRAMMING | 800 | [
"constructive algorithms",
"math"
] | null | null | Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally. | The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission. | Output "First" if the first player wins and "Second" otherwise. | [
"2 2 1 2\n",
"2 1 1 1\n"
] | [
"Second\n",
"First\n"
] | Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely. | 3 | [
{
"input": "2 2 1 2",
"output": "Second"
},
{
"input": "2 1 1 1",
"output": "First"
},
{
"input": "5 7 4 1",
"output": "Second"
},
{
"input": "5 7 1 4",
"output": "Second"
},
{
"input": "5 7 10 10",
"output": "Second"
},
{
"input": "5 7 1 10",
"output": "Second"
},
{
"input": "1 1 1 1",
"output": "Second"
},
{
"input": "50 50 50 50",
"output": "Second"
},
{
"input": "50 49 1 2",
"output": "First"
},
{
"input": "50 48 3 1",
"output": "First"
},
{
"input": "48 50 12 11",
"output": "Second"
},
{
"input": "49 50 11 12",
"output": "Second"
},
{
"input": "49 49 4 1",
"output": "Second"
},
{
"input": "49 49 3 3",
"output": "Second"
},
{
"input": "1 50 1 50",
"output": "Second"
},
{
"input": "1 50 50 50",
"output": "Second"
},
{
"input": "50 1 1 1",
"output": "First"
},
{
"input": "50 1 1 50",
"output": "First"
},
{
"input": "32 31 10 9",
"output": "First"
},
{
"input": "32 4 17 3",
"output": "First"
}
] | 1,652,696,391 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | a,b,c,d=input().split()
a=int(a)
b=int(b)
c=int(c)
d=int(d)
if a<=b:
print("Second")
else:
print("First")
#,loh,bhopgfbgfmpobmdfopbhopgftbmpgftrmbnogtopp
| Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally.
Input Specification:
The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
Output Specification:
Output "First" if the first player wins and "Second" otherwise.
Demo Input:
['2 2 1 2\n', '2 1 1 1\n']
Demo Output:
['Second\n', 'First\n']
Note:
Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely. | ```python
a,b,c,d=input().split()
a=int(a)
b=int(b)
c=int(c)
d=int(d)
if a<=b:
print("Second")
else:
print("First")
#,loh,bhopgfbgfmpobmdfopbhopgftbmpgftrmbnogtopp
``` | -1 |
|
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,688,269,726 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | s=str(input())
l=len(s)
print(s[0],end='')
print(l-2,end='')
print(s[-1]) | Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
s=str(input())
l=len(s)
print(s[0],end='')
print(l-2,end='')
print(s[-1])
``` | 0 |
784 | E | Twisted Circuit | PROGRAMMING | 2,000 | [
"*special",
"brute force",
"implementation"
] | null | null | The input consists of four lines, each line containing a single digit 0 or 1. | Output a single digit, 0 or 1. | [
"0\n1\n1\n0\n"
] | [
"0\n"
] | none | 0 | [
{
"input": "0\n1\n1\n0",
"output": "0"
},
{
"input": "0\n0\n0\n0",
"output": "0"
},
{
"input": "1\n0\n0\n0",
"output": "1"
},
{
"input": "0\n1\n0\n0",
"output": "0"
},
{
"input": "1\n1\n0\n0",
"output": "1"
},
{
"input": "0\n0\n1\n0",
"output": "0"
},
{
"input": "1\n0\n1\n0",
"output": "0"
},
{
"input": "1\n1\n1\n0",
"output": "1"
},
{
"input": "0\n0\n0\n1",
"output": "1"
},
{
"input": "1\n0\n0\n1",
"output": "1"
},
{
"input": "0\n1\n0\n1",
"output": "0"
},
{
"input": "1\n1\n0\n1",
"output": "0"
},
{
"input": "0\n0\n1\n1",
"output": "1"
},
{
"input": "1\n0\n1\n1",
"output": "1"
},
{
"input": "0\n1\n1\n1",
"output": "0"
},
{
"input": "1\n1\n1\n1",
"output": "1"
}
] | 1,491,825,009 | 6,309 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 5,529,600 | a = int(input())
b = int(input())
c = int(input())
d = int(input())
n = a*8+b*4+c*2+d
# 0123456789012345
a = "1111110101111111"
print(a[n]) | Title: Twisted Circuit
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Input Specification:
The input consists of four lines, each line containing a single digit 0 or 1.
Output Specification:
Output a single digit, 0 or 1.
Demo Input:
['0\n1\n1\n0\n']
Demo Output:
['0\n']
Note:
none | ```python
a = int(input())
b = int(input())
c = int(input())
d = int(input())
n = a*8+b*4+c*2+d
# 0123456789012345
a = "1111110101111111"
print(a[n])
``` | 0 |
||
776 | C | Molly's Chemicals | PROGRAMMING | 1,800 | [
"binary search",
"brute force",
"data structures",
"implementation",
"math"
] | null | null | Molly Hooper has *n* different kinds of chemicals arranged in a line. Each of the chemicals has an affection value, The *i*-th of them has affection value *a**i*.
Molly wants Sherlock to fall in love with her. She intends to do this by mixing a contiguous segment of chemicals together to make a love potion with total affection value as a non-negative integer power of *k*. Total affection value of a continuous segment of chemicals is the sum of affection values of each chemical in that segment.
Help her to do so in finding the total number of such segments. | The first line of input contains two integers, *n* and *k*, the number of chemicals and the number, such that the total affection value is a non-negative power of this number *k*. (1<=≤<=*n*<=≤<=105, 1<=≤<=|*k*|<=≤<=10).
Next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109) — affection values of chemicals. | Output a single integer — the number of valid segments. | [
"4 2\n2 2 2 2\n",
"4 -3\n3 -6 -3 12\n"
] | [
"8\n",
"3\n"
] | Do keep in mind that *k*<sup class="upper-index">0</sup> = 1.
In the first sample, Molly can get following different affection values:
- 2: segments [1, 1], [2, 2], [3, 3], [4, 4]; - 4: segments [1, 2], [2, 3], [3, 4]; - 6: segments [1, 3], [2, 4]; - 8: segments [1, 4].
Out of these, 2, 4 and 8 are powers of *k* = 2. Therefore, the answer is 8.
In the second sample, Molly can choose segments [1, 2], [3, 3], [3, 4]. | 1,500 | [
{
"input": "4 2\n2 2 2 2",
"output": "8"
},
{
"input": "4 -3\n3 -6 -3 12",
"output": "3"
},
{
"input": "14 -9\n-2 -4 62 53 90 41 35 21 85 74 85 57 10 39",
"output": "0"
},
{
"input": "20 9\n90 21 -6 -61 14 -21 -17 -65 -84 -75 -48 56 67 -50 16 65 -79 -61 92 85",
"output": "1"
},
{
"input": "89 -7\n5972 4011 3914 670 3727 2913 6935 6927 2118 6645 7141 3585 9811 2859 459 8870 6578 8667 468 5152 3241 7455 7323 8817 4866 1040 5102 9146 621 5002 396 4967 9822 4200 3899 4416 5225 9415 9606 4802 5589 1798 9094 5453 7163 264 1026 6187 3918 4237 -17 4306 8960 3321 2927 9205 6248 7607 564 364 3503 8149 2235 8278 6249 3987 524 5718 9359 3549 1474 9204 3870 6996 3932 8295 612 6310 4461 1129 6441 3465 4654 7583 3274 6309 4831 4918 558",
"output": "0"
},
{
"input": "10 2\n2 4 8 16 32 64 128 256 512 1024",
"output": "10"
},
{
"input": "10 1\n-1 1 -1 1 -1 1 -1 1 -1 1",
"output": "15"
},
{
"input": "32 2\n8 16384 32768 65536 32 8388608 1048576 16777216 65536 8 16384 128 2097152 1024 16777216 4 8192 8388608 65536 1024 1024 16 8 16 128 2 1024 128 8 33554432 32768 2048",
"output": "33"
},
{
"input": "1 2\n2",
"output": "1"
},
{
"input": "14 2\n2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "45"
},
{
"input": "2 1\n1 1",
"output": "2"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "10"
},
{
"input": "4 1\n1 1 1 1",
"output": "4"
},
{
"input": "3 1\n1 1 1",
"output": "3"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "10 -1\n1 0 -1 1 0 -1 1 0 -1 1",
"output": "28"
},
{
"input": "4 1\n-1 -2 3 1",
"output": "3"
},
{
"input": "26 -1\n0 0 1 1 -1 -1 0 0 1 0 0 0 -1 1 0 0 -1 1 -1 1 -1 1 0 0 5 -4",
"output": "168"
},
{
"input": "1 -1\n-1",
"output": "1"
},
{
"input": "10 1\n1 2 3 4 5 6 7 8 9 10",
"output": "1"
},
{
"input": "1 2\n1048576",
"output": "1"
},
{
"input": "4 -1\n1 1 1 1",
"output": "4"
},
{
"input": "5 -1\n1 1 1 1 1",
"output": "5"
},
{
"input": "33 2\n536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912 536870912",
"output": "141"
},
{
"input": "1 1\n-1",
"output": "0"
}
] | 1,694,868,271 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | a=list(map(int,input().split()))
d={0:1}
p=[]
c=0
for i in range(47):
p.append(pow(k,i))
ps=0
for i in a:
ps+=i
d[ps]=d.get(ps,0)+1
for x in p:
c+=d.get(ps -x,0)
print(c) | Title: Molly's Chemicals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Molly Hooper has *n* different kinds of chemicals arranged in a line. Each of the chemicals has an affection value, The *i*-th of them has affection value *a**i*.
Molly wants Sherlock to fall in love with her. She intends to do this by mixing a contiguous segment of chemicals together to make a love potion with total affection value as a non-negative integer power of *k*. Total affection value of a continuous segment of chemicals is the sum of affection values of each chemical in that segment.
Help her to do so in finding the total number of such segments.
Input Specification:
The first line of input contains two integers, *n* and *k*, the number of chemicals and the number, such that the total affection value is a non-negative power of this number *k*. (1<=≤<=*n*<=≤<=105, 1<=≤<=|*k*|<=≤<=10).
Next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109) — affection values of chemicals.
Output Specification:
Output a single integer — the number of valid segments.
Demo Input:
['4 2\n2 2 2 2\n', '4 -3\n3 -6 -3 12\n']
Demo Output:
['8\n', '3\n']
Note:
Do keep in mind that *k*<sup class="upper-index">0</sup> = 1.
In the first sample, Molly can get following different affection values:
- 2: segments [1, 1], [2, 2], [3, 3], [4, 4]; - 4: segments [1, 2], [2, 3], [3, 4]; - 6: segments [1, 3], [2, 4]; - 8: segments [1, 4].
Out of these, 2, 4 and 8 are powers of *k* = 2. Therefore, the answer is 8.
In the second sample, Molly can choose segments [1, 2], [3, 3], [3, 4]. | ```python
a=list(map(int,input().split()))
d={0:1}
p=[]
c=0
for i in range(47):
p.append(pow(k,i))
ps=0
for i in a:
ps+=i
d[ps]=d.get(ps,0)+1
for x in p:
c+=d.get(ps -x,0)
print(c)
``` | -1 |
|
994 | B | Knights of a Polygonal Table | PROGRAMMING | 1,400 | [
"greedy",
"implementation",
"sortings"
] | null | null | Unlike Knights of a Round Table, Knights of a Polygonal Table deprived of nobility and happy to kill each other. But each knight has some power and a knight can kill another knight if and only if his power is greater than the power of victim. However, even such a knight will torment his conscience, so he can kill no more than $k$ other knights. Also, each knight has some number of coins. After a kill, a knight can pick up all victim's coins.
Now each knight ponders: how many coins he can have if only he kills other knights?
You should answer this question for each knight. | The first line contains two integers $n$ and $k$ $(1 \le n \le 10^5, 0 \le k \le \min(n-1,10))$ — the number of knights and the number $k$ from the statement.
The second line contains $n$ integers $p_1, p_2 ,\ldots,p_n$ $(1 \le p_i \le 10^9)$ — powers of the knights. All $p_i$ are distinct.
The third line contains $n$ integers $c_1, c_2 ,\ldots,c_n$ $(0 \le c_i \le 10^9)$ — the number of coins each knight has. | Print $n$ integers — the maximum number of coins each knight can have it only he kills other knights. | [
"4 2\n4 5 9 7\n1 2 11 33\n",
"5 1\n1 2 3 4 5\n1 2 3 4 5\n",
"1 0\n2\n3\n"
] | [
"1 3 46 36 ",
"1 3 5 7 9 ",
"3 "
] | Consider the first example.
- The first knight is the weakest, so he can't kill anyone. That leaves him with the only coin he initially has. - The second knight can kill the first knight and add his coin to his own two. - The third knight is the strongest, but he can't kill more than $k = 2$ other knights. It is optimal to kill the second and the fourth knights: $2+11+33 = 46$. - The fourth knight should kill the first and the second knights: $33+1+2 = 36$.
In the second example the first knight can't kill anyone, while all the others should kill the one with the index less by one than their own.
In the third example there is only one knight, so he can't kill anyone. | 1,000 | [
{
"input": "4 2\n4 5 9 7\n1 2 11 33",
"output": "1 3 46 36 "
},
{
"input": "5 1\n1 2 3 4 5\n1 2 3 4 5",
"output": "1 3 5 7 9 "
},
{
"input": "1 0\n2\n3",
"output": "3 "
},
{
"input": "7 1\n2 3 4 5 7 8 9\n0 3 7 9 5 8 9",
"output": "0 3 10 16 14 17 18 "
},
{
"input": "7 2\n2 4 6 7 8 9 10\n10 8 4 8 4 5 9",
"output": "10 18 22 26 22 23 27 "
},
{
"input": "11 10\n1 2 3 4 5 6 7 8 9 10 11\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1000000000 2000000000 3000000000 4000000000 5000000000 6000000000 7000000000 8000000000 9000000000 10000000000 11000000000 "
},
{
"input": "2 0\n2 3\n3 3",
"output": "3 3 "
},
{
"input": "7 3\n1 2 3 4 5 6 7\n3 3 3 4 5 6 7",
"output": "3 6 9 13 15 18 22 "
},
{
"input": "3 0\n3 2 1\n1 2 3",
"output": "1 2 3 "
},
{
"input": "5 3\n4 5 7 9 11\n10 10 10 10 10",
"output": "10 20 30 40 40 "
},
{
"input": "4 0\n4 5 9 7\n1 2 11 33",
"output": "1 2 11 33 "
},
{
"input": "7 3\n1 2 3 4 5 6 7\n3 3 3 8 8 8 8",
"output": "3 6 9 17 22 27 32 "
},
{
"input": "3 0\n1 2 3\n5 5 5",
"output": "5 5 5 "
},
{
"input": "4 2\n4 5 9 7\n2 2 11 33",
"output": "2 4 46 37 "
},
{
"input": "6 3\n1 2 3 4 5 6\n1 1 1 1 1 1",
"output": "1 2 3 4 4 4 "
},
{
"input": "10 5\n1 2 3 4 5 6 7 8 9 10\n1 1 1 1 1 1 1 1 1 1",
"output": "1 2 3 4 5 6 6 6 6 6 "
},
{
"input": "3 2\n1 2 3\n1 1 1",
"output": "1 2 3 "
},
{
"input": "3 0\n1 2 3\n10 20 30",
"output": "10 20 30 "
},
{
"input": "4 0\n4 5 9 7\n1 2 3 4",
"output": "1 2 3 4 "
},
{
"input": "5 4\n1 2 3 4 5\n1 1 1 1 1",
"output": "1 2 3 4 5 "
},
{
"input": "4 3\n1 2 3 4\n5 5 5 5",
"output": "5 10 15 20 "
},
{
"input": "5 3\n1 2 3 4 5\n7 7 7 7 7",
"output": "7 14 21 28 28 "
}
] | 1,529,171,420 | 4,520 | Python 3 | WRONG_ANSWER | PRETESTS | 4 | 77 | 0 | some_chars=input()
sc=some_chars.split()
knight=int(sc[0])
limit=int(sc[1])
Knight_Force=['']*knight
money=['']*knight
ans=['']*knight
Knight_Force_str=input()
money_str=input()
money=money_str.split()
Knight_Force=Knight_Force_str.split()
looted=[0]*(knight-1)
ind=0
for a in range(knight):
for i in range(knight):
if a!=i and Knight_Force[i]<Knight_Force[a]:
looted[ind]=int(money[i])
ind+=1
ind=0
i=0
looted.sort()
looted.reverse()
conscience=limit
ans[a]=int(money[a])
h=0
while conscience!=0:
if conscience!=0 :
ans[a]=int(ans[a])+looted[h]
h+=1
conscience-=1
looted.clear()
looted=[0]*(knight-1)
j=knight-1
while j>=0:
ans[j]=str(ans[j])
j-=1
ans_str=' '.join(ans)
print(ans_str)
| Title: Knights of a Polygonal Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Unlike Knights of a Round Table, Knights of a Polygonal Table deprived of nobility and happy to kill each other. But each knight has some power and a knight can kill another knight if and only if his power is greater than the power of victim. However, even such a knight will torment his conscience, so he can kill no more than $k$ other knights. Also, each knight has some number of coins. After a kill, a knight can pick up all victim's coins.
Now each knight ponders: how many coins he can have if only he kills other knights?
You should answer this question for each knight.
Input Specification:
The first line contains two integers $n$ and $k$ $(1 \le n \le 10^5, 0 \le k \le \min(n-1,10))$ — the number of knights and the number $k$ from the statement.
The second line contains $n$ integers $p_1, p_2 ,\ldots,p_n$ $(1 \le p_i \le 10^9)$ — powers of the knights. All $p_i$ are distinct.
The third line contains $n$ integers $c_1, c_2 ,\ldots,c_n$ $(0 \le c_i \le 10^9)$ — the number of coins each knight has.
Output Specification:
Print $n$ integers — the maximum number of coins each knight can have it only he kills other knights.
Demo Input:
['4 2\n4 5 9 7\n1 2 11 33\n', '5 1\n1 2 3 4 5\n1 2 3 4 5\n', '1 0\n2\n3\n']
Demo Output:
['1 3 46 36 ', '1 3 5 7 9 ', '3 ']
Note:
Consider the first example.
- The first knight is the weakest, so he can't kill anyone. That leaves him with the only coin he initially has. - The second knight can kill the first knight and add his coin to his own two. - The third knight is the strongest, but he can't kill more than $k = 2$ other knights. It is optimal to kill the second and the fourth knights: $2+11+33 = 46$. - The fourth knight should kill the first and the second knights: $33+1+2 = 36$.
In the second example the first knight can't kill anyone, while all the others should kill the one with the index less by one than their own.
In the third example there is only one knight, so he can't kill anyone. | ```python
some_chars=input()
sc=some_chars.split()
knight=int(sc[0])
limit=int(sc[1])
Knight_Force=['']*knight
money=['']*knight
ans=['']*knight
Knight_Force_str=input()
money_str=input()
money=money_str.split()
Knight_Force=Knight_Force_str.split()
looted=[0]*(knight-1)
ind=0
for a in range(knight):
for i in range(knight):
if a!=i and Knight_Force[i]<Knight_Force[a]:
looted[ind]=int(money[i])
ind+=1
ind=0
i=0
looted.sort()
looted.reverse()
conscience=limit
ans[a]=int(money[a])
h=0
while conscience!=0:
if conscience!=0 :
ans[a]=int(ans[a])+looted[h]
h+=1
conscience-=1
looted.clear()
looted=[0]*(knight-1)
j=knight-1
while j>=0:
ans[j]=str(ans[j])
j-=1
ans_str=' '.join(ans)
print(ans_str)
``` | 0 |
|
92 | A | Chips | PROGRAMMING | 800 | [
"implementation",
"math"
] | A. Chips | 2 | 256 | There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. | Print the number of chips the presenter ended up with. | [
"4 11\n",
"17 107\n",
"3 8\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | 500 | [
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "25 9712",
"output": "11"
},
{
"input": "9 7601",
"output": "5"
},
{
"input": "1 9058",
"output": "0"
},
{
"input": "29 7772",
"output": "26"
},
{
"input": "45 9465",
"output": "14"
},
{
"input": "46 866",
"output": "5"
},
{
"input": "29 1241",
"output": "20"
},
{
"input": "17 4248",
"output": "12"
},
{
"input": "20 8082",
"output": "11"
},
{
"input": "50 9555",
"output": "0"
},
{
"input": "4 7455",
"output": "2"
},
{
"input": "36 880",
"output": "4"
},
{
"input": "24 7440",
"output": "9"
},
{
"input": "44 7888",
"output": "12"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "50 10000",
"output": "40"
},
{
"input": "1 10000",
"output": "0"
},
{
"input": "50 1",
"output": "0"
},
{
"input": "50 50",
"output": "5"
}
] | 1,618,188,790 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 124 | 0 | n,m = map(int,input().split())
k = m%(n*(n+1)//2)
c = 1
while k-c >= 0:
k = k - c
c+=1
print(k) | Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | ```python
n,m = map(int,input().split())
k = m%(n*(n+1)//2)
c = 1
while k-c >= 0:
k = k - c
c+=1
print(k)
``` | 3.969 |
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"
}
] | 1,658,842,577 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 9 | 62 | 0 | n,m,a=map(int,input().split())
if n*m<=a*a:
print(1)
else:
count=0
count+=m//a
ch=n//a
count+=((n//a)-1 if ch else 0)*m
if m%a!=0: count+=1
if n%a!=0: count+=(m//a)+1
print(count)
| 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())
if n*m<=a*a:
print(1)
else:
count=0
count+=m//a
ch=n//a
count+=((n//a)-1 if ch else 0)*m
if m%a!=0: count+=1
if n%a!=0: count+=(m//a)+1
print(count)
``` | 0 |
678 | A | Johny Likes Numbers | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*. | The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109). | Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*. | [
"5 3\n",
"25 13\n",
"26 13\n"
] | [
"6\n",
"26\n",
"39\n"
] | none | 0 | [
{
"input": "5 3",
"output": "6"
},
{
"input": "25 13",
"output": "26"
},
{
"input": "26 13",
"output": "39"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "8 8",
"output": "16"
},
{
"input": "14 15",
"output": "15"
},
{
"input": "197 894",
"output": "894"
},
{
"input": "6058 8581",
"output": "8581"
},
{
"input": "97259 41764",
"output": "125292"
},
{
"input": "453145 333625",
"output": "667250"
},
{
"input": "2233224 4394826",
"output": "4394826"
},
{
"input": "76770926 13350712",
"output": "80104272"
},
{
"input": "687355301 142098087",
"output": "710490435"
},
{
"input": "1000000000 999999999",
"output": "1999999998"
},
{
"input": "1000000000 1000000000",
"output": "2000000000"
},
{
"input": "999999999 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 1",
"output": "1000000001"
},
{
"input": "1000000000 2",
"output": "1000000002"
},
{
"input": "999999999 1",
"output": "1000000000"
},
{
"input": "100000000 1",
"output": "100000001"
},
{
"input": "999999999 500000000",
"output": "1000000000"
},
{
"input": "999999990 10",
"output": "1000000000"
},
{
"input": "1000000000 999999997",
"output": "1999999994"
},
{
"input": "999999999 2",
"output": "1000000000"
},
{
"input": "999999984 1",
"output": "999999985"
},
{
"input": "999999983 1",
"output": "999999984"
},
{
"input": "666666666 1",
"output": "666666667"
},
{
"input": "1000000000 990000000",
"output": "1980000000"
},
{
"input": "41 48",
"output": "48"
},
{
"input": "123456 2",
"output": "123458"
},
{
"input": "111 111",
"output": "222"
},
{
"input": "878787 1",
"output": "878788"
},
{
"input": "121 1",
"output": "122"
},
{
"input": "114514 114514",
"output": "229028"
},
{
"input": "500000001 1000000000",
"output": "1000000000"
},
{
"input": "999999997 1",
"output": "999999998"
},
{
"input": "100000000 10",
"output": "100000010"
}
] | 1,618,393,706 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 62 | 0 | n,k=map(int,input().split())
a=n//k
print((a+1)*k)
| Title: Johny Likes Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
Output Specification:
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
Demo Input:
['5 3\n', '25 13\n', '26 13\n']
Demo Output:
['6\n', '26\n', '39\n']
Note:
none | ```python
n,k=map(int,input().split())
a=n//k
print((a+1)*k)
``` | 3 |
|
81 | A | Plug-in | PROGRAMMING | 1,400 | [
"implementation"
] | A. Plug-in | 1 | 256 | Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result. | The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters. | Print the given string after it is processed. It is guaranteed that the result will contain at least one character. | [
"hhoowaaaareyyoouu\n",
"reallazy\n",
"abacabaabacabaa\n"
] | [
"wre",
"rezy",
"a"
] | none | 500 | [
{
"input": "hhoowaaaareyyoouu",
"output": "wre"
},
{
"input": "reallazy",
"output": "rezy"
},
{
"input": "abacabaabacabaa",
"output": "a"
},
{
"input": "xraccabccbry",
"output": "xy"
},
{
"input": "a",
"output": "a"
},
{
"input": "b",
"output": "b"
},
{
"input": "il",
"output": "il"
},
{
"input": "gfj",
"output": "gfj"
},
{
"input": "babbbbbababa",
"output": "babababa"
},
{
"input": "babbbbabbabbbababbabbbbbbabaabaababaaabbbbbabbbbaaaaabbaaabbaabaabbbbabbbababbabaaabbababaaababbbaaa",
"output": "babababababababababababa"
},
{
"input": "aab",
"output": "b"
},
{
"input": "abb",
"output": "a"
},
{
"input": "aba",
"output": "aba"
},
{
"input": "ab",
"output": "ab"
}
] | 1,574,981,906 | 2,147,483,647 | PyPy 3 | OK | TESTS | 32 | 434 | 15,974,400 | from collections import deque
s = list(input())
a = deque()
for i in s:
if len(a):
if i == a[-1]:
a.pop()
else:
a.append(i)
else:
a.append(i)
print(''.join(map(str, a)))
| Title: Plug-in
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input Specification:
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output Specification:
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Demo Input:
['hhoowaaaareyyoouu\n', 'reallazy\n', 'abacabaabacabaa\n']
Demo Output:
['wre', 'rezy', 'a']
Note:
none | ```python
from collections import deque
s = list(input())
a = deque()
for i in s:
if len(a):
if i == a[-1]:
a.pop()
else:
a.append(i)
else:
a.append(i)
print(''.join(map(str, a)))
``` | 3.753245 |
913 | A | Modular Exponentiation | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | The following problem is well-known: given integers *n* and *m*, calculate
where 2*n*<==<=2·2·...·2 (*n* factors), and denotes the remainder of division of *x* by *y*.
You are asked to solve the "reverse" problem. Given integers *n* and *m*, calculate | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=108).
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=108). | Output a single integer — the value of . | [
"4\n42\n",
"1\n58\n",
"98765432\n23456789\n"
] | [
"10\n",
"0\n",
"23456789\n"
] | In the first example, the remainder of division of 42 by 2<sup class="upper-index">4</sup> = 16 is equal to 10.
In the second example, 58 is divisible by 2<sup class="upper-index">1</sup> = 2 without remainder, and the answer is 0. | 500 | [
{
"input": "4\n42",
"output": "10"
},
{
"input": "1\n58",
"output": "0"
},
{
"input": "98765432\n23456789",
"output": "23456789"
},
{
"input": "8\n88127381",
"output": "149"
},
{
"input": "32\n92831989",
"output": "92831989"
},
{
"input": "92831989\n25",
"output": "25"
},
{
"input": "100000000\n100000000",
"output": "100000000"
},
{
"input": "7\n1234",
"output": "82"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n100000000",
"output": "0"
},
{
"input": "100000000\n1",
"output": "1"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "2\n1",
"output": "1"
},
{
"input": "2\n2",
"output": "2"
},
{
"input": "2\n3",
"output": "3"
},
{
"input": "2\n4",
"output": "0"
},
{
"input": "2\n5",
"output": "1"
},
{
"input": "25\n33554432",
"output": "0"
},
{
"input": "26\n33554432",
"output": "33554432"
},
{
"input": "25\n67108864",
"output": "0"
},
{
"input": "26\n67108864",
"output": "0"
},
{
"input": "25\n92831989",
"output": "25723125"
},
{
"input": "26\n92831989",
"output": "25723125"
},
{
"input": "27\n92831989",
"output": "92831989"
},
{
"input": "29\n92831989",
"output": "92831989"
},
{
"input": "30\n92831989",
"output": "92831989"
},
{
"input": "31\n92831989",
"output": "92831989"
},
{
"input": "99812399\n93172892",
"output": "93172892"
},
{
"input": "67108864\n12345678",
"output": "12345678"
},
{
"input": "3\n98391849",
"output": "1"
},
{
"input": "13\n1231232",
"output": "2432"
},
{
"input": "18\n9842194",
"output": "142866"
},
{
"input": "24\n71237177",
"output": "4128313"
},
{
"input": "11\n65535",
"output": "2047"
},
{
"input": "11\n65536",
"output": "0"
},
{
"input": "12\n123",
"output": "123"
},
{
"input": "123456\n12",
"output": "12"
},
{
"input": "88\n888",
"output": "888"
},
{
"input": "888\n88",
"output": "88"
},
{
"input": "8888\n8888",
"output": "8888"
},
{
"input": "64\n128",
"output": "128"
},
{
"input": "128\n64",
"output": "64"
},
{
"input": "60\n9999999",
"output": "9999999"
},
{
"input": "200\n100000",
"output": "100000"
},
{
"input": "3\n8",
"output": "0"
},
{
"input": "10000000\n25024886",
"output": "25024886"
},
{
"input": "100\n100000",
"output": "100000"
},
{
"input": "99999999\n100000000",
"output": "100000000"
},
{
"input": "70\n12345678",
"output": "12345678"
}
] | 1,603,695,554 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 1,000 | 46,284,800 | n = int(input())
m = int(input())
p = (2 ** (n // 2))
p = p * p
if n%2==1:
p*=2
r = m%(p)
print(r) | Title: Modular Exponentiation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The following problem is well-known: given integers *n* and *m*, calculate
where 2*n*<==<=2·2·...·2 (*n* factors), and denotes the remainder of division of *x* by *y*.
You are asked to solve the "reverse" problem. Given integers *n* and *m*, calculate
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=108).
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=108).
Output Specification:
Output a single integer — the value of .
Demo Input:
['4\n42\n', '1\n58\n', '98765432\n23456789\n']
Demo Output:
['10\n', '0\n', '23456789\n']
Note:
In the first example, the remainder of division of 42 by 2<sup class="upper-index">4</sup> = 16 is equal to 10.
In the second example, 58 is divisible by 2<sup class="upper-index">1</sup> = 2 without remainder, and the answer is 0. | ```python
n = int(input())
m = int(input())
p = (2 ** (n // 2))
p = p * p
if n%2==1:
p*=2
r = m%(p)
print(r)
``` | 0 |
|
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"
}
] | 1,685,057,660 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 46 | 3,072,000 |
n = int(input())
counts = [0]*(10**5+2)
for num in map(int, input().split()):
counts[num] += 1
dp = [0]*(10**5+2)
for i in range(2, 10**5+2):
dp[i] = max(dp[i-1], dp[i-2] + counts[i]*i)
print(dp[10**5]) | 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())
counts = [0]*(10**5+2)
for num in map(int, input().split()):
counts[num] += 1
dp = [0]*(10**5+2)
for i in range(2, 10**5+2):
dp[i] = max(dp[i-1], dp[i-2] + counts[i]*i)
print(dp[10**5])
``` | 0 |
|
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"
}
] | 1,589,993,187 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 218 | 0 | # http://codeforces.com/contest/9/problem/A
x, y = list(map(int, input().split()))
num, den = 7 - max(x, y), 6
gcd = lambda m, n: m if not n else gcd(n, m % n)
div = gcd(num, den)
print(str(num // div) + '/' + str(den // div)) | 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
# http://codeforces.com/contest/9/problem/A
x, y = list(map(int, input().split()))
num, den = 7 - max(x, y), 6
gcd = lambda m, n: m if not n else gcd(n, m % n)
div = gcd(num, den)
print(str(num // div) + '/' + str(den // div))
``` | 3.891 |
453 | A | Little Pony and Expected Maximum | PROGRAMMING | 1,600 | [
"probabilities"
] | null | null | Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times. | A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105). | Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4. | [
"6 1\n",
"6 3\n",
"2 2\n"
] | [
"3.500000000000\n",
"4.958333333333\n",
"1.750000000000\n"
] | Consider the third test example. If you've made two tosses:
1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value | 500 | [
{
"input": "6 1",
"output": "3.500000000000"
},
{
"input": "6 3",
"output": "4.958333333333"
},
{
"input": "2 2",
"output": "1.750000000000"
},
{
"input": "5 4",
"output": "4.433600000000"
},
{
"input": "5 8",
"output": "4.814773760000"
},
{
"input": "3 10",
"output": "2.982641534996"
},
{
"input": "3 6",
"output": "2.910836762689"
},
{
"input": "1 8",
"output": "1.000000000000"
},
{
"input": "24438 9",
"output": "21994.699969310015"
},
{
"input": "94444 9",
"output": "85000.099992058866"
},
{
"input": "8 66716",
"output": "8.000000000000"
},
{
"input": "4 25132",
"output": "4.000000000000"
},
{
"input": "51520 73331",
"output": "51519.682650242677"
},
{
"input": "54230 31747",
"output": "54228.743352775018"
},
{
"input": "24236 90163",
"output": "24235.975171545670"
},
{
"input": "26946 99523",
"output": "26945.974480086279"
},
{
"input": "50323 7",
"output": "44033.124988408454"
},
{
"input": "53033 3",
"output": "39775.249995286234"
},
{
"input": "55743 5",
"output": "46452.999992525307"
},
{
"input": "59964 79",
"output": "59214.949890211828"
},
{
"input": "1 1",
"output": "1.000000000000"
},
{
"input": "1 1",
"output": "1.000000000000"
},
{
"input": "3 1",
"output": "2.000000000000"
},
{
"input": "1 2",
"output": "1.000000000000"
},
{
"input": "53513 34040",
"output": "53511.875329020870"
},
{
"input": "100000 100000",
"output": "99999.418033254507"
},
{
"input": "1 100000",
"output": "1.000000000000"
},
{
"input": "100000 1",
"output": "50000.499999999935"
},
{
"input": "2 100000",
"output": "2.000000000000"
},
{
"input": "100000 2",
"output": "66667.166665000332"
},
{
"input": "50000 100000",
"output": "49999.843487110789"
},
{
"input": "99999 1111",
"output": "99909.571915885972"
},
{
"input": "99999 99999",
"output": "99998.418033254609"
},
{
"input": "1000 1000",
"output": "999.419018443269"
},
{
"input": "50000 50000",
"output": "49999.418043215679"
},
{
"input": "88888 88888",
"output": "88887.418034499773"
},
{
"input": "99999 100000",
"output": "99998.418042461126"
},
{
"input": "100000 555",
"output": "99820.643422392372"
},
{
"input": "10000 10000",
"output": "9999.418122897887"
},
{
"input": "100000 5001",
"output": "99980.503829474910"
},
{
"input": "100000 1000",
"output": "99900.599066768002"
}
] | 1,609,194,745 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 1,000 | 819,200 | def binpow (a, n):
res = 1
while n:
if (n & 1):
res *= a
a *= a
n >>= 1
return res
m, n = list(map(int, input().split()))
res = 0
for k in range(1, m+1):
res += k * (binpow(k,n)-binpow((k-1),n))
res /= binpow(m,n)
print(res)
| Title: Little Pony and Expected Maximum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has *m* faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the *m*-th face contains *m* dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice *n* times.
Input Specification:
A single line contains two integers *m* and *n* (1<=≤<=*m*,<=*n*<=≤<=105).
Output Specification:
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=<=-<=4.
Demo Input:
['6 1\n', '6 3\n', '2 2\n']
Demo Output:
['3.500000000000\n', '4.958333333333\n', '1.750000000000\n']
Note:
Consider the third test example. If you've made two tosses:
1. You can get 1 in the first toss, and 2 in the second. Maximum equals to 2. 1. You can get 1 in the first toss, and 1 in the second. Maximum equals to 1. 1. You can get 2 in the first toss, and 1 in the second. Maximum equals to 2. 1. You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value | ```python
def binpow (a, n):
res = 1
while n:
if (n & 1):
res *= a
a *= a
n >>= 1
return res
m, n = list(map(int, input().split()))
res = 0
for k in range(1, m+1):
res += k * (binpow(k,n)-binpow((k-1),n))
res /= binpow(m,n)
print(res)
``` | 0 |
|
494 | A | Treasure | PROGRAMMING | 1,500 | [
"greedy"
] | null | null | Malek has recently found a treasure map. While he was looking for a treasure he found a locked door. There was a string *s* written on the door consisting of characters '(', ')' and '#'. Below there was a manual on how to open the door. After spending a long time Malek managed to decode the manual and found out that the goal is to replace each '#' with one or more ')' characters so that the final string becomes beautiful.
Below there was also written that a string is called beautiful if for each *i* (1<=≤<=*i*<=≤<=|*s*|) there are no more ')' characters than '(' characters among the first *i* characters of *s* and also the total number of '(' characters is equal to the total number of ')' characters.
Help Malek open the door by telling him for each '#' character how many ')' characters he must replace it with. | The first line of the input contains a string *s* (1<=≤<=|*s*|<=≤<=105). Each character of this string is one of the characters '(', ')' or '#'. It is guaranteed that *s* contains at least one '#' character. | If there is no way of replacing '#' characters which leads to a beautiful string print <=-<=1. Otherwise for each character '#' print a separate line containing a positive integer, the number of ')' characters this character must be replaced with.
If there are several possible answers, you may output any of them. | [
"(((#)((#)\n",
"()((#((#(#()\n",
"#\n",
"(#)\n"
] | [
"1\n2\n",
"2\n2\n1",
"-1\n",
"-1\n"
] | |*s*| denotes the length of the string *s*. | 500 | [
{
"input": "(((#)((#)",
"output": "1\n2"
},
{
"input": "()((#((#(#()",
"output": "1\n1\n3"
},
{
"input": "#",
"output": "-1"
},
{
"input": "(#)",
"output": "-1"
},
{
"input": "(((((#(#(#(#()",
"output": "1\n1\n1\n5"
},
{
"input": "#))))",
"output": "-1"
},
{
"input": "((#(()#(##",
"output": "1\n1\n1\n1"
},
{
"input": "##((((((()",
"output": "-1"
},
{
"input": "(((((((((((((((((((###################",
"output": "1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1"
},
{
"input": "((#)(",
"output": "-1"
},
{
"input": "((#)((#)((#)((#)((#)((#)((#)((#)((#)((#)((#)((#)((#)((#)((#)((##",
"output": "1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1"
},
{
"input": ")((##((###",
"output": "-1"
},
{
"input": "(#))(#(#)((((#(##((#(#((((#(##((((((#((()(()(())((()#((((#((()((((#(((((#(##)(##()((((()())(((((#(((",
"output": "-1"
},
{
"input": "#(#(#((##((()))(((#)(#()#(((()()(()#(##(((()(((()))#(((((()(((((((()#((#((()(#(((()(()##(()(((()((#(",
"output": "-1"
},
{
"input": "((#(",
"output": "-1"
},
{
"input": "()#(#())()()#)(#)()##)#((()#)((#)()#())((#((((((((#)()()(()()(((((#)#(#((((#((##()(##(((#(()(#((#))#",
"output": "-1"
},
{
"input": "(())((((#)",
"output": "3"
},
{
"input": "(#(",
"output": "-1"
},
{
"input": "((#)(",
"output": "-1"
},
{
"input": "(((()#(#)(",
"output": "-1"
},
{
"input": "#((#",
"output": "-1"
},
{
"input": "(#((((()",
"output": "-1"
},
{
"input": "(#((",
"output": "-1"
},
{
"input": ")(((())#",
"output": "-1"
}
] | 1,584,357,149 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 13 | 171 | 5,734,400 | s = input()
ret = [None] * len(s)
bal = [None] * len(s)
b = 0
last = None
for i, c in enumerate(s):
if c == '(':
b += 1
elif c == ')':
b -= 1
else:
ret[i] = 1
b -= 1
last = i
if b < 0:
print(-1)
exit()
bal[i] = b
ret[last] += bal[-1]
print('\n'.join(map(str, filter(lambda x: x, ret))) if bal[last] - bal[-1] >= 0 else -1) | Title: Treasure
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Malek has recently found a treasure map. While he was looking for a treasure he found a locked door. There was a string *s* written on the door consisting of characters '(', ')' and '#'. Below there was a manual on how to open the door. After spending a long time Malek managed to decode the manual and found out that the goal is to replace each '#' with one or more ')' characters so that the final string becomes beautiful.
Below there was also written that a string is called beautiful if for each *i* (1<=≤<=*i*<=≤<=|*s*|) there are no more ')' characters than '(' characters among the first *i* characters of *s* and also the total number of '(' characters is equal to the total number of ')' characters.
Help Malek open the door by telling him for each '#' character how many ')' characters he must replace it with.
Input Specification:
The first line of the input contains a string *s* (1<=≤<=|*s*|<=≤<=105). Each character of this string is one of the characters '(', ')' or '#'. It is guaranteed that *s* contains at least one '#' character.
Output Specification:
If there is no way of replacing '#' characters which leads to a beautiful string print <=-<=1. Otherwise for each character '#' print a separate line containing a positive integer, the number of ')' characters this character must be replaced with.
If there are several possible answers, you may output any of them.
Demo Input:
['(((#)((#)\n', '()((#((#(#()\n', '#\n', '(#)\n']
Demo Output:
['1\n2\n', '2\n2\n1', '-1\n', '-1\n']
Note:
|*s*| denotes the length of the string *s*. | ```python
s = input()
ret = [None] * len(s)
bal = [None] * len(s)
b = 0
last = None
for i, c in enumerate(s):
if c == '(':
b += 1
elif c == ')':
b -= 1
else:
ret[i] = 1
b -= 1
last = i
if b < 0:
print(-1)
exit()
bal[i] = b
ret[last] += bal[-1]
print('\n'.join(map(str, filter(lambda x: x, ret))) if bal[last] - bal[-1] >= 0 else -1)
``` | 0 |
|
242 | B | Big Segment | PROGRAMMING | 1,100 | [
"implementation",
"sortings"
] | null | null | A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment.
It is guaranteed that no two segments coincide. | Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input. | [
"3\n1 1\n2 2\n3 3\n",
"6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n"
] | [
"-1\n",
"3\n"
] | none | 1,000 | [
{
"input": "3\n1 1\n2 2\n3 3",
"output": "-1"
},
{
"input": "6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10",
"output": "3"
},
{
"input": "4\n1 5\n2 2\n2 4\n2 5",
"output": "1"
},
{
"input": "5\n3 3\n1 3\n2 2\n2 3\n1 2",
"output": "2"
},
{
"input": "7\n7 7\n8 8\n3 7\n1 6\n1 7\n4 7\n2 8",
"output": "-1"
},
{
"input": "3\n2 5\n3 4\n2 3",
"output": "1"
},
{
"input": "16\n15 15\n8 12\n6 9\n15 16\n8 14\n3 12\n7 19\n9 13\n5 16\n9 17\n10 15\n9 14\n9 9\n18 19\n5 15\n6 19",
"output": "-1"
},
{
"input": "9\n1 10\n7 8\n6 7\n1 4\n5 9\n2 8\n3 10\n1 1\n2 3",
"output": "1"
},
{
"input": "1\n1 100000",
"output": "1"
},
{
"input": "6\n2 2\n3 3\n3 5\n4 5\n1 1\n1 5",
"output": "6"
},
{
"input": "33\n2 18\n4 14\n2 16\n10 12\n4 6\n9 17\n2 8\n4 12\n8 20\n1 10\n11 14\n11 17\n8 15\n3 16\n3 4\n6 9\n6 19\n4 17\n17 19\n6 16\n3 12\n1 7\n6 20\n8 16\n12 19\n1 3\n12 18\n6 11\n7 20\n16 18\n4 15\n3 15\n15 19",
"output": "-1"
},
{
"input": "34\n3 8\n5 9\n2 9\n1 4\n3 7\n3 3\n8 9\n6 10\n4 7\n6 7\n5 8\n5 10\n1 5\n8 8\n2 5\n3 5\n7 7\n2 8\n4 5\n1 1\n7 9\n5 6\n2 3\n1 2\n2 4\n8 10\n7 8\n1 3\n4 8\n9 10\n1 7\n10 10\n2 2\n1 8",
"output": "-1"
},
{
"input": "55\n3 4\n6 8\n9 10\n3 9\n9 9\n2 5\n4 8\n3 8\n8 10\n1 1\n4 9\n10 10\n6 6\n8 8\n1 8\n5 5\n4 5\n5 9\n2 2\n3 10\n4 6\n3 6\n1 6\n1 7\n6 10\n2 6\n3 7\n2 4\n4 4\n5 10\n1 4\n2 9\n1 3\n7 9\n7 8\n1 9\n1 10\n2 8\n8 9\n6 7\n1 2\n6 9\n7 7\n4 7\n3 3\n2 7\n4 10\n7 10\n2 3\n2 10\n5 7\n3 5\n5 8\n1 5\n5 6",
"output": "37"
},
{
"input": "1\n999999999 1000000000",
"output": "1"
},
{
"input": "3\n1 20\n2 22\n3 18",
"output": "-1"
},
{
"input": "1\n1000000000 1000000000",
"output": "1"
},
{
"input": "2\n100001 100008\n100005 100006",
"output": "1"
},
{
"input": "1\n1000000 10000000",
"output": "1"
},
{
"input": "3\n3 6\n2 4\n1 5",
"output": "-1"
},
{
"input": "2\n3 5\n1 2",
"output": "-1"
}
] | 1,549,416,616 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 18 | 2,000 | 5,222,400 |
a = []
b = []
n = int(input())
for i in range (n): #[first, last)
x, y = map(int, input().split())
a.append(x)
b.append(y)
answer = False
for i in range (n):
if a[i] == min(a) and b[i] == max(b):
print(i+1)
answer = True
if answer == False:
print(-1)
| Title: Big Segment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment.
It is guaranteed that no two segments coincide.
Output Specification:
Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n']
Demo Output:
['-1\n', '3\n']
Note:
none | ```python
a = []
b = []
n = int(input())
for i in range (n): #[first, last)
x, y = map(int, input().split())
a.append(x)
b.append(y)
answer = False
for i in range (n):
if a[i] == min(a) and b[i] == max(b):
print(i+1)
answer = True
if answer == False:
print(-1)
``` | 0 |
|
175 | A | Robot Bicorn Attack | PROGRAMMING | 1,400 | [
"brute force",
"implementation"
] | null | null | Vasya plays Robot Bicorn Attack.
The game consists of three rounds. For each one a non-negative integer amount of points is given. The result of the game is the sum of obtained points. Vasya has already played three rounds and wrote obtained points one by one (without leading zeros) into the string *s*. Vasya decided to brag about his achievement to the friends. However, he has forgotten how many points he got for each round. The only thing he remembers is the string *s*.
Help Vasya to find out what is the maximum amount of points he could get. Take into account that Vasya played Robot Bicorn Attack for the first time, so he could not get more than 1000000 (106) points for one round. | The only line of input contains non-empty string *s* obtained by Vasya. The string consists of digits only. The string length does not exceed 30 characters. | Print the only number — the maximum amount of points Vasya could get. If Vasya is wrong and the string could not be obtained according to the rules then output number -1. | [
"1234\n",
"9000\n",
"0009\n"
] | [
"37\n",
"90\n",
"-1\n"
] | In the first example the string must be split into numbers 1, 2 and 34.
In the second example the string must be split into numbers 90, 0 and 0.
In the third example the string is incorrect, because after splitting the string into 3 numbers number 00 or 09 will be obtained, but numbers cannot have leading zeroes. | 500 | [
{
"input": "1234",
"output": "37"
},
{
"input": "9000",
"output": "90"
},
{
"input": "0009",
"output": "-1"
},
{
"input": "100000010000001000000",
"output": "3000000"
},
{
"input": "1000000011",
"output": "1000011"
},
{
"input": "9991",
"output": "109"
},
{
"input": "1000001999",
"output": "101000"
},
{
"input": "100000010000011000000",
"output": "-1"
},
{
"input": "100000010000001000001",
"output": "-1"
},
{
"input": "102",
"output": "3"
},
{
"input": "100",
"output": "1"
},
{
"input": "000",
"output": "0"
},
{
"input": "001",
"output": "1"
},
{
"input": "090",
"output": "9"
},
{
"input": "999999999999999999999999999999",
"output": "-1"
},
{
"input": "12345678901234567",
"output": "947035"
},
{
"input": "12635000000127683",
"output": "-1"
},
{
"input": "428595000000042345353",
"output": "-1"
},
{
"input": "48726340000000",
"output": "-1"
},
{
"input": "93246310000000",
"output": "1932463"
},
{
"input": "00123456",
"output": "123456"
},
{
"input": "001234567",
"output": "-1"
},
{
"input": "0891249843934",
"output": "1735183"
},
{
"input": "04234581000000",
"output": "1423458"
},
{
"input": "10000000999999",
"output": "1999999"
},
{
"input": "69284626624",
"output": "935092"
},
{
"input": "6061357",
"output": "61363"
},
{
"input": "1215",
"output": "27"
},
{
"input": "5305167573651040691",
"output": "-1"
},
{
"input": "948245431759126577",
"output": "1506581"
},
{
"input": "3699951264723380749",
"output": "-1"
},
{
"input": "97862382",
"output": "978633"
},
{
"input": "4073734152",
"output": "737433"
},
{
"input": "9123396793",
"output": "913138"
},
{
"input": "52027725398636318",
"output": "1413743"
},
{
"input": "03990796938958",
"output": "-1"
},
{
"input": "460657644093588",
"output": "1105338"
},
{
"input": "8793853284967905230",
"output": "-1"
},
{
"input": "298036933890712",
"output": "1232638"
},
{
"input": "80796461750373352",
"output": "1498819"
},
{
"input": "2376843019069559343",
"output": "-1"
},
{
"input": "7464340065209674219",
"output": "-1"
},
{
"input": "1551627800131899383",
"output": "-1"
},
{
"input": "20973541540356666047",
"output": "-1"
},
{
"input": "31587335978612",
"output": "1565978"
},
{
"input": "744503034",
"output": "744537"
},
{
"input": "15533274860535679",
"output": "939616"
},
{
"input": "58308242321837",
"output": "1006337"
},
{
"input": "9929120076123816",
"output": "323868"
},
{
"input": "623650711335",
"output": "773700"
},
{
"input": "1066168753173",
"output": "981936"
},
{
"input": "5353645507305053",
"output": "1091147"
},
{
"input": "111111111111111111111111111111",
"output": "-1"
},
{
"input": "1000000999888777666",
"output": "2777554"
},
{
"input": "1000001999888777666",
"output": "-1"
},
{
"input": "9998881000001777666",
"output": "-1"
},
{
"input": "9999991000000999999",
"output": "2999998"
},
{
"input": "9999999999991000000",
"output": "2999998"
},
{
"input": "1000000999999999999",
"output": "2999998"
},
{
"input": "123",
"output": "6"
},
{
"input": "132",
"output": "6"
},
{
"input": "213",
"output": "6"
},
{
"input": "231",
"output": "6"
},
{
"input": "312",
"output": "6"
},
{
"input": "321",
"output": "6"
},
{
"input": "666",
"output": "18"
},
{
"input": "012894600",
"output": "894612"
},
{
"input": "01289400",
"output": "289401"
},
{
"input": "429496729742949672974294967297",
"output": "-1"
},
{
"input": "429496729700",
"output": "959414"
},
{
"input": "429496729701",
"output": "959415"
},
{
"input": "1",
"output": "-1"
},
{
"input": "55",
"output": "-1"
},
{
"input": "99",
"output": "-1"
},
{
"input": "9991999999999999",
"output": "2009989"
},
{
"input": "999199999999099",
"output": "1999297"
},
{
"input": "9991990199999999",
"output": "2000189"
},
{
"input": "99999900999999",
"output": "1999998"
},
{
"input": "4444440044444",
"output": "488888"
},
{
"input": "0089",
"output": "89"
},
{
"input": "00243",
"output": "243"
},
{
"input": "008743",
"output": "8743"
},
{
"input": "0042764",
"output": "42764"
},
{
"input": "00912838",
"output": "912838"
},
{
"input": "001000000",
"output": "1000000"
},
{
"input": "001000001",
"output": "-1"
},
{
"input": "12435900",
"output": "435903"
},
{
"input": "28492300",
"output": "849232"
},
{
"input": "99999900",
"output": "999999"
},
{
"input": "931863000",
"output": "863094"
},
{
"input": "23566000",
"output": "566005"
},
{
"input": "100000000",
"output": "1000000"
},
{
"input": "100000100",
"output": "100010"
},
{
"input": "3246870032",
"output": "870362"
},
{
"input": "04354640947",
"output": "645301"
},
{
"input": "0734620342343",
"output": "1076963"
},
{
"input": "09999990",
"output": "999999"
},
{
"input": "010000000",
"output": "1000000"
},
{
"input": "010000001",
"output": "1000001"
},
{
"input": "01000001000000",
"output": "1100000"
},
{
"input": "1844674407370955161600",
"output": "-1"
},
{
"input": "184467440737105516141111",
"output": "-1"
},
{
"input": "1844674407371055161600",
"output": "-1"
},
{
"input": "1111111111111111111111111",
"output": "-1"
},
{
"input": "999999999999999999999",
"output": "-1"
},
{
"input": "100000110000011000001",
"output": "-1"
},
{
"input": "0018446744073709551616",
"output": "-1"
},
{
"input": "429596729610000001000",
"output": "-1"
},
{
"input": "123456789123456789123456789123",
"output": "-1"
},
{
"input": "011",
"output": "2"
},
{
"input": "10000100001000",
"output": "21000"
},
{
"input": "091",
"output": "10"
},
{
"input": "1234567891012131416123222443",
"output": "-1"
},
{
"input": "429496729642949672964294967296",
"output": "-1"
},
{
"input": "214748364821474836482147483648",
"output": "-1"
},
{
"input": "10000001073741824001000000",
"output": "-1"
},
{
"input": "5000000000",
"output": "-1"
},
{
"input": "999955555555555555555555",
"output": "-1"
},
{
"input": "1844674407371055161699",
"output": "-1"
},
{
"input": "99999999999999999999999999999",
"output": "-1"
},
{
"input": "009",
"output": "9"
},
{
"input": "1844674407370955161611",
"output": "-1"
},
{
"input": "000000999",
"output": "-1"
},
{
"input": "003",
"output": "3"
},
{
"input": "0000000009000000000",
"output": "-1"
},
{
"input": "010",
"output": "1"
},
{
"input": "0",
"output": "-1"
},
{
"input": "000009000000",
"output": "-1"
},
{
"input": "21371283283242384239423423123",
"output": "-1"
},
{
"input": "100000010004294967300",
"output": "-1"
},
{
"input": "429596729600",
"output": "969314"
},
{
"input": "429496729510000001000000",
"output": "-1"
},
{
"input": "2147483648010000010000",
"output": "-1"
},
{
"input": "349967407479541078587181768161",
"output": "-1"
}
] | 1,556,048,800 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 109 | 0 | x = str(input())
y = len(x)
s=0
p=0
for i in range(y-1):
if(int(x[s] + x[s+1]) > p):
p = int(x[s] + x[s+1])
s+=1
x = x.replace(str(p), '')
aw=sum(map(int,str(x))) + p
print(aw) | Title: Robot Bicorn Attack
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya plays Robot Bicorn Attack.
The game consists of three rounds. For each one a non-negative integer amount of points is given. The result of the game is the sum of obtained points. Vasya has already played three rounds and wrote obtained points one by one (without leading zeros) into the string *s*. Vasya decided to brag about his achievement to the friends. However, he has forgotten how many points he got for each round. The only thing he remembers is the string *s*.
Help Vasya to find out what is the maximum amount of points he could get. Take into account that Vasya played Robot Bicorn Attack for the first time, so he could not get more than 1000000 (106) points for one round.
Input Specification:
The only line of input contains non-empty string *s* obtained by Vasya. The string consists of digits only. The string length does not exceed 30 characters.
Output Specification:
Print the only number — the maximum amount of points Vasya could get. If Vasya is wrong and the string could not be obtained according to the rules then output number -1.
Demo Input:
['1234\n', '9000\n', '0009\n']
Demo Output:
['37\n', '90\n', '-1\n']
Note:
In the first example the string must be split into numbers 1, 2 and 34.
In the second example the string must be split into numbers 90, 0 and 0.
In the third example the string is incorrect, because after splitting the string into 3 numbers number 00 or 09 will be obtained, but numbers cannot have leading zeroes. | ```python
x = str(input())
y = len(x)
s=0
p=0
for i in range(y-1):
if(int(x[s] + x[s+1]) > p):
p = int(x[s] + x[s+1])
s+=1
x = x.replace(str(p), '')
aw=sum(map(int,str(x))) + p
print(aw)
``` | 0 |
|
292 | A | SMSC | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Some large corporation where Polycarpus works has its own short message service center (SMSC). The center's task is to send all sorts of crucial information. Polycarpus decided to check the efficiency of the SMSC.
For that, he asked to give him the statistics of the performance of the SMSC for some period of time. In the end, Polycarpus got a list of *n* tasks that went to the SMSC of the corporation. Each task was described by the time it was received by the SMSC and the number of text messages to send. More formally, the *i*-th task was described by two integers *t**i* and *c**i* — the receiving time (the second) and the number of the text messages, correspondingly.
Polycarpus knows that the SMSC cannot send more than one text message per second. The SMSC uses a queue to organize its work. Consider a time moment *x*, the SMSC work at that moment as follows:
1. If at the time moment *x* the queue is not empty, then SMSC sends one message from the queue (SMSC gets the message from the head of the queue). Otherwise it doesn't send messages at the time moment *x*. 1. If at the time moment *x* SMSC receives a task, then it adds to the queue all the messages from this task (SMSC adds messages to the tail of the queue). Note, that the messages from the task cannot be send at time moment *x*. That's because the decision about sending message or not is made at point 1 before adding these messages to the queue.
Given the information about all *n* tasks, Polycarpus wants to count two values: the time when the last text message was sent and the maximum size of the queue at some time. Help him count these two characteristics he needs to evaluate the efficiency of the SMSC. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=103) — the number of tasks of the SMSC. Next *n* lines contain the tasks' descriptions: the *i*-th line contains two space-separated integers *t**i* and *c**i* (1<=≤<=*t**i*,<=*c**i*<=≤<=106) — the time (the second) when the *i*-th task was received and the number of messages to send, correspondingly.
It is guaranteed that all tasks were received at different moments of time. It is guaranteed that the tasks are sorted in the chronological order, that is, *t**i*<=<<=*t**i*<=+<=1 for all integer *i* (1<=≤<=*i*<=<<=*n*). | In a single line print two space-separated integers — the time when the last text message was sent and the maximum queue size at a certain moment of time. | [
"2\n1 1\n2 1\n",
"1\n1000000 10\n",
"3\n3 3\n4 3\n5 3\n"
] | [
"3 1\n",
"1000010 10\n",
"12 7\n"
] | In the first test sample:
- second 1: the first message has appeared in the queue, the queue's size is 1; - second 2: the first message is sent, the second message has been received, the queue's size is 1; - second 3: the second message is sent, the queue's size is 0,
Thus, the maximum size of the queue is 1, the last message was sent at the second 3. | 1,000 | [
{
"input": "2\n1 1\n2 1",
"output": "3 1"
},
{
"input": "1\n1000000 10",
"output": "1000010 10"
},
{
"input": "3\n3 3\n4 3\n5 3",
"output": "12 7"
},
{
"input": "1\n1 1",
"output": "2 1"
},
{
"input": "2\n1 11\n100 10",
"output": "110 11"
},
{
"input": "4\n1 10\n2 9\n3 8\n40 3",
"output": "43 25"
},
{
"input": "5\n2 1\n5 2\n6 1\n7 1\n8 1",
"output": "10 2"
},
{
"input": "4\n10 1000\n99998 20\n99999 10\n1000000 100",
"output": "1000100 1000"
},
{
"input": "6\n10 10\n100 500\n200 500\n500 1\n999995 4\n999996 15",
"output": "1000014 900"
},
{
"input": "10\n1 5\n2 5\n3 10\n4 8\n5 5\n6 4\n7 8\n8 9\n9 2\n10 10",
"output": "67 57"
},
{
"input": "10\n26 4\n85 97\n86 62\n87 74\n92 8\n93 81\n97 12\n98 25\n99 31\n100 3",
"output": "478 378"
},
{
"input": "10\n964416 3980\n987048 334\n999576 6922\n999684 2385\n999896 6558\n999948 3515\n999966 1517\n999984 2233\n999988 7242\n999994 91",
"output": "1030039 30045"
},
{
"input": "50\n3 16\n25 5\n35 6\n39 9\n48 9\n50 12\n57 9\n58 2\n59 9\n60 16\n61 8\n62 7\n63 15\n64 3\n65 16\n66 12\n67 12\n68 8\n69 5\n70 11\n71 4\n72 6\n73 12\n74 20\n75 12\n76 11\n77 5\n78 15\n79 12\n80 5\n81 4\n82 17\n83 14\n84 11\n85 6\n86 10\n87 12\n88 8\n89 9\n90 2\n91 17\n92 14\n93 7\n94 15\n95 8\n96 3\n97 14\n98 13\n99 18\n100 7",
"output": "515 415"
},
{
"input": "40\n313 97063\n372 23668\n377 56079\n428 88458\n435 57330\n454 88869\n456 75553\n466 87607\n468 4542\n469 30243\n471 8958\n472 59745\n473 90985\n474 32247\n475 71576\n476 14016\n477 91574\n478 38345\n479 93280\n480 89902\n481 41828\n482 3233\n483 66725\n484 23523\n485 46772\n486 95688\n487 19864\n488 10295\n489 83431\n490 51687\n491 4362\n492 85411\n493 30044\n494 97895\n495 6379\n496 79232\n497 341\n498 82766\n499 77612\n500 76069",
"output": "2183510 2183010"
},
{
"input": "5\n987640 52\n994481 69\n995526 50\n996631 75\n999763 22",
"output": "999785 75"
},
{
"input": "30\n1227 3920\n2007 7796\n3074 5424\n5172 6174\n5314 8761\n5329 5644\n5352 5880\n5395 2721\n5403 4406\n5420 3835\n5421 2692\n5423 8241\n5425 6433\n5426 2900\n5427 2209\n5428 8672\n5429 3317\n5430 4857\n5431 5735\n5432 2784\n5433 1564\n5434 2094\n5435 6258\n5436 101\n5437 3745\n5438 7124\n5439 1484\n5440 8593\n5441 7198\n5442 8318",
"output": "150107 144665"
},
{
"input": "23\n5 1045\n12 703\n16 26\n23 3384\n28 4563\n30 4501\n34 1033\n35 1393\n36 4095\n37 1279\n38 1787\n39 770\n40 5362\n41 4569\n42 3148\n43 2619\n44 5409\n45 3919\n46 732\n47 1297\n48 4512\n49 3231\n50 5169",
"output": "64551 64501"
}
] | 1,574,444,576 | 2,147,483,647 | Python 3 | OK | TESTS | 38 | 218 | 204,800 | n = int(input())
message = 0
m = 0
l = 0
for _ in range(n):
t, c = map(int, input().split())
message = max(0, message-(t-l))
message += c
m = max(message, m)
l = t
print(l+message, m)
| Title: SMSC
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some large corporation where Polycarpus works has its own short message service center (SMSC). The center's task is to send all sorts of crucial information. Polycarpus decided to check the efficiency of the SMSC.
For that, he asked to give him the statistics of the performance of the SMSC for some period of time. In the end, Polycarpus got a list of *n* tasks that went to the SMSC of the corporation. Each task was described by the time it was received by the SMSC and the number of text messages to send. More formally, the *i*-th task was described by two integers *t**i* and *c**i* — the receiving time (the second) and the number of the text messages, correspondingly.
Polycarpus knows that the SMSC cannot send more than one text message per second. The SMSC uses a queue to organize its work. Consider a time moment *x*, the SMSC work at that moment as follows:
1. If at the time moment *x* the queue is not empty, then SMSC sends one message from the queue (SMSC gets the message from the head of the queue). Otherwise it doesn't send messages at the time moment *x*. 1. If at the time moment *x* SMSC receives a task, then it adds to the queue all the messages from this task (SMSC adds messages to the tail of the queue). Note, that the messages from the task cannot be send at time moment *x*. That's because the decision about sending message or not is made at point 1 before adding these messages to the queue.
Given the information about all *n* tasks, Polycarpus wants to count two values: the time when the last text message was sent and the maximum size of the queue at some time. Help him count these two characteristics he needs to evaluate the efficiency of the SMSC.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=103) — the number of tasks of the SMSC. Next *n* lines contain the tasks' descriptions: the *i*-th line contains two space-separated integers *t**i* and *c**i* (1<=≤<=*t**i*,<=*c**i*<=≤<=106) — the time (the second) when the *i*-th task was received and the number of messages to send, correspondingly.
It is guaranteed that all tasks were received at different moments of time. It is guaranteed that the tasks are sorted in the chronological order, that is, *t**i*<=<<=*t**i*<=+<=1 for all integer *i* (1<=≤<=*i*<=<<=*n*).
Output Specification:
In a single line print two space-separated integers — the time when the last text message was sent and the maximum queue size at a certain moment of time.
Demo Input:
['2\n1 1\n2 1\n', '1\n1000000 10\n', '3\n3 3\n4 3\n5 3\n']
Demo Output:
['3 1\n', '1000010 10\n', '12 7\n']
Note:
In the first test sample:
- second 1: the first message has appeared in the queue, the queue's size is 1; - second 2: the first message is sent, the second message has been received, the queue's size is 1; - second 3: the second message is sent, the queue's size is 0,
Thus, the maximum size of the queue is 1, the last message was sent at the second 3. | ```python
n = int(input())
message = 0
m = 0
l = 0
for _ in range(n):
t, c = map(int, input().split())
message = max(0, message-(t-l))
message += c
m = max(message, m)
l = t
print(l+message, m)
``` | 3 |
|
678 | A | Johny Likes Numbers | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*. | The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109). | Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*. | [
"5 3\n",
"25 13\n",
"26 13\n"
] | [
"6\n",
"26\n",
"39\n"
] | none | 0 | [
{
"input": "5 3",
"output": "6"
},
{
"input": "25 13",
"output": "26"
},
{
"input": "26 13",
"output": "39"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "8 8",
"output": "16"
},
{
"input": "14 15",
"output": "15"
},
{
"input": "197 894",
"output": "894"
},
{
"input": "6058 8581",
"output": "8581"
},
{
"input": "97259 41764",
"output": "125292"
},
{
"input": "453145 333625",
"output": "667250"
},
{
"input": "2233224 4394826",
"output": "4394826"
},
{
"input": "76770926 13350712",
"output": "80104272"
},
{
"input": "687355301 142098087",
"output": "710490435"
},
{
"input": "1000000000 999999999",
"output": "1999999998"
},
{
"input": "1000000000 1000000000",
"output": "2000000000"
},
{
"input": "999999999 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 1",
"output": "1000000001"
},
{
"input": "1000000000 2",
"output": "1000000002"
},
{
"input": "999999999 1",
"output": "1000000000"
},
{
"input": "100000000 1",
"output": "100000001"
},
{
"input": "999999999 500000000",
"output": "1000000000"
},
{
"input": "999999990 10",
"output": "1000000000"
},
{
"input": "1000000000 999999997",
"output": "1999999994"
},
{
"input": "999999999 2",
"output": "1000000000"
},
{
"input": "999999984 1",
"output": "999999985"
},
{
"input": "999999983 1",
"output": "999999984"
},
{
"input": "666666666 1",
"output": "666666667"
},
{
"input": "1000000000 990000000",
"output": "1980000000"
},
{
"input": "41 48",
"output": "48"
},
{
"input": "123456 2",
"output": "123458"
},
{
"input": "111 111",
"output": "222"
},
{
"input": "878787 1",
"output": "878788"
},
{
"input": "121 1",
"output": "122"
},
{
"input": "114514 114514",
"output": "229028"
},
{
"input": "500000001 1000000000",
"output": "1000000000"
},
{
"input": "999999997 1",
"output": "999999998"
},
{
"input": "100000000 10",
"output": "100000010"
}
] | 1,478,525,383 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 62 | 0 | n, k = [int(x) for x in input().split(' ')]
print(n + k-n%k)
| Title: Johny Likes Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
Output Specification:
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
Demo Input:
['5 3\n', '25 13\n', '26 13\n']
Demo Output:
['6\n', '26\n', '39\n']
Note:
none | ```python
n, k = [int(x) for x in input().split(' ')]
print(n + k-n%k)
``` | 3 |
|
604 | A | Uncowed Forces | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.
Codeforces scores are computed as follows: If the maximum point value of a problem is *x*, and Kevin submitted correctly at minute *m* but made *w* wrong submissions, then his score on that problem is . His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points for each unsuccessful hack.
All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer. | The first line of the input contains five space-separated integers *m*1, *m*2, *m*3, *m*4, *m*5, where *m**i* (0<=≤<=*m**i*<=≤<=119) is the time of Kevin's last submission for problem *i*. His last submission is always correct and gets accepted.
The second line contains five space-separated integers *w*1, *w*2, *w*3, *w*4, *w*5, where *w**i* (0<=≤<=*w**i*<=≤<=10) is Kevin's number of wrong submissions on problem *i*.
The last line contains two space-separated integers *h**s* and *h**u* (0<=≤<=*h**s*,<=*h**u*<=≤<=20), denoting the Kevin's numbers of successful and unsuccessful hacks, respectively. | Print a single integer, the value of Kevin's final score. | [
"20 40 60 80 100\n0 1 2 3 4\n1 0\n",
"119 119 119 119 119\n0 0 0 0 0\n10 0\n"
] | [
"4900\n",
"4930\n"
] | In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/42158dc2bc78cd21fa679530ae9ef8b9ea298d15.png" style="max-width: 100.0%;max-height: 100.0%;"/> of the points on each problem. So his score from solving problems is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/fdf392d8508500b57f8057ac0c4c892ab5f925a2.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Adding in 10·100 = 1000 points from hacks, his total score becomes 3930 + 1000 = 4930. | 500 | [
{
"input": "20 40 60 80 100\n0 1 2 3 4\n1 0",
"output": "4900"
},
{
"input": "119 119 119 119 119\n0 0 0 0 0\n10 0",
"output": "4930"
},
{
"input": "3 6 13 38 60\n6 10 10 3 8\n9 9",
"output": "5088"
},
{
"input": "21 44 11 68 75\n6 2 4 8 4\n2 8",
"output": "4522"
},
{
"input": "16 112 50 114 68\n1 4 8 4 9\n19 11",
"output": "5178"
},
{
"input": "55 66 75 44 47\n6 0 6 6 10\n19 0",
"output": "6414"
},
{
"input": "47 11 88 5 110\n6 10 4 2 3\n10 6",
"output": "5188"
},
{
"input": "5 44 61 103 92\n9 0 10 4 8\n15 7",
"output": "4914"
},
{
"input": "115 53 96 62 110\n7 8 1 7 9\n7 16",
"output": "3416"
},
{
"input": "102 83 26 6 11\n3 4 1 8 3\n17 14",
"output": "6704"
},
{
"input": "36 102 73 101 19\n5 9 2 2 6\n4 13",
"output": "4292"
},
{
"input": "40 115 93 107 113\n5 7 2 6 8\n6 17",
"output": "2876"
},
{
"input": "53 34 53 107 81\n4 3 1 10 8\n7 7",
"output": "4324"
},
{
"input": "113 37 4 84 66\n2 0 10 3 0\n20 19",
"output": "6070"
},
{
"input": "10 53 101 62 1\n8 0 9 7 9\n0 11",
"output": "4032"
},
{
"input": "45 45 75 36 76\n6 2 2 0 0\n8 17",
"output": "5222"
},
{
"input": "47 16 44 78 111\n7 9 8 0 2\n1 19",
"output": "3288"
},
{
"input": "7 54 39 102 31\n6 0 2 10 1\n18 3",
"output": "6610"
},
{
"input": "0 46 86 72 40\n1 5 5 5 9\n6 5",
"output": "4924"
},
{
"input": "114 4 45 78 113\n0 4 8 10 2\n10 12",
"output": "4432"
},
{
"input": "56 56 96 105 107\n4 9 10 4 8\n2 1",
"output": "3104"
},
{
"input": "113 107 59 50 56\n3 7 10 6 3\n10 12",
"output": "4586"
},
{
"input": "96 104 9 94 84\n6 10 7 8 3\n14 11",
"output": "4754"
},
{
"input": "98 15 116 43 55\n4 3 0 9 3\n10 7",
"output": "5400"
},
{
"input": "0 26 99 108 35\n0 4 3 0 10\n9 5",
"output": "5388"
},
{
"input": "89 24 51 49 84\n5 6 2 2 9\n2 14",
"output": "4066"
},
{
"input": "57 51 76 45 96\n1 0 4 3 6\n12 15",
"output": "5156"
},
{
"input": "79 112 37 36 116\n2 8 4 7 5\n4 12",
"output": "3872"
},
{
"input": "71 42 60 20 7\n7 1 1 10 6\n1 7",
"output": "5242"
},
{
"input": "86 10 66 80 55\n0 2 5 10 5\n15 6",
"output": "5802"
},
{
"input": "66 109 22 22 62\n3 1 5 4 5\n10 5",
"output": "5854"
},
{
"input": "97 17 43 84 58\n2 8 3 8 6\n10 7",
"output": "5028"
},
{
"input": "109 83 5 114 104\n6 0 3 9 5\n5 2",
"output": "4386"
},
{
"input": "94 18 24 91 105\n2 0 7 10 3\n1 4",
"output": "4118"
},
{
"input": "64 17 86 59 45\n8 0 10 2 2\n4 4",
"output": "5144"
},
{
"input": "70 84 31 57 2\n7 0 0 2 7\n12 5",
"output": "6652"
},
{
"input": "98 118 117 86 4\n2 10 9 7 5\n11 15",
"output": "4476"
},
{
"input": "103 110 101 97 70\n4 2 1 0 5\n7 5",
"output": "4678"
},
{
"input": "78 96 6 97 62\n7 7 9 2 9\n10 3",
"output": "4868"
},
{
"input": "95 28 3 31 115\n1 9 0 7 3\n10 13",
"output": "5132"
},
{
"input": "45 17 116 58 3\n8 8 7 6 4\n3 19",
"output": "3992"
},
{
"input": "19 12 0 113 77\n3 0 10 9 2\n8 6",
"output": "5040"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0",
"output": "7500"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n20 0",
"output": "9500"
},
{
"input": "119 119 119 119 119\n10 10 10 10 10\n0 20",
"output": "1310"
},
{
"input": "0 0 0 0 0\n10 10 10 10 10\n0 20",
"output": "4150"
},
{
"input": "119 0 0 0 0\n10 0 0 0 0\n5 5",
"output": "7400"
},
{
"input": "0 119 0 0 0\n0 10 0 0 0\n5 5",
"output": "7050"
},
{
"input": "0 0 119 0 0\n0 0 10 0 0\n0 0",
"output": "6450"
},
{
"input": "0 0 0 119 0\n0 0 0 10 0\n5 5",
"output": "6350"
},
{
"input": "0 0 0 0 119\n0 0 0 0 10\n5 5",
"output": "6060"
},
{
"input": "119 0 0 0 0\n2 0 0 0 0\n5 5",
"output": "7412"
},
{
"input": "0 119 0 0 0\n0 2 0 0 0\n5 5",
"output": "7174"
},
{
"input": "0 0 119 0 0\n0 0 2 0 0\n5 5",
"output": "6936"
},
{
"input": "0 0 0 119 0\n0 0 0 2 0\n5 5",
"output": "6698"
},
{
"input": "0 0 0 0 119\n0 0 0 0 2\n5 5",
"output": "6460"
},
{
"input": "119 0 0 0 0\n0 0 0 0 0\n4 9",
"output": "7212"
}
] | 1,448,984,502 | 402 | Python 3 | OK | TESTS | 57 | 61 | 0 | max_scores = [ 500, 1000, 1500, 2000, 2500 ]
corects = [ int(c) for c in input().split() ]
wrongs = [ int(w) for w in input().split() ]
shacks, unshacks = map(int, input().split())
final_score = shacks*100 - unshacks*50
def score(x, m, w):
return max(0.3*x, (1-(m/250))*x-50*w)
for i in range(5):
final_score += score(max_scores[i], corects[i], wrongs[i])
print(int(final_score)) | Title: Uncowed Forces
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.
Codeforces scores are computed as follows: If the maximum point value of a problem is *x*, and Kevin submitted correctly at minute *m* but made *w* wrong submissions, then his score on that problem is . His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points for each unsuccessful hack.
All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer.
Input Specification:
The first line of the input contains five space-separated integers *m*1, *m*2, *m*3, *m*4, *m*5, where *m**i* (0<=≤<=*m**i*<=≤<=119) is the time of Kevin's last submission for problem *i*. His last submission is always correct and gets accepted.
The second line contains five space-separated integers *w*1, *w*2, *w*3, *w*4, *w*5, where *w**i* (0<=≤<=*w**i*<=≤<=10) is Kevin's number of wrong submissions on problem *i*.
The last line contains two space-separated integers *h**s* and *h**u* (0<=≤<=*h**s*,<=*h**u*<=≤<=20), denoting the Kevin's numbers of successful and unsuccessful hacks, respectively.
Output Specification:
Print a single integer, the value of Kevin's final score.
Demo Input:
['20 40 60 80 100\n0 1 2 3 4\n1 0\n', '119 119 119 119 119\n0 0 0 0 0\n10 0\n']
Demo Output:
['4900\n', '4930\n']
Note:
In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/42158dc2bc78cd21fa679530ae9ef8b9ea298d15.png" style="max-width: 100.0%;max-height: 100.0%;"/> of the points on each problem. So his score from solving problems is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/fdf392d8508500b57f8057ac0c4c892ab5f925a2.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Adding in 10·100 = 1000 points from hacks, his total score becomes 3930 + 1000 = 4930. | ```python
max_scores = [ 500, 1000, 1500, 2000, 2500 ]
corects = [ int(c) for c in input().split() ]
wrongs = [ int(w) for w in input().split() ]
shacks, unshacks = map(int, input().split())
final_score = shacks*100 - unshacks*50
def score(x, m, w):
return max(0.3*x, (1-(m/250))*x-50*w)
for i in range(5):
final_score += score(max_scores[i], corects[i], wrongs[i])
print(int(final_score))
``` | 3 |
|
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,687,993,101 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | x = input()
if not x.isdigit():
if len(x) > 10:
print((x[0]) + str(len(x)-2) + (x[-1]))
else:
print(x) | Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
x = input()
if not x.isdigit():
if len(x) > 10:
print((x[0]) + str(len(x)-2) + (x[-1]))
else:
print(x)
``` | 0 |
834 | A | The Useless Toy | PROGRAMMING | 900 | [
"implementation"
] | null | null | Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.
Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):
After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.
Slastyona managed to have spinner rotating for exactly *n* seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this. | There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.
In the second strings, a single number *n* is given (0<=≤<=*n*<=≤<=109) – the duration of the rotation.
It is guaranteed that the ending position of a spinner is a result of a *n* second spin in any of the directions, assuming the given starting position. | Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise. | [
"^ >\n1\n",
"< ^\n3\n",
"^ v\n6\n"
] | [
"cw\n",
"ccw\n",
"undefined\n"
] | none | 500 | [
{
"input": "^ >\n1",
"output": "cw"
},
{
"input": "< ^\n3",
"output": "ccw"
},
{
"input": "^ v\n6",
"output": "undefined"
},
{
"input": "^ >\n999999999",
"output": "ccw"
},
{
"input": "> v\n1",
"output": "cw"
},
{
"input": "v <\n1",
"output": "cw"
},
{
"input": "< ^\n1",
"output": "cw"
},
{
"input": "v <\n422435957",
"output": "cw"
},
{
"input": "v >\n139018901",
"output": "ccw"
},
{
"input": "v ^\n571728018",
"output": "undefined"
},
{
"input": "^ ^\n0",
"output": "undefined"
},
{
"input": "< >\n2",
"output": "undefined"
},
{
"input": "> >\n1000000000",
"output": "undefined"
},
{
"input": "v v\n8",
"output": "undefined"
},
{
"input": "< <\n1568",
"output": "undefined"
},
{
"input": "^ v\n2",
"output": "undefined"
},
{
"input": "^ <\n1",
"output": "ccw"
},
{
"input": "< v\n1",
"output": "ccw"
},
{
"input": "v >\n1",
"output": "ccw"
},
{
"input": "> ^\n1",
"output": "ccw"
},
{
"input": "v <\n422435957",
"output": "cw"
},
{
"input": "v v\n927162384",
"output": "undefined"
},
{
"input": "v ^\n571728018",
"output": "undefined"
},
{
"input": "^ <\n467441155",
"output": "cw"
},
{
"input": "^ >\n822875521",
"output": "cw"
},
{
"input": "^ <\n821690113",
"output": "ccw"
},
{
"input": "^ <\n171288453",
"output": "ccw"
},
{
"input": "^ <\n110821381",
"output": "ccw"
},
{
"input": "^ ^\n539580280",
"output": "undefined"
},
{
"input": "^ >\n861895563",
"output": "ccw"
},
{
"input": "v v\n4",
"output": "undefined"
},
{
"input": "^ ^\n4",
"output": "undefined"
},
{
"input": "> >\n4",
"output": "undefined"
},
{
"input": "< <\n8",
"output": "undefined"
},
{
"input": "v v\n0",
"output": "undefined"
},
{
"input": "^ <\n11",
"output": "cw"
},
{
"input": "< <\n4",
"output": "undefined"
},
{
"input": "< <\n0",
"output": "undefined"
},
{
"input": "< v\n3",
"output": "cw"
},
{
"input": "^ <\n3",
"output": "cw"
},
{
"input": "^ <\n7",
"output": "cw"
},
{
"input": "< >\n6",
"output": "undefined"
},
{
"input": "v >\n3",
"output": "cw"
},
{
"input": "> >\n300",
"output": "undefined"
},
{
"input": "> >\n0",
"output": "undefined"
},
{
"input": "v <\n3",
"output": "ccw"
},
{
"input": "> >\n12",
"output": "undefined"
}
] | 1,501,750,893 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 93 | 23,142,400 | dict = {'v': 0, '<': 1, '^': 2, '>': 3}
i, j = input().split()
num = int(input())
l = dict[i]
r = dict[j]
s = (l-r if l>r else r-l) % 4
if s == 0 or s == 2:
print("undefined")
elif s == 1:
print("cw")
else:
print("ccw")
| Title: The Useless Toy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.
Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):
After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.
Slastyona managed to have spinner rotating for exactly *n* seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this.
Input Specification:
There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.
In the second strings, a single number *n* is given (0<=≤<=*n*<=≤<=109) – the duration of the rotation.
It is guaranteed that the ending position of a spinner is a result of a *n* second spin in any of the directions, assuming the given starting position.
Output Specification:
Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise.
Demo Input:
['^ >\n1\n', '< ^\n3\n', '^ v\n6\n']
Demo Output:
['cw\n', 'ccw\n', 'undefined\n']
Note:
none | ```python
dict = {'v': 0, '<': 1, '^': 2, '>': 3}
i, j = input().split()
num = int(input())
l = dict[i]
r = dict[j]
s = (l-r if l>r else r-l) % 4
if s == 0 or s == 2:
print("undefined")
elif s == 1:
print("cw")
else:
print("ccw")
``` | 0 |
|
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"
}
] | 1,655,134,274 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 0 | 0 | X,Y=map(int,input().split())
res=int(A*B/2)
print(res) | 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
X,Y=map(int,input().split())
res=int(A*B/2)
print(res)
``` | -1 |
0 | none | none | none | 0 | [
"none"
] | null | null | Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built *n* towers in a row. The *i*-th tower is made of *h**i* identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109) — sizes of towers. | Print the number of operations needed to destroy all towers. | [
"6\n2 1 4 6 2 2\n",
"7\n3 3 3 1 3 3 3\n"
] | [
"3\n",
"2\n"
] | The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. | 0 | [
{
"input": "6\n2 1 4 6 2 2",
"output": "3"
},
{
"input": "7\n3 3 3 1 3 3 3",
"output": "2"
},
{
"input": "7\n5128 5672 5805 5452 5882 5567 5032",
"output": "4"
},
{
"input": "10\n1 2 2 3 5 5 5 4 2 1",
"output": "5"
},
{
"input": "14\n20 20 20 20 20 20 3 20 20 20 20 20 20 20",
"output": "5"
},
{
"input": "50\n3 2 4 3 5 3 4 5 3 2 3 3 3 4 5 4 2 2 3 3 4 4 3 2 3 3 2 3 4 4 5 2 5 2 3 5 4 4 2 2 3 5 2 5 2 2 5 4 5 4",
"output": "4"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n1049 1098",
"output": "1"
},
{
"input": "2\n100 100",
"output": "1"
},
{
"input": "5\n1 2 3 2 1",
"output": "3"
},
{
"input": "15\n2 2 1 1 2 2 2 2 2 2 2 2 2 1 2",
"output": "2"
},
{
"input": "28\n415546599 415546599 415546599 415546599 415546599 415546599 415546599 415546599 415546599 2 802811737 802811737 802811737 802811737 802811737 802811737 802811737 802811737 1 550595901 550595901 550595901 550595901 550595901 550595901 550595901 550595901 550595901",
"output": "6"
},
{
"input": "45\n3 12 13 11 13 13 10 11 14 15 15 13 14 12 13 11 14 10 10 14 14 11 10 12 11 11 13 14 10 11 14 13 14 11 11 11 12 15 1 10 15 12 14 14 14",
"output": "13"
},
{
"input": "84\n1 3 4 5 6 5 6 7 8 9 7 4 5 4 2 5 1 1 1 3 2 7 7 8 10 9 5 6 5 2 3 3 3 3 3 2 4 8 6 5 8 9 8 7 9 3 4 4 4 2 2 1 6 4 9 5 9 9 10 7 10 4 5 4 2 4 3 3 4 4 6 6 6 9 10 12 7 5 9 8 5 3 3 2",
"output": "8"
},
{
"input": "170\n1 2 1 2 1 1 1 1 2 3 2 1 1 2 2 1 2 1 2 1 1 2 3 3 2 1 1 1 1 1 1 1 1 2 1 2 3 3 2 1 2 2 1 2 3 2 1 1 2 3 2 1 2 1 1 1 2 3 3 2 1 2 1 2 1 1 1 2 1 2 1 1 2 2 1 1 2 1 2 2 1 2 1 2 2 1 2 1 2 3 2 1 1 2 3 4 4 3 2 1 2 1 2 1 2 3 3 2 1 2 1 1 1 1 1 1 1 2 2 1 1 2 1 1 1 1 2 1 1 2 3 2 1 2 2 1 2 1 1 1 2 2 1 2 1 2 3 2 1 2 1 1 1 2 3 4 5 4 3 2 1 1 2 1 2 3 4 3 2 1",
"output": "5"
},
{
"input": "1\n5",
"output": "1"
}
] | 1,500,489,881 | 6,281 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
#define mp make_pair
#define pb push_back
#define F first
#define S second
const int MAX_SIZE = 1e5 + 10;
const ll inf = 1e14;
vector<pll > graph[MAX_SIZE];
priority_queue< pll, vector<pll > , greater< pll > > pq;
vector<ll> dist(MAX_SIZE);
int n;
void dijkstra(int s)
{
for(int i = 0; i <= n + 1; i++)
dist[i] = inf;
dist[s] = 0;
pq.push({0, s});
while(!pq.empty())
{
int u = pq.top().S;
pq.pop();
for(auto &e: graph[u])
{
int v = e.F;
ll w = e.S;
if(dist[v] > w + dist[u])
{
dist[v] = w + dist[u];
pq.push({dist[v], v});
}
}
}
}
int main()
{
ll h;
cin >> n;
for(int i = 1; i <= n; i++)
{
cin >> h;
graph[n + 2].pb({i, h});
}
graph[n + 2].pb({0, 0});
graph[n + 2].pb({n + 1, 0});
for(int i = 1; i <= n + 1; i++)
{
graph[i].pb({i - 1, 1});
graph[i - 1].pb({i, 1});
}
dijkstra(n + 2);
ll ans = 0;
for(int i = 1; i <= n; i++)
{
ans = max(ans, dist[i]);
}
cout << ans << endl;
return 0;
} | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built *n* towers in a row. The *i*-th tower is made of *h**i* identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109) — sizes of towers.
Output Specification:
Print the number of operations needed to destroy all towers.
Demo Input:
['6\n2 1 4 6 2 2\n', '7\n3 3 3 1 3 3 3\n']
Demo Output:
['3\n', '2\n']
Note:
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. | ```python
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
#define mp make_pair
#define pb push_back
#define F first
#define S second
const int MAX_SIZE = 1e5 + 10;
const ll inf = 1e14;
vector<pll > graph[MAX_SIZE];
priority_queue< pll, vector<pll > , greater< pll > > pq;
vector<ll> dist(MAX_SIZE);
int n;
void dijkstra(int s)
{
for(int i = 0; i <= n + 1; i++)
dist[i] = inf;
dist[s] = 0;
pq.push({0, s});
while(!pq.empty())
{
int u = pq.top().S;
pq.pop();
for(auto &e: graph[u])
{
int v = e.F;
ll w = e.S;
if(dist[v] > w + dist[u])
{
dist[v] = w + dist[u];
pq.push({dist[v], v});
}
}
}
}
int main()
{
ll h;
cin >> n;
for(int i = 1; i <= n; i++)
{
cin >> h;
graph[n + 2].pb({i, h});
}
graph[n + 2].pb({0, 0});
graph[n + 2].pb({n + 1, 0});
for(int i = 1; i <= n + 1; i++)
{
graph[i].pb({i - 1, 1});
graph[i - 1].pb({i, 1});
}
dijkstra(n + 2);
ll ans = 0;
for(int i = 1; i <= n; i++)
{
ans = max(ans, dist[i]);
}
cout << ans << endl;
return 0;
}
``` | -1 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | Stepan has the newest electronic device with a display. Different digits can be shown on it. Each digit is shown on a seven-section indicator like it is shown on the picture below.
So, for example, to show the digit 3 on the display, 5 sections must be highlighted; and for the digit 6, 6 sections must be highlighted.
The battery of the newest device allows to highlight at most *n* sections on the display.
Stepan wants to know the maximum possible integer number which can be shown on the display of his newest device. Your task is to determine this number. Note that this number must not contain leading zeros. Assume that the size of the display is enough to show any integer. | The first line contains the integer *n* (2<=≤<=*n*<=≤<=100<=000) — the maximum number of sections which can be highlighted on the display. | Print the maximum integer which can be shown on the display of Stepan's newest device. | [
"2\n",
"3\n"
] | [
"1\n",
"7\n"
] | none | 0 | [
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "7"
},
{
"input": "4",
"output": "11"
},
{
"input": "5",
"output": "71"
},
{
"input": "6",
"output": "111"
},
{
"input": "85651",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "85666",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99999",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "100000",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99998",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "9",
"output": "7111"
},
{
"input": "99997",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99996",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99995",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "156",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "255",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "4568",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "5431",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "6782",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "8343",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "9514",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
}
] | 1,491,409,983 | 3,483 | Python 3 | OK | TESTS | 21 | 62 | 7,782,400 | n = int(input())
number = [1] * (n // 2)
if n % 2:
number[0] = 7
print(''.join(map(str, number)))
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Stepan has the newest electronic device with a display. Different digits can be shown on it. Each digit is shown on a seven-section indicator like it is shown on the picture below.
So, for example, to show the digit 3 on the display, 5 sections must be highlighted; and for the digit 6, 6 sections must be highlighted.
The battery of the newest device allows to highlight at most *n* sections on the display.
Stepan wants to know the maximum possible integer number which can be shown on the display of his newest device. Your task is to determine this number. Note that this number must not contain leading zeros. Assume that the size of the display is enough to show any integer.
Input Specification:
The first line contains the integer *n* (2<=≤<=*n*<=≤<=100<=000) — the maximum number of sections which can be highlighted on the display.
Output Specification:
Print the maximum integer which can be shown on the display of Stepan's newest device.
Demo Input:
['2\n', '3\n']
Demo Output:
['1\n', '7\n']
Note:
none | ```python
n = int(input())
number = [1] * (n // 2)
if n % 2:
number[0] = 7
print(''.join(map(str, number)))
``` | 3 |
|
903 | A | Hungry Student Problem | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.
CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly *x* chunks. Now he wonders whether he can buy exactly this amount of chicken.
Formally, Ivan wants to know if he can choose two non-negative integers *a* and *b* in such a way that *a* small portions and *b* large ones contain exactly *x* chunks.
Help Ivan to answer this question for several values of *x*! | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of testcases.
The *i*-th of the following *n* lines contains one integer *x**i* (1<=≤<=*x**i*<=≤<=100) — the number of chicken chunks Ivan wants to eat. | Print *n* lines, in *i*-th line output YES if Ivan can buy exactly *x**i* chunks. Otherwise, print NO. | [
"2\n6\n5\n"
] | [
"YES\nNO\n"
] | In the first example Ivan can buy two small portions.
In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much. | 0 | [
{
"input": "2\n6\n5",
"output": "YES\nNO"
},
{
"input": "100\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n20\n21\n22\n23\n24\n25\n26\n27\n28\n29\n30\n31\n32\n33\n34\n35\n36\n37\n38\n39\n40\n41\n42\n43\n44\n45\n46\n47\n48\n49\n50\n51\n52\n53\n54\n55\n56\n57\n58\n59\n60\n61\n62\n63\n64\n65\n66\n67\n68\n69\n70\n71\n72\n73\n74\n75\n76\n77\n78\n79\n80\n81\n82\n83\n84\n85\n86\n87\n88\n89\n90\n91\n92\n93\n94\n95\n96\n97\n98\n99\n100",
"output": "NO\nNO\nYES\nNO\nNO\nYES\nYES\nNO\nYES\nYES\nNO\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
},
{
"input": "3\n6\n6\n6",
"output": "YES\nYES\nYES"
},
{
"input": "47\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "3\n1\n52\n76",
"output": "NO\nYES\nYES"
},
{
"input": "87\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "YES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
},
{
"input": "3\n3\n2\n1",
"output": "YES\nNO\nNO"
},
{
"input": "100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "YES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
}
] | 1,585,770,848 | 2,147,483,647 | Python 3 | OK | TESTS | 8 | 93 | 0 | t=int(input())
s=[]
for hjbbh in range(0,t):
x=int(input())
f=int(x//7)
h=0
for i in range(0,f+1):
if (x-i*7)%3==0:
s.append('YES')
h+=1
break
if h==0:
s.append('NO')
for i in s:
print(i)
| Title: Hungry Student Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.
CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly *x* chunks. Now he wonders whether he can buy exactly this amount of chicken.
Formally, Ivan wants to know if he can choose two non-negative integers *a* and *b* in such a way that *a* small portions and *b* large ones contain exactly *x* chunks.
Help Ivan to answer this question for several values of *x*!
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of testcases.
The *i*-th of the following *n* lines contains one integer *x**i* (1<=≤<=*x**i*<=≤<=100) — the number of chicken chunks Ivan wants to eat.
Output Specification:
Print *n* lines, in *i*-th line output YES if Ivan can buy exactly *x**i* chunks. Otherwise, print NO.
Demo Input:
['2\n6\n5\n']
Demo Output:
['YES\nNO\n']
Note:
In the first example Ivan can buy two small portions.
In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much. | ```python
t=int(input())
s=[]
for hjbbh in range(0,t):
x=int(input())
f=int(x//7)
h=0
for i in range(0,f+1):
if (x-i*7)%3==0:
s.append('YES')
h+=1
break
if h==0:
s.append('NO')
for i in s:
print(i)
``` | 3 |
|
245 | A | System Administrator | PROGRAMMING | 800 | [
"implementation"
] | null | null | Polycarpus is a system administrator. There are two servers under his strict guidance — *a* and *b*. To stay informed about the servers' performance, Polycarpus executes commands "ping a" and "ping b". Each ping command sends exactly ten packets to the server specified in the argument of the command. Executing a program results in two integers *x* and *y* (*x*<=+<=*y*<==<=10; *x*,<=*y*<=≥<=0). These numbers mean that *x* packets successfully reached the corresponding server through the network and *y* packets were lost.
Today Polycarpus has performed overall *n* ping commands during his workday. Now for each server Polycarpus wants to know whether the server is "alive" or not. Polycarpus thinks that the server is "alive", if at least half of the packets that we send to this server reached it successfully along the network.
Help Polycarpus, determine for each server, whether it is "alive" or not by the given commands and their results. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of commands Polycarpus has fulfilled. Each of the following *n* lines contains three integers — the description of the commands. The *i*-th of these lines contains three space-separated integers *t**i*, *x**i*, *y**i* (1<=≤<=*t**i*<=≤<=2; *x**i*,<=*y**i*<=≥<=0; *x**i*<=+<=*y**i*<==<=10). If *t**i*<==<=1, then the *i*-th command is "ping a", otherwise the *i*-th command is "ping b". Numbers *x**i*, *y**i* represent the result of executing this command, that is, *x**i* packets reached the corresponding server successfully and *y**i* packets were lost.
It is guaranteed that the input has at least one "ping a" command and at least one "ping b" command. | In the first line print string "LIVE" (without the quotes) if server *a* is "alive", otherwise print "DEAD" (without the quotes).
In the second line print the state of server *b* in the similar format. | [
"2\n1 5 5\n2 6 4\n",
"3\n1 0 10\n2 0 10\n1 10 0\n"
] | [
"LIVE\nLIVE\n",
"LIVE\nDEAD\n"
] | Consider the first test case. There 10 packets were sent to server *a*, 5 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall there were 10 packets sent to server *b*, 6 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network.
Consider the second test case. There were overall 20 packages sent to server *a*, 10 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall 10 packets were sent to server *b*, 0 of them reached it. Therefore, less than half of all packets sent to this server successfully reached it through the network. | 0 | [
{
"input": "2\n1 5 5\n2 6 4",
"output": "LIVE\nLIVE"
},
{
"input": "3\n1 0 10\n2 0 10\n1 10 0",
"output": "LIVE\nDEAD"
},
{
"input": "10\n1 3 7\n2 4 6\n1 2 8\n2 5 5\n2 10 0\n2 10 0\n1 8 2\n2 2 8\n2 10 0\n1 1 9",
"output": "DEAD\nLIVE"
},
{
"input": "11\n1 8 2\n1 6 4\n1 9 1\n1 7 3\n2 0 10\n2 0 10\n1 8 2\n2 2 8\n2 6 4\n2 7 3\n2 9 1",
"output": "LIVE\nDEAD"
},
{
"input": "12\n1 5 5\n1 0 10\n1 4 6\n1 2 8\n1 2 8\n1 5 5\n1 9 1\n2 9 1\n1 5 5\n1 1 9\n2 9 1\n2 7 3",
"output": "DEAD\nLIVE"
},
{
"input": "13\n1 8 2\n1 4 6\n1 5 5\n1 5 5\n2 10 0\n2 9 1\n1 3 7\n2 6 4\n2 6 4\n2 5 5\n1 7 3\n2 3 7\n2 9 1",
"output": "LIVE\nLIVE"
},
{
"input": "14\n1 7 3\n1 0 10\n1 7 3\n1 1 9\n2 2 8\n2 0 10\n1 1 9\n2 8 2\n2 6 4\n1 3 7\n1 3 7\n2 6 4\n2 1 9\n2 7 3",
"output": "DEAD\nDEAD"
}
] | 1,580,924,129 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 280 | 0 | n = int(input())
a,b,c,d=0,0,0,0
for i in range(n):
a1,b1,c1 = map(int,input().split())
if a1 ==1:
a += b1
c += c1
if a1 ==2:
b += b1
d += c1
if a >= c:
print("LIVE")
if b >= d:
print("LIVE")
else:
print("DEAD")
| Title: System Administrator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus is a system administrator. There are two servers under his strict guidance — *a* and *b*. To stay informed about the servers' performance, Polycarpus executes commands "ping a" and "ping b". Each ping command sends exactly ten packets to the server specified in the argument of the command. Executing a program results in two integers *x* and *y* (*x*<=+<=*y*<==<=10; *x*,<=*y*<=≥<=0). These numbers mean that *x* packets successfully reached the corresponding server through the network and *y* packets were lost.
Today Polycarpus has performed overall *n* ping commands during his workday. Now for each server Polycarpus wants to know whether the server is "alive" or not. Polycarpus thinks that the server is "alive", if at least half of the packets that we send to this server reached it successfully along the network.
Help Polycarpus, determine for each server, whether it is "alive" or not by the given commands and their results.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of commands Polycarpus has fulfilled. Each of the following *n* lines contains three integers — the description of the commands. The *i*-th of these lines contains three space-separated integers *t**i*, *x**i*, *y**i* (1<=≤<=*t**i*<=≤<=2; *x**i*,<=*y**i*<=≥<=0; *x**i*<=+<=*y**i*<==<=10). If *t**i*<==<=1, then the *i*-th command is "ping a", otherwise the *i*-th command is "ping b". Numbers *x**i*, *y**i* represent the result of executing this command, that is, *x**i* packets reached the corresponding server successfully and *y**i* packets were lost.
It is guaranteed that the input has at least one "ping a" command and at least one "ping b" command.
Output Specification:
In the first line print string "LIVE" (without the quotes) if server *a* is "alive", otherwise print "DEAD" (without the quotes).
In the second line print the state of server *b* in the similar format.
Demo Input:
['2\n1 5 5\n2 6 4\n', '3\n1 0 10\n2 0 10\n1 10 0\n']
Demo Output:
['LIVE\nLIVE\n', 'LIVE\nDEAD\n']
Note:
Consider the first test case. There 10 packets were sent to server *a*, 5 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall there were 10 packets sent to server *b*, 6 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network.
Consider the second test case. There were overall 20 packages sent to server *a*, 10 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall 10 packets were sent to server *b*, 0 of them reached it. Therefore, less than half of all packets sent to this server successfully reached it through the network. | ```python
n = int(input())
a,b,c,d=0,0,0,0
for i in range(n):
a1,b1,c1 = map(int,input().split())
if a1 ==1:
a += b1
c += c1
if a1 ==2:
b += b1
d += c1
if a >= c:
print("LIVE")
if b >= d:
print("LIVE")
else:
print("DEAD")
``` | 0 |
|
811 | A | Vladik and Courtesy | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy. | Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively. | Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise. | [
"1 1\n",
"7 6\n"
] | [
"Valera\n",
"Vladik\n"
] | Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 500 | [
{
"input": "1 1",
"output": "Valera"
},
{
"input": "7 6",
"output": "Vladik"
},
{
"input": "25 38",
"output": "Vladik"
},
{
"input": "8311 2468",
"output": "Valera"
},
{
"input": "250708 857756",
"output": "Vladik"
},
{
"input": "957985574 24997558",
"output": "Valera"
},
{
"input": "999963734 999994456",
"output": "Vladik"
},
{
"input": "1000000000 1000000000",
"output": "Vladik"
},
{
"input": "946 879",
"output": "Valera"
},
{
"input": "10819 45238",
"output": "Vladik"
},
{
"input": "101357 236928",
"output": "Vladik"
},
{
"input": "1033090 7376359",
"output": "Vladik"
},
{
"input": "9754309 9525494",
"output": "Valera"
},
{
"input": "90706344 99960537",
"output": "Vladik"
},
{
"input": "965161805 908862070",
"output": "Valera"
},
{
"input": "9 11",
"output": "Valera"
},
{
"input": "3 2",
"output": "Vladik"
},
{
"input": "6 6",
"output": "Vladik"
},
{
"input": "4 4",
"output": "Valera"
},
{
"input": "5 5",
"output": "Valera"
},
{
"input": "5 4",
"output": "Valera"
},
{
"input": "12345680 1",
"output": "Valera"
},
{
"input": "9 10",
"output": "Valera"
},
{
"input": "678 76687",
"output": "Vladik"
},
{
"input": "1 678",
"output": "Vladik"
},
{
"input": "45 1678",
"output": "Vladik"
},
{
"input": "3 3",
"output": "Vladik"
},
{
"input": "10 11",
"output": "Valera"
},
{
"input": "2 1",
"output": "Valera"
},
{
"input": "1 2",
"output": "Vladik"
},
{
"input": "2 2",
"output": "Vladik"
},
{
"input": "4 5",
"output": "Valera"
},
{
"input": "9 6",
"output": "Valera"
},
{
"input": "1 5",
"output": "Vladik"
},
{
"input": "7 8",
"output": "Vladik"
},
{
"input": "1000000000 999982505",
"output": "Valera"
},
{
"input": "12 12",
"output": "Vladik"
},
{
"input": "1000 950",
"output": "Valera"
},
{
"input": "10 9",
"output": "Valera"
},
{
"input": "100 9",
"output": "Valera"
},
{
"input": "1000 996",
"output": "Vladik"
},
{
"input": "9 5",
"output": "Valera"
}
] | 1,677,840,145 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 42 | 62 | 1,331,200 | n, k = map(int, input().split())
d = 1
while True:
if n-d >= 0:
n = n -d
d += 1
else:
s = "Vladik"
break
if k-d >= 0:
k = k -d
d += 1
else:
s = "Valera"
break
print(s) | Title: Vladik and Courtesy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
Input Specification:
Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively.
Output Specification:
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
Demo Input:
['1 1\n', '7 6\n']
Demo Output:
['Valera\n', 'Vladik\n']
Note:
Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
n, k = map(int, input().split())
d = 1
while True:
if n-d >= 0:
n = n -d
d += 1
else:
s = "Vladik"
break
if k-d >= 0:
k = k -d
d += 1
else:
s = "Valera"
break
print(s)
``` | 3 |
|
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
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] | 1,593,221,090 | 2,147,483,647 | PyPy 3 | OK | TESTS | 102 | 171 | 20,172,800 | f = input()
m = input()
z = ""
for i in range(len(f)):
if f[i]!=m[i]:
z+="1"
else:
z+="0"
print(z) | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
f = input()
m = input()
z = ""
for i in range(len(f)):
if f[i]!=m[i]:
z+="1"
else:
z+="0"
print(z)
``` | 3.919675 |
295 | A | Greg and Array | PROGRAMMING | 1,400 | [
"data structures",
"implementation"
] | null | null | Greg has an array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n* and *m* operations. Each operation looks as: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). To apply operation *i* to the array means to increase all array elements with numbers *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i* by value *d**i*.
Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). That means that one should apply operations with numbers *x**i*,<=*x**i*<=+<=1,<=...,<=*y**i* to the array.
Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg. | The first line contains integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105). The second line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105) — the initial array.
Next *m* lines contain operations, the operation number *i* is written as three integers: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), (0<=≤<=*d**i*<=≤<=105).
Next *k* lines contain the queries, the query number *i* is written as two integers: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*).
The numbers in the lines are separated by single spaces. | On a single line print *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the array after executing all the queries. Separate the printed numbers by spaces.
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. | [
"3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3\n",
"1 1 1\n1\n1 1 1\n1 1\n",
"4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3\n"
] | [
"9 18 17\n",
"2\n",
"5 18 31 20\n"
] | none | 500 | [
{
"input": "3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3",
"output": "9 18 17"
},
{
"input": "1 1 1\n1\n1 1 1\n1 1",
"output": "2"
},
{
"input": "4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3",
"output": "5 18 31 20"
},
{
"input": "1 1 1\n0\n1 1 0\n1 1",
"output": "0"
}
] | 1,659,458,504 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 23 | 1,500 | 12,492,800 | """
செயல் பேசும் ஆழம் இங்கே சொற்கள் பேசுமா?
Focus, Determination and Sheer-Will
The woods are lovely, dark and deep,
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep.
"""
import os, sys, math, cmath, time, collections
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import nsmallest, nlargest, heapify, heappop, heappush, heapreplace
from math import ceil, floor, log, log2, sqrt, gcd, factorial, pow, pi
from bisect import bisect_left, bisect_right
from functools import reduce
# SOME GENERAL HELPER
def input_as_array():
return list(map(int, input().split()))
def debug():
if os.path.exists("data.in"):
print("*" * 10)
def debug2(value):
if os.path.exists("data.in"):
print(value)
def debug3():
if os.path.exists("data.in"):
print("-" * 10)
def debug4(msg, value):
if os.path.exists("data.in"):
print(msg, value)
def debug5(msg):
if os.path.exists("data.in"):
print(msg)
start_time = time.time()
def solve(A, n, ops, m, Q, k):
diff = [0] * (n+2)
for i in range(n):
if i == 0:
diff[i] = A[i]
else:
diff[i] = A[i] - A[i - 1]
for a, b in Q:
for i in range(a, b + 1):
L, R, D = ops[i][0], ops[i][1], ops[i][2]
diff[L] += D
diff[R+1] -= D
# Finally updating the initial array
for i in range(n):
if i == 0:
A[i] = diff[i]
else:
A[i] = diff[i] + A[i - 1]
print(*A)
def main():
n, m, k = input_as_array()
A = input_as_array()
i = 0
ops = []
while i < m:
l, r, d = input_as_array()
l, r = l - 1, r - 1
ops.append((l, r, d))
i += 1
Q = []
for _ in range(k):
x, y = input_as_array()
x, y = x - 1, y - 1
Q.append((x, y))
solve(A, n, ops, m, Q, k)
debug3()
if __name__ == "__main__":
if os.path.exists("data.in"):
sys.stdin = open("data.in", "r")
sys.stdout = open("data.out", "w")
testcases = 1
for i in range(testcases):
main()
# If it's local - Print this O/P
if os.path.exists("data.in"):
print(f"Time Elapsed: {time.time() - start_time} seconds")
sys.stdout.close()
| Title: Greg and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg has an array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n* and *m* operations. Each operation looks as: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). To apply operation *i* to the array means to increase all array elements with numbers *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i* by value *d**i*.
Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). That means that one should apply operations with numbers *x**i*,<=*x**i*<=+<=1,<=...,<=*y**i* to the array.
Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg.
Input Specification:
The first line contains integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105). The second line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105) — the initial array.
Next *m* lines contain operations, the operation number *i* is written as three integers: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), (0<=≤<=*d**i*<=≤<=105).
Next *k* lines contain the queries, the query number *i* is written as two integers: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*).
The numbers in the lines are separated by single spaces.
Output Specification:
On a single line print *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the array after executing all the queries. Separate the printed numbers by spaces.
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:
['3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3\n', '1 1 1\n1\n1 1 1\n1 1\n', '4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3\n']
Demo Output:
['9 18 17\n', '2\n', '5 18 31 20\n']
Note:
none | ```python
"""
செயல் பேசும் ஆழம் இங்கே சொற்கள் பேசுமா?
Focus, Determination and Sheer-Will
The woods are lovely, dark and deep,
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep.
"""
import os, sys, math, cmath, time, collections
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import nsmallest, nlargest, heapify, heappop, heappush, heapreplace
from math import ceil, floor, log, log2, sqrt, gcd, factorial, pow, pi
from bisect import bisect_left, bisect_right
from functools import reduce
# SOME GENERAL HELPER
def input_as_array():
return list(map(int, input().split()))
def debug():
if os.path.exists("data.in"):
print("*" * 10)
def debug2(value):
if os.path.exists("data.in"):
print(value)
def debug3():
if os.path.exists("data.in"):
print("-" * 10)
def debug4(msg, value):
if os.path.exists("data.in"):
print(msg, value)
def debug5(msg):
if os.path.exists("data.in"):
print(msg)
start_time = time.time()
def solve(A, n, ops, m, Q, k):
diff = [0] * (n+2)
for i in range(n):
if i == 0:
diff[i] = A[i]
else:
diff[i] = A[i] - A[i - 1]
for a, b in Q:
for i in range(a, b + 1):
L, R, D = ops[i][0], ops[i][1], ops[i][2]
diff[L] += D
diff[R+1] -= D
# Finally updating the initial array
for i in range(n):
if i == 0:
A[i] = diff[i]
else:
A[i] = diff[i] + A[i - 1]
print(*A)
def main():
n, m, k = input_as_array()
A = input_as_array()
i = 0
ops = []
while i < m:
l, r, d = input_as_array()
l, r = l - 1, r - 1
ops.append((l, r, d))
i += 1
Q = []
for _ in range(k):
x, y = input_as_array()
x, y = x - 1, y - 1
Q.append((x, y))
solve(A, n, ops, m, Q, k)
debug3()
if __name__ == "__main__":
if os.path.exists("data.in"):
sys.stdin = open("data.in", "r")
sys.stdout = open("data.out", "w")
testcases = 1
for i in range(testcases):
main()
# If it's local - Print this O/P
if os.path.exists("data.in"):
print(f"Time Elapsed: {time.time() - start_time} seconds")
sys.stdout.close()
``` | 0 |
|
119 | A | Epic Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game. | The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. | If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). | [
"3 5 9\n",
"1 1 100\n"
] | [
"0",
"1"
] | The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | 500 | [
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "41 12 65",
"output": "1"
},
{
"input": "13 61 100",
"output": "1"
},
{
"input": "100 100 10",
"output": "0"
},
{
"input": "12 24 26",
"output": "1"
},
{
"input": "73 21 96",
"output": "1"
},
{
"input": "17 22 81",
"output": "1"
},
{
"input": "14 88 97",
"output": "1"
},
{
"input": "42 81 17",
"output": "0"
},
{
"input": "34 36 19",
"output": "1"
},
{
"input": "21 87 12",
"output": "1"
},
{
"input": "97 92 98",
"output": "1"
},
{
"input": "49 25 62",
"output": "1"
},
{
"input": "29 17 44",
"output": "1"
},
{
"input": "10 1 10",
"output": "0"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "100 100 1",
"output": "0"
},
{
"input": "11 99 55",
"output": "0"
},
{
"input": "8 13 21",
"output": "0"
},
{
"input": "34 55 89",
"output": "0"
},
{
"input": "21 34 55",
"output": "1"
},
{
"input": "13 21 34",
"output": "1"
},
{
"input": "34 21 55",
"output": "0"
},
{
"input": "10 20 84",
"output": "1"
},
{
"input": "4 2 5",
"output": "0"
},
{
"input": "3 2 3",
"output": "0"
},
{
"input": "31 23 1",
"output": "0"
},
{
"input": "100 100 100",
"output": "0"
},
{
"input": "100 1 100",
"output": "0"
},
{
"input": "1 100 100",
"output": "1"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "17 23 100",
"output": "1"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "1 1 2",
"output": "1"
},
{
"input": "2 1 1",
"output": "0"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "2 1 2",
"output": "0"
},
{
"input": "10 10 3",
"output": "1"
},
{
"input": "2 2 1",
"output": "0"
},
{
"input": "9 5 1",
"output": "0"
}
] | 1,664,375,703 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 0 | a, b, n = map( int, input().split() )
def nod(x, y):
while(y):
x, x = y, x % y
return x
last_move = True
while n > 1:
n -= nod(a, n) if last_move else nod(b, n)
last_move = False if last_move else True
print(0) if last_move else print(1)
| Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | ```python
a, b, n = map( int, input().split() )
def nod(x, y):
while(y):
x, x = y, x % y
return x
last_move = True
while n > 1:
n -= nod(a, n) if last_move else nod(b, n)
last_move = False if last_move else True
print(0) if last_move else print(1)
``` | 0 |
|
45 | A | Codecraft III | PROGRAMMING | 900 | [
"implementation"
] | A. Codecraft III | 2 | 256 | Today Vasya visited a widely known site and learned that the continuation of his favourite game Codecraft II will appear after exactly *k* months. He looked at the calendar and learned that at the moment is the month number *s*. Vasya immediately got interested in what month Codecraft III will appear. Help him understand that.
All the twelve months in Vasya's calendar are named using their usual English names: January, February, March, April, May, June, July, August, September, October, November, December. | The first input line contains the name of the current month. It is guaranteed that it is a proper English name of one of twelve months. The first letter is uppercase, the rest are lowercase. The second line contains integer *k* (0<=≤<=*k*<=≤<=100) — the number of months left till the appearance of Codecraft III. | Print starting from an uppercase letter the name of the month in which the continuation of Codeforces II will appear. The printed name must be contained in the list January, February, March, April, May, June, July, August, September, October, November, December. | [
"November\n3\n",
"May\n24\n"
] | [
"February\n",
"May\n"
] | none | 0 | [
{
"input": "November\n3",
"output": "February"
},
{
"input": "May\n24",
"output": "May"
},
{
"input": "April\n0",
"output": "April"
},
{
"input": "September\n0",
"output": "September"
},
{
"input": "August\n0",
"output": "August"
},
{
"input": "June\n1",
"output": "July"
},
{
"input": "July\n2",
"output": "September"
},
{
"input": "September\n3",
"output": "December"
},
{
"input": "July\n4",
"output": "November"
},
{
"input": "August\n24",
"output": "August"
},
{
"input": "May\n48",
"output": "May"
},
{
"input": "November\n47",
"output": "October"
},
{
"input": "December\n49",
"output": "January"
},
{
"input": "June\n99",
"output": "September"
},
{
"input": "March\n100",
"output": "July"
},
{
"input": "December\n1",
"output": "January"
},
{
"input": "January\n11",
"output": "December"
},
{
"input": "December\n0",
"output": "December"
},
{
"input": "January\n0",
"output": "January"
},
{
"input": "July\n77",
"output": "December"
},
{
"input": "February\n11",
"output": "January"
},
{
"input": "February\n22",
"output": "December"
},
{
"input": "July\n33",
"output": "April"
},
{
"input": "May\n44",
"output": "January"
},
{
"input": "June\n97",
"output": "July"
}
] | 1,670,696,300 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 92 | 0 | s = 'January, February, March, April, May, June, July, August, September, October, November, December'
s = s.split(', ')
a =input()
n = int(input())
p = s.index(a)
p = (p + n) % 12
print (s[p]) | Title: Codecraft III
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Today Vasya visited a widely known site and learned that the continuation of his favourite game Codecraft II will appear after exactly *k* months. He looked at the calendar and learned that at the moment is the month number *s*. Vasya immediately got interested in what month Codecraft III will appear. Help him understand that.
All the twelve months in Vasya's calendar are named using their usual English names: January, February, March, April, May, June, July, August, September, October, November, December.
Input Specification:
The first input line contains the name of the current month. It is guaranteed that it is a proper English name of one of twelve months. The first letter is uppercase, the rest are lowercase. The second line contains integer *k* (0<=≤<=*k*<=≤<=100) — the number of months left till the appearance of Codecraft III.
Output Specification:
Print starting from an uppercase letter the name of the month in which the continuation of Codeforces II will appear. The printed name must be contained in the list January, February, March, April, May, June, July, August, September, October, November, December.
Demo Input:
['November\n3\n', 'May\n24\n']
Demo Output:
['February\n', 'May\n']
Note:
none | ```python
s = 'January, February, March, April, May, June, July, August, September, October, November, December'
s = s.split(', ')
a =input()
n = int(input())
p = s.index(a)
p = (p + n) % 12
print (s[p])
``` | 3.977 |
155 | A | I_love_\%username\% | PROGRAMMING | 800 | [
"brute force"
] | null | null | Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him. | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000. | Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests. | [
"5\n100 50 200 150 200\n",
"10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n"
] | [
"2\n",
"4\n"
] | In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | 500 | [
{
"input": "5\n100 50 200 150 200",
"output": "2"
},
{
"input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242",
"output": "4"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n100 36 53 7 81",
"output": "2"
},
{
"input": "5\n7 36 53 81 100",
"output": "4"
},
{
"input": "5\n100 81 53 36 7",
"output": "4"
},
{
"input": "10\n8 6 3 4 9 10 7 7 1 3",
"output": "5"
},
{
"input": "10\n1627 1675 1488 1390 1812 1137 1746 1324 1952 1862",
"output": "6"
},
{
"input": "10\n1 3 3 4 6 7 7 8 9 10",
"output": "7"
},
{
"input": "10\n1952 1862 1812 1746 1675 1627 1488 1390 1324 1137",
"output": "9"
},
{
"input": "25\n1448 4549 2310 2725 2091 3509 1565 2475 2232 3989 4231 779 2967 2702 608 3739 721 1552 2767 530 3114 665 1940 48 4198",
"output": "5"
},
{
"input": "33\n1097 1132 1091 1104 1049 1038 1023 1080 1104 1029 1035 1061 1049 1060 1088 1106 1105 1087 1063 1076 1054 1103 1047 1041 1028 1120 1126 1063 1117 1110 1044 1093 1101",
"output": "5"
},
{
"input": "34\n821 5536 2491 6074 7216 9885 764 1603 778 8736 8987 771 617 1587 8943 7922 439 7367 4115 8886 7878 6899 8811 5752 3184 3401 9760 9400 8995 4681 1323 6637 6554 6498",
"output": "7"
},
{
"input": "68\n6764 6877 6950 6768 6839 6755 6726 6778 6699 6805 6777 6985 6821 6801 6791 6805 6940 6761 6677 6999 6911 6699 6959 6933 6903 6843 6972 6717 6997 6756 6789 6668 6735 6852 6735 6880 6723 6834 6810 6694 6780 6679 6698 6857 6826 6896 6979 6968 6957 6988 6960 6700 6919 6892 6984 6685 6813 6678 6715 6857 6976 6902 6780 6686 6777 6686 6842 6679",
"output": "9"
},
{
"input": "60\n9000 9014 9034 9081 9131 9162 9174 9199 9202 9220 9221 9223 9229 9235 9251 9260 9268 9269 9270 9298 9307 9309 9313 9323 9386 9399 9407 9495 9497 9529 9531 9544 9614 9615 9627 9627 9643 9654 9656 9657 9685 9699 9701 9736 9745 9758 9799 9827 9843 9845 9854 9854 9885 9891 9896 9913 9942 9963 9986 9992",
"output": "57"
},
{
"input": "100\n7 61 12 52 41 16 34 99 30 44 48 89 31 54 21 1 48 52 61 15 35 87 21 76 64 92 44 81 16 93 84 92 32 15 68 76 53 39 26 4 11 26 7 4 99 99 61 65 55 85 65 67 47 39 2 74 63 49 98 87 5 94 22 30 25 42 31 84 49 23 89 60 16 26 92 27 9 57 75 61 94 35 83 47 99 100 63 24 91 88 79 10 15 45 22 64 3 11 89 83",
"output": "4"
},
{
"input": "100\n9999 9999 9999 9998 9998 9998 9997 9996 9996 9995 9993 9993 9991 9990 9989 9986 9984 9984 9983 9981 9981 9980 9980 9980 9979 9977 9977 9977 9977 9977 9976 9976 9975 9975 9973 9972 9972 9972 9972 9971 9969 9969 9969 9968 9967 9965 9962 9962 9962 9962 9960 9958 9958 9955 9954 9954 9954 9954 9950 9949 9949 9947 9943 9943 9940 9940 9939 9937 9937 9937 9936 9935 9932 9932 9931 9929 9927 9927 9926 9923 9919 9919 9918 9918 9918 9917 9917 9916 9916 9915 9914 9911 9910 9907 9907 9906 9905 9905 9904 9901",
"output": "57"
},
{
"input": "100\n158 159 157 156 155 154 160 153 161 152 162 163 151 164 150 165 149 148 147 166 146 167 145 144 143 142 168 141 169 170 140 139 138 137 171 136 135 134 172 173 174 133 175 132 177 178 131 179 180 130 129 181 128 127 182 126 183 125 124 184 123 122 121 120 119 118 117 185 186 187 188 189 116 190 115 191 192 193 114 113 112 111 110 109 108 194 195 107 106 105 196 197 104 198 199 103 102 200 101 100",
"output": "99"
},
{
"input": "2\n0 10000",
"output": "1"
},
{
"input": "2\n5 5",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n10 10",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
}
] | 1,698,315,256 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 92 | 0 | n=int(input())
s=0
a=list(map(int,input().split()))
i=0
mini= a[0]
maxi=a[0]
while i<n:
if a[i]<mini:
mini=a[i]
s+=1
elif a[i]>maxi:
maxi=a[i]
s+=1
i+=1
print(s)
| Title: I_love_\%username\%
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
Output Specification:
Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.
Demo Input:
['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | ```python
n=int(input())
s=0
a=list(map(int,input().split()))
i=0
mini= a[0]
maxi=a[0]
while i<n:
if a[i]<mini:
mini=a[i]
s+=1
elif a[i]>maxi:
maxi=a[i]
s+=1
i+=1
print(s)
``` | 3 |
|
313 | B | Ilya and Queries | PROGRAMMING | 1,100 | [
"dp",
"implementation"
] | null | null | Ilya the Lion wants to help all his friends with passing exams. They need to solve the following problem to pass the IT exam.
You've got string *s*<==<=*s*1*s*2... *s**n* (*n* is the length of the string), consisting only of characters "." and "#" and *m* queries. Each query is described by a pair of integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*). The answer to the query *l**i*,<=*r**i* is the number of such integers *i* (*l**i*<=≤<=*i*<=<<=*r**i*), that *s**i*<==<=*s**i*<=+<=1.
Ilya the Lion wants to help his friends but is there anyone to help him? Help Ilya, solve the problem. | The first line contains string *s* of length *n* (2<=≤<=*n*<=≤<=105). It is guaranteed that the given string only consists of characters "." and "#".
The next line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains the description of the corresponding query. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*). | Print *m* integers — the answers to the queries in the order in which they are given in the input. | [
"......\n4\n3 4\n2 3\n1 6\n2 6\n",
"#..###\n5\n1 3\n5 6\n1 5\n3 6\n3 4\n"
] | [
"1\n1\n5\n4\n",
"1\n1\n2\n2\n0\n"
] | none | 1,000 | [
{
"input": "......\n4\n3 4\n2 3\n1 6\n2 6",
"output": "1\n1\n5\n4"
},
{
"input": "#..###\n5\n1 3\n5 6\n1 5\n3 6\n3 4",
"output": "1\n1\n2\n2\n0"
},
{
"input": ".#...#..\n6\n1 5\n2 3\n6 7\n2 4\n2 5\n1 3",
"output": "2\n0\n0\n1\n2\n0"
},
{
"input": "#.#.#..\n5\n3 4\n4 5\n5 7\n5 7\n1 3",
"output": "0\n0\n1\n1\n0"
},
{
"input": "#.##.##.\n7\n1 8\n2 6\n2 6\n6 8\n3 5\n2 4\n2 5",
"output": "2\n1\n1\n1\n1\n1\n1"
},
{
"input": "#..#\n1\n1 4",
"output": "1"
},
{
"input": "#..##...#.\n7\n5 9\n6 10\n1 7\n5 8\n3 5\n2 10\n3 4",
"output": "2\n2\n3\n2\n1\n4\n0"
},
{
"input": "#.#.#\n7\n1 2\n3 4\n3 5\n2 3\n3 5\n1 5\n1 3",
"output": "0\n0\n0\n0\n0\n0\n0"
},
{
"input": "###..#...#\n2\n2 4\n1 2",
"output": "1\n1"
},
{
"input": "..\n1\n1 2",
"output": "1"
},
{
"input": "##\n1\n1 2",
"output": "1"
},
{
"input": ".#\n1\n1 2",
"output": "0"
},
{
"input": "#.\n1\n1 2",
"output": "0"
},
{
"input": "...\n2\n1 2\n1 2",
"output": "1\n1"
}
] | 1,699,761,894 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 810 | 16,588,800 | s=input()
same={i:0 for i in range(len(s))}
answers=[]
for i in range(len(s)-1):
if s[i] == s[i+1]:
same[i+1]=same[i]+1
else:
same[i+1]=same[i]
for _ in range(int(input())):
r,l=map(int,input().split())
answers.append(same[l-1]-same[r-1])
for a in answers:
print(a) | Title: Ilya and Queries
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya the Lion wants to help all his friends with passing exams. They need to solve the following problem to pass the IT exam.
You've got string *s*<==<=*s*1*s*2... *s**n* (*n* is the length of the string), consisting only of characters "." and "#" and *m* queries. Each query is described by a pair of integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*). The answer to the query *l**i*,<=*r**i* is the number of such integers *i* (*l**i*<=≤<=*i*<=<<=*r**i*), that *s**i*<==<=*s**i*<=+<=1.
Ilya the Lion wants to help his friends but is there anyone to help him? Help Ilya, solve the problem.
Input Specification:
The first line contains string *s* of length *n* (2<=≤<=*n*<=≤<=105). It is guaranteed that the given string only consists of characters "." and "#".
The next line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains the description of the corresponding query. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*).
Output Specification:
Print *m* integers — the answers to the queries in the order in which they are given in the input.
Demo Input:
['......\n4\n3 4\n2 3\n1 6\n2 6\n', '#..###\n5\n1 3\n5 6\n1 5\n3 6\n3 4\n']
Demo Output:
['1\n1\n5\n4\n', '1\n1\n2\n2\n0\n']
Note:
none | ```python
s=input()
same={i:0 for i in range(len(s))}
answers=[]
for i in range(len(s)-1):
if s[i] == s[i+1]:
same[i+1]=same[i]+1
else:
same[i+1]=same[i]
for _ in range(int(input())):
r,l=map(int,input().split())
answers.append(same[l-1]-same[r-1])
for a in answers:
print(a)
``` | 3 |
|
389 | A | Fox and Number Game | PROGRAMMING | 1,000 | [
"greedy",
"math"
] | null | null | Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum. | The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100). | Output a single integer — the required minimal sum. | [
"2\n1 2\n",
"3\n2 4 6\n",
"2\n12 18\n",
"5\n45 12 27 30 18\n"
] | [
"2\n",
"6\n",
"12\n",
"15\n"
] | In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>. | 500 | [
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n2 4 6",
"output": "6"
},
{
"input": "2\n12 18",
"output": "12"
},
{
"input": "5\n45 12 27 30 18",
"output": "15"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "2\n100 100",
"output": "200"
},
{
"input": "2\n87 58",
"output": "58"
},
{
"input": "39\n52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52",
"output": "2028"
},
{
"input": "59\n96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96",
"output": "5664"
},
{
"input": "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": "10000"
},
{
"input": "100\n70 70 77 42 98 84 56 91 35 21 7 70 77 77 56 63 14 84 56 14 77 77 63 70 14 7 28 91 63 49 21 84 98 56 77 98 98 84 98 14 7 56 49 28 91 98 7 56 14 91 14 98 49 28 98 14 98 98 14 70 35 28 63 28 49 63 63 56 91 98 35 42 42 35 63 35 42 14 63 21 77 56 42 77 35 91 56 21 28 84 56 70 70 91 98 70 84 63 21 98",
"output": "700"
},
{
"input": "39\n63 21 21 42 21 63 21 84 42 21 84 63 42 63 84 84 84 42 42 84 21 63 42 63 42 42 63 42 42 63 84 42 21 84 21 63 42 21 42",
"output": "819"
},
{
"input": "59\n70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70",
"output": "4130"
},
{
"input": "87\n44 88 88 88 88 66 88 22 22 88 88 44 88 22 22 22 88 88 88 88 66 22 88 88 88 88 66 66 44 88 44 44 66 22 88 88 22 44 66 44 88 66 66 22 22 22 22 88 22 22 44 66 88 22 22 88 66 66 88 22 66 88 66 88 66 44 88 44 22 44 44 22 44 88 44 44 44 44 22 88 88 88 66 66 88 44 22",
"output": "1914"
},
{
"input": "15\n63 63 63 63 63 63 63 63 63 63 63 63 63 63 63",
"output": "945"
},
{
"input": "39\n63 77 21 14 14 35 21 21 70 42 21 70 28 77 28 77 7 42 63 7 98 49 98 84 35 70 70 91 14 42 98 7 42 7 98 42 56 35 91",
"output": "273"
},
{
"input": "18\n18 18 18 36 36 36 54 72 54 36 72 54 36 36 36 36 18 36",
"output": "324"
},
{
"input": "46\n71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71",
"output": "3266"
},
{
"input": "70\n66 11 66 11 44 11 44 99 55 22 88 11 11 22 55 44 22 77 44 77 77 22 44 55 88 11 99 99 88 22 77 77 66 11 11 66 99 55 55 44 66 44 77 44 44 55 33 55 44 88 77 77 22 66 33 44 11 22 55 44 22 66 77 33 33 44 44 44 22 33",
"output": "770"
},
{
"input": "10\n60 12 96 48 60 24 60 36 60 60",
"output": "120"
},
{
"input": "20\n51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51",
"output": "1020"
},
{
"input": "50\n58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58",
"output": "2900"
},
{
"input": "98\n70 60 100 30 70 20 30 50 50 30 90 40 30 40 60 80 60 60 80 50 10 80 20 10 20 10 50 70 30 80 30 50 60 90 90 100 60 30 90 20 30 60 90 80 60 60 10 90 10 50 40 40 80 90 100 40 70 40 30 50 60 50 60 30 40 20 90 60 20 20 20 70 60 70 50 100 90 50 20 40 80 60 10 60 50 40 40 10 50 10 40 10 80 100 100 90 10 90",
"output": "980"
},
{
"input": "100\n82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82",
"output": "8200"
},
{
"input": "100\n11 87 77 93 3 54 21 93 9 71 37 23 69 85 74 3 48 99 51 31 56 19 21 96 39 6 4 4 29 69 100 42 1 22 81 53 48 49 81 61 10 7 40 61 7 71 51 59 79 44 50 35 95 80 83 8 98 40 18 94 84 49 52 74 66 69 39 37 100 44 38 62 2 80 46 31 35 53 5 60 21 49 63 55 20 53 80 53 66 34 23 92 77 50 86 63 65 24 12 70",
"output": "100"
},
{
"input": "2\n100 1",
"output": "2"
},
{
"input": "2\n18 30",
"output": "12"
},
{
"input": "2\n3 5",
"output": "2"
},
{
"input": "2\n1 10",
"output": "2"
},
{
"input": "2\n8 5",
"output": "2"
},
{
"input": "5\n2 3 5 8 18",
"output": "5"
},
{
"input": "5\n2 4 1 6 8",
"output": "5"
},
{
"input": "3\n12 10 5",
"output": "3"
},
{
"input": "3\n6 10 15",
"output": "3"
}
] | 1,680,033,864 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 34 | 62 | 0 | from math import gcd
n = int(input())
lst = list(map(int, input().split()))
lst.sort()
r = lst[0]
for x in lst[1:]:
r = gcd(r, x)
print(n * r) | Title: Fox and Number Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100).
Output Specification:
Output a single integer — the required minimal sum.
Demo Input:
['2\n1 2\n', '3\n2 4 6\n', '2\n12 18\n', '5\n45 12 27 30 18\n']
Demo Output:
['2\n', '6\n', '12\n', '15\n']
Note:
In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>. | ```python
from math import gcd
n = int(input())
lst = list(map(int, input().split()))
lst.sort()
r = lst[0]
for x in lst[1:]:
r = gcd(r, x)
print(n * r)
``` | 3 |
|
665 | C | Simple Strings | PROGRAMMING | 1,300 | [
"dp",
"greedy",
"strings"
] | null | null | zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task! | The only line contains the string *s* (1<=≤<=|*s*|<=≤<=2·105) — the string given to zscoder. The string *s* consists of only lowercase English letters. | Print the simple string *s*' — the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string *s*' should also consist of only lowercase English letters. | [
"aab\n",
"caaab\n",
"zscoder\n"
] | [
"bab\n",
"cabab\n",
"zscoder\n"
] | none | 0 | [
{
"input": "aab",
"output": "bab"
},
{
"input": "caaab",
"output": "cabab"
},
{
"input": "zscoder",
"output": "zscoder"
},
{
"input": "u",
"output": "u"
},
{
"input": "h",
"output": "h"
},
{
"input": "dtottttotd",
"output": "dtotataotd"
},
{
"input": "rxxxrrxrxxxxxrrrrrxxxxrrrrxrxxrxxrxrxrrrxrrxrrxrxxxrxrrxrrxrxrxxxxxrxxxxrrrxrxxrxxrxxxrrrrrxrrxrrxrr",
"output": "rxaxraxrxaxaxrararxaxararaxrxarxarxrxrarxraxraxrxaxrxraxraxrxrxaxaxrxaxararxrxarxarxaxrararxraxraxra"
},
{
"input": "aazz",
"output": "baza"
},
{
"input": "zz",
"output": "za"
},
{
"input": "gg",
"output": "ga"
},
{
"input": "qasdasd",
"output": "qasdasd"
},
{
"input": "aa",
"output": "ba"
},
{
"input": "ab",
"output": "ab"
},
{
"input": "zza",
"output": "zba"
},
{
"input": "g",
"output": "g"
},
{
"input": "nnop",
"output": "naop"
},
{
"input": "xx",
"output": "xa"
}
] | 1,487,820,189 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 1,903 | 7,372,800 | s=list(input())
dah='abcdefghijklmnopqrstuvwxyz'
for i in range(len(s)-1):
if s[i]==s[i+1]:
if i+2<len(s):
for ch in dah:
if ch!=s[i+2] and ch!=s[i]:
s[i+1]=ch
else:
for ch in dah:
if ch!=s[i]:
s[i+1]=ch
for i in range(len(s)):
print(s[i],end='')
| Title: Simple Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task!
Input Specification:
The only line contains the string *s* (1<=≤<=|*s*|<=≤<=2·105) — the string given to zscoder. The string *s* consists of only lowercase English letters.
Output Specification:
Print the simple string *s*' — the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string *s*' should also consist of only lowercase English letters.
Demo Input:
['aab\n', 'caaab\n', 'zscoder\n']
Demo Output:
['bab\n', 'cabab\n', 'zscoder\n']
Note:
none | ```python
s=list(input())
dah='abcdefghijklmnopqrstuvwxyz'
for i in range(len(s)-1):
if s[i]==s[i+1]:
if i+2<len(s):
for ch in dah:
if ch!=s[i+2] and ch!=s[i]:
s[i+1]=ch
else:
for ch in dah:
if ch!=s[i]:
s[i+1]=ch
for i in range(len(s)):
print(s[i],end='')
``` | 3 |
|
1,008 | A | Romaji | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese. | The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters. | Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower). | [
"sumimasen\n",
"ninja\n",
"codeforces\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese. | 500 | [
{
"input": "sumimasen",
"output": "YES"
},
{
"input": "ninja",
"output": "YES"
},
{
"input": "codeforces",
"output": "NO"
},
{
"input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen",
"output": "YES"
},
{
"input": "n",
"output": "YES"
},
{
"input": "necnei",
"output": "NO"
},
{
"input": "nternn",
"output": "NO"
},
{
"input": "aucunuohja",
"output": "NO"
},
{
"input": "a",
"output": "YES"
},
{
"input": "b",
"output": "NO"
},
{
"input": "nn",
"output": "YES"
},
{
"input": "nnnzaaa",
"output": "YES"
},
{
"input": "zn",
"output": "NO"
},
{
"input": "ab",
"output": "NO"
},
{
"input": "aaaaaaaaaa",
"output": "YES"
},
{
"input": "aaaaaaaaab",
"output": "NO"
},
{
"input": "aaaaaaaaan",
"output": "YES"
},
{
"input": "baaaaaaaaa",
"output": "YES"
},
{
"input": "naaaaaaaaa",
"output": "YES"
},
{
"input": "nbaaaaaaaa",
"output": "YES"
},
{
"input": "bbaaaaaaaa",
"output": "NO"
},
{
"input": "bnaaaaaaaa",
"output": "NO"
},
{
"input": "eonwonojannonnufimiiniewuqaienokacevecinfuqihatenhunliquuyebayiaenifuexuanenuaounnboancaeowonu",
"output": "YES"
},
{
"input": "uixinnepnlinqaingieianndeakuniooudidonnnqeaituioeneiroionxuowudiooonayenfeonuino",
"output": "NO"
},
{
"input": "nnnnnyigaveteononnnnxaalenxuiiwannntoxonyoqonlejuoxuoconnnentoinnul",
"output": "NO"
},
{
"input": "ndonneasoiunhomuunnhuitonnntunntoanerekonoupunanuauenu",
"output": "YES"
},
{
"input": "anujemogawautiedoneobninnibonuunaoennnyoorufonxionntinimiboonununnnnnleenqunminzayoutanlalo",
"output": "NO"
},
{
"input": "y",
"output": "NO"
},
{
"input": "by",
"output": "NO"
},
{
"input": "yy",
"output": "NO"
},
{
"input": "nbn",
"output": "NO"
},
{
"input": "nz",
"output": "NO"
},
{
"input": "king",
"output": "NO"
},
{
"input": "g",
"output": "NO"
},
{
"input": "az",
"output": "NO"
},
{
"input": "x",
"output": "NO"
},
{
"input": "z",
"output": "NO"
},
{
"input": "yn",
"output": "NO"
},
{
"input": "aeo",
"output": "YES"
},
{
"input": "nb",
"output": "NO"
},
{
"input": "npn",
"output": "NO"
},
{
"input": "kini",
"output": "YES"
},
{
"input": "pya",
"output": "NO"
},
{
"input": "m",
"output": "NO"
},
{
"input": "p",
"output": "NO"
},
{
"input": "aaaaaaaak",
"output": "NO"
},
{
"input": "aab",
"output": "NO"
},
{
"input": "d",
"output": "NO"
},
{
"input": "at",
"output": "NO"
},
{
"input": "aaaaaak",
"output": "NO"
},
{
"input": "aaz",
"output": "NO"
},
{
"input": "aaab",
"output": "NO"
},
{
"input": "s",
"output": "NO"
},
{
"input": "nzzen",
"output": "NO"
},
{
"input": "aeionnhhhn",
"output": "NO"
},
{
"input": "h",
"output": "NO"
},
{
"input": "ny",
"output": "NO"
}
] | 1,636,957,367 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 4 | 31 | 0 | s = input()
s = s.lower()
ans = 0
if len(s) >= 1 and len(s) <=100:
if len(s)==1 and s not in 'aeioun':
ans = 1
else:
for i in range(len(s)-1):
if s[i] not in 'aeioun':
if s[i+1] not in 'aeiou':
ans = 1
if i == len(s)-2:
if s[-1] not in 'aeion':
ans = 1
if ans != 1:
print('YES')
elif ans == 1:
print('NO') | Title: Romaji
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
Input Specification:
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
Output Specification:
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['sumimasen\n', 'ninja\n', 'codeforces\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese. | ```python
s = input()
s = s.lower()
ans = 0
if len(s) >= 1 and len(s) <=100:
if len(s)==1 and s not in 'aeioun':
ans = 1
else:
for i in range(len(s)-1):
if s[i] not in 'aeioun':
if s[i+1] not in 'aeiou':
ans = 1
if i == len(s)-2:
if s[-1] not in 'aeion':
ans = 1
if ans != 1:
print('YES')
elif ans == 1:
print('NO')
``` | -1 |
|
964 | B | Messages | PROGRAMMING | 1,300 | [
"math"
] | null | null | There are *n* incoming messages for Vasya. The *i*-th message is going to be received after *t**i* minutes. Each message has a cost, which equals to *A* initially. After being received, the cost of a message decreases by *B* each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives *C*·*k*, where *k* is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after *T* minutes.
Determine the maximum amount of money Vasya's bank account can hold after *T* minutes. | The first line contains five integers *n*, *A*, *B*, *C* and *T* (1<=≤<=*n*,<=*A*,<=*B*,<=*C*,<=*T*<=≤<=1000).
The second string contains *n* integers *t**i* (1<=≤<=*t**i*<=≤<=*T*). | Output one integer — the answer to the problem. | [
"4 5 5 3 5\n1 5 5 4\n",
"5 3 1 1 3\n2 2 2 1 1\n",
"5 5 3 4 5\n1 2 3 4 5\n"
] | [
"20\n",
"15\n",
"35\n"
] | In the first sample the messages must be read immediately after receiving, Vasya receives *A* points for each message, *n*·*A* = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | 1,000 | [
{
"input": "4 5 5 3 5\n1 5 5 4",
"output": "20"
},
{
"input": "5 3 1 1 3\n2 2 2 1 1",
"output": "15"
},
{
"input": "5 5 3 4 5\n1 2 3 4 5",
"output": "35"
},
{
"input": "1 6 4 3 9\n2",
"output": "6"
},
{
"input": "10 9 7 5 3\n3 3 3 3 2 3 2 2 3 3",
"output": "90"
},
{
"input": "44 464 748 420 366\n278 109 293 161 336 9 194 203 13 226 303 303 300 131 134 47 235 110 263 67 185 337 360 253 270 97 162 190 143 267 18 311 329 138 322 167 324 33 3 104 290 260 349 89",
"output": "20416"
},
{
"input": "80 652 254 207 837\n455 540 278 38 19 781 686 110 733 40 434 581 77 381 818 236 444 615 302 251 762 676 771 483 767 479 326 214 316 551 544 95 157 828 813 201 103 502 751 410 84 733 431 90 261 326 731 374 730 748 303 83 302 673 50 822 46 590 248 751 345 579 689 616 331 593 428 344 754 777 178 80 602 268 776 234 637 780 712 539",
"output": "52160"
},
{
"input": "62 661 912 575 6\n3 5 6 6 5 6 6 6 3 2 3 1 4 3 2 5 3 6 1 4 2 5 1 2 6 4 6 6 5 5 4 3 4 1 4 2 4 4 2 6 4 6 3 5 3 4 1 5 3 6 5 6 4 1 2 1 6 5 5 4 2 3",
"output": "40982"
},
{
"input": "49 175 330 522 242\n109 81 215 5 134 185 60 242 154 148 14 221 146 229 45 120 142 43 202 176 231 105 212 69 109 219 58 103 53 211 128 138 157 95 96 122 69 109 35 46 122 118 132 135 224 150 178 134 28",
"output": "1083967"
},
{
"input": "27 27 15 395 590\n165 244 497 107 546 551 232 177 428 237 209 186 135 162 511 514 408 132 11 364 16 482 279 246 30 103 152",
"output": "3347009"
},
{
"input": "108 576 610 844 573\n242 134 45 515 430 354 405 179 174 366 155 4 300 176 96 36 508 70 75 316 118 563 55 340 128 214 138 511 507 437 454 478 341 443 421 573 270 362 208 107 256 471 436 378 336 507 383 352 450 411 297 34 179 551 119 524 141 288 387 9 283 241 304 214 503 559 416 447 495 61 169 228 479 568 368 441 467 401 467 542 370 243 371 315 65 67 161 383 19 144 283 5 369 242 122 396 276 488 401 387 256 128 87 425 124 226 335 238",
"output": "6976440"
},
{
"input": "67 145 951 829 192\n2 155 41 125 20 70 43 47 120 190 141 8 37 183 72 141 52 168 185 71 36 12 31 3 151 98 95 82 148 110 64 10 67 54 176 130 116 5 61 90 24 43 156 49 70 186 165 109 56 11 148 119 139 120 138 124 3 159 75 173 4 101 190 64 90 176 176",
"output": "9715"
},
{
"input": "67 322 317 647 99\n68 33 75 39 10 60 93 40 77 71 90 14 67 26 54 87 91 67 60 76 83 7 20 47 39 79 54 43 35 9 19 39 77 56 83 31 95 15 40 37 56 88 7 89 11 49 72 48 85 95 50 78 12 1 81 53 94 97 9 26 78 62 57 23 18 19 4",
"output": "1066024"
},
{
"input": "32 2 74 772 674\n598 426 358 191 471 667 412 44 183 358 436 654 572 489 79 191 374 33 1 627 154 132 101 236 443 112 77 93 553 53 260 498",
"output": "8161080"
},
{
"input": "111 907 98 340 279\n187 200 223 12 179 57 81 195 250 139 2 190 21 91 145 251 113 41 18 55 235 123 99 154 179 81 59 20 145 244 131 210 76 6 198 43 71 267 60 92 101 265 55 63 231 232 74 233 246 265 102 92 78 111 107 37 51 135 38 62 156 112 70 37 227 25 111 263 175 114 4 128 50 276 226 119 130 33 134 38 48 229 108 88 53 142 233 86 214 173 136 68 2 202 132 49 73 205 208 224 99 96 116 5 74 179 63 197 58 68 50",
"output": "4297441"
}
] | 1,556,887,397 | 2,147,483,647 | Python 3 | OK | TESTS | 60 | 109 | 0 | n, A, B, C, T = map(int, input().split())
t = sum(map(int, input().split()))
print(max(n * A - (T * n - t) * B + (n * T - t) * C, n * A))
| Title: Messages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* incoming messages for Vasya. The *i*-th message is going to be received after *t**i* minutes. Each message has a cost, which equals to *A* initially. After being received, the cost of a message decreases by *B* each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.
Also, each minute Vasya's bank account receives *C*·*k*, where *k* is the amount of received but unread messages.
Vasya's messages are very important to him, and because of that he wants to have all messages read after *T* minutes.
Determine the maximum amount of money Vasya's bank account can hold after *T* minutes.
Input Specification:
The first line contains five integers *n*, *A*, *B*, *C* and *T* (1<=≤<=*n*,<=*A*,<=*B*,<=*C*,<=*T*<=≤<=1000).
The second string contains *n* integers *t**i* (1<=≤<=*t**i*<=≤<=*T*).
Output Specification:
Output one integer — the answer to the problem.
Demo Input:
['4 5 5 3 5\n1 5 5 4\n', '5 3 1 1 3\n2 2 2 1 1\n', '5 5 3 4 5\n1 2 3 4 5\n']
Demo Output:
['20\n', '15\n', '35\n']
Note:
In the first sample the messages must be read immediately after receiving, Vasya receives *A* points for each message, *n*·*A* = 20 in total.
In the second sample the messages can be read at any integer moment.
In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 = - 5 points. This is 35 in total. | ```python
n, A, B, C, T = map(int, input().split())
t = sum(map(int, input().split()))
print(max(n * A - (T * n - t) * B + (n * T - t) * C, n * A))
``` | 3 |
|
859 | B | Lazy Security Guard | PROGRAMMING | 1,000 | [
"brute force",
"geometry",
"math"
] | null | null | Your security guard friend recently got a new job at a new security company. The company requires him to patrol an area of the city encompassing exactly *N* city blocks, but they let him choose which blocks. That is, your friend must walk the perimeter of a region whose area is exactly *N* blocks. Your friend is quite lazy and would like your help to find the shortest possible route that meets the requirements. The city is laid out in a square grid pattern, and is large enough that for the sake of the problem it can be considered infinite. | Input will consist of a single integer *N* (1<=≤<=*N*<=≤<=106), the number of city blocks that must be enclosed by the route. | Print the minimum perimeter that can be achieved. | [
"4\n",
"11\n",
"22\n"
] | [
"8\n",
"14\n",
"20\n"
] | Here are some possible shapes for the examples:
<img class="tex-graphics" src="https://espresso.codeforces.com/e11bef2cf82b55dd583cfc97d12b5aee5e483a65.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 750 | [
{
"input": "4",
"output": "8"
},
{
"input": "11",
"output": "14"
},
{
"input": "22",
"output": "20"
},
{
"input": "3",
"output": "8"
},
{
"input": "1024",
"output": "128"
},
{
"input": "101",
"output": "42"
},
{
"input": "30",
"output": "22"
},
{
"input": "1000000",
"output": "4000"
},
{
"input": "1",
"output": "4"
},
{
"input": "2",
"output": "6"
},
{
"input": "5",
"output": "10"
},
{
"input": "6",
"output": "10"
},
{
"input": "7",
"output": "12"
},
{
"input": "8",
"output": "12"
},
{
"input": "9",
"output": "12"
},
{
"input": "10",
"output": "14"
},
{
"input": "999000",
"output": "3998"
},
{
"input": "999001",
"output": "4000"
},
{
"input": "999999",
"output": "4000"
},
{
"input": "933206",
"output": "3866"
},
{
"input": "718351",
"output": "3392"
},
{
"input": "607443",
"output": "3118"
},
{
"input": "347887",
"output": "2360"
},
{
"input": "246206",
"output": "1986"
},
{
"input": "151375",
"output": "1558"
},
{
"input": "12639",
"output": "450"
},
{
"input": "3751",
"output": "246"
},
{
"input": "3607",
"output": "242"
},
{
"input": "124",
"output": "46"
},
{
"input": "64",
"output": "32"
},
{
"input": "31",
"output": "24"
},
{
"input": "23",
"output": "20"
},
{
"input": "15",
"output": "16"
},
{
"input": "19",
"output": "18"
},
{
"input": "59637",
"output": "978"
}
] | 1,586,158,037 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 109 | 0 | #----Kuzlyaev-Nikita-Codeforces-----
#------------06.04.2020-------------
alph="abcdefghijklmnopqrstuvwxyz"
#-----------------------------------
n=int(input())
r=int(n**0.5)
h=int(n/r)+int(n%r!=0)
print(2*(h+r)) | Title: Lazy Security Guard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your security guard friend recently got a new job at a new security company. The company requires him to patrol an area of the city encompassing exactly *N* city blocks, but they let him choose which blocks. That is, your friend must walk the perimeter of a region whose area is exactly *N* blocks. Your friend is quite lazy and would like your help to find the shortest possible route that meets the requirements. The city is laid out in a square grid pattern, and is large enough that for the sake of the problem it can be considered infinite.
Input Specification:
Input will consist of a single integer *N* (1<=≤<=*N*<=≤<=106), the number of city blocks that must be enclosed by the route.
Output Specification:
Print the minimum perimeter that can be achieved.
Demo Input:
['4\n', '11\n', '22\n']
Demo Output:
['8\n', '14\n', '20\n']
Note:
Here are some possible shapes for the examples:
<img class="tex-graphics" src="https://espresso.codeforces.com/e11bef2cf82b55dd583cfc97d12b5aee5e483a65.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
#----Kuzlyaev-Nikita-Codeforces-----
#------------06.04.2020-------------
alph="abcdefghijklmnopqrstuvwxyz"
#-----------------------------------
n=int(input())
r=int(n**0.5)
h=int(n/r)+int(n%r!=0)
print(2*(h+r))
``` | 3 |
|
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"
}
] | 1,602,211,003 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 248 | 0 | # -*- coding: utf-8 -*-
"""
Created on Fri Oct 9 09:21:45 2020
@author: 86198
"""
def add(x,y):
return x+y
n=int(input())
a=0
m=[0,0,0]
for i in (0,n):
l=[int(i)for i in input().split()]
m=map(add,m,l)
for i in list(m):
if i !=0:
a=a+1
else:
a=0
if a!=0:
print('NO')
else:
print('YES')
| 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
# -*- coding: utf-8 -*-
"""
Created on Fri Oct 9 09:21:45 2020
@author: 86198
"""
def add(x,y):
return x+y
n=int(input())
a=0
m=[0,0,0]
for i in (0,n):
l=[int(i)for i in input().split()]
m=map(add,m,l)
for i in list(m):
if i !=0:
a=a+1
else:
a=0
if a!=0:
print('NO')
else:
print('YES')
``` | 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"
}
] | 1,650,568,872 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | s = str(input())
upper = sum(map(str.isupper, s))
low = sum(map(str.islower, s))
if upper > low:
t = s.upper()
else:
t = s.lower()
print(t) | 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 = str(input())
upper = sum(map(str.isupper, s))
low = sum(map(str.islower, s))
if upper > low:
t = s.upper()
else:
t = s.lower()
print(t)
``` | 3.977 |
18 | B | Platforms | PROGRAMMING | 1,700 | [
"brute force",
"math"
] | B. Platforms | 2 | 64 | In one one-dimensional world there are *n* platforms. Platform with index *k* (platforms are numbered from 1) is a segment with coordinates [(*k*<=-<=1)*m*,<=(*k*<=-<=1)*m*<=+<=*l*], and *l*<=<<=*m*. Grasshopper Bob starts to jump along the platforms from point 0, with each jump he moves exactly *d* units right. Find out the coordinate of the point, where Bob will fall down. The grasshopper falls down, if he finds himself not on the platform, but if he finds himself on the edge of the platform, he doesn't fall down. | The first input line contains 4 integer numbers *n*, *d*, *m*, *l* (1<=≤<=*n*,<=*d*,<=*m*,<=*l*<=≤<=106,<=*l*<=<<=*m*) — respectively: amount of platforms, length of the grasshopper Bob's jump, and numbers *m* and *l* needed to find coordinates of the *k*-th platform: [(*k*<=-<=1)*m*,<=(*k*<=-<=1)*m*<=+<=*l*]. | Output the coordinates of the point, where the grosshopper will fall down. Don't forget that if Bob finds himself on the platform edge, he doesn't fall down. | [
"2 2 5 3\n",
"5 4 11 8\n"
] | [
"4\n",
"20\n"
] | none | 0 | [
{
"input": "2 2 5 3",
"output": "4"
},
{
"input": "5 4 11 8",
"output": "20"
},
{
"input": "228385 744978 699604 157872",
"output": "2979912"
},
{
"input": "773663 427904 329049 243542",
"output": "1283712"
},
{
"input": "835293 627183 442142 361649",
"output": "1254366"
},
{
"input": "896922 310109 71587 16487",
"output": "310109"
},
{
"input": "958552 993036 701031 109903",
"output": "993036"
},
{
"input": "20182 192314 814124 268107",
"output": "384628"
},
{
"input": "81812 875240 443569 287155",
"output": "875240"
},
{
"input": "3 6 6 3",
"output": "18"
},
{
"input": "3 16 6 3",
"output": "16"
},
{
"input": "3 4 6 3",
"output": "4"
},
{
"input": "680892 333996 619800 374820",
"output": "1001988"
},
{
"input": "658990 366800 43771 676",
"output": "366800"
},
{
"input": "637089 915955 984094 706836",
"output": "915955"
},
{
"input": "615188 948759 924417 924407",
"output": "183286007415"
},
{
"input": "593287 497915 864740 864733",
"output": "82319789035"
},
{
"input": "87738 530718 805063 805047",
"output": "11387616126"
},
{
"input": "65837 79874 229034 229024",
"output": "1636218890"
},
{
"input": "755991 187301 743241 743232",
"output": "2217831141"
},
{
"input": "402841 635488 123613 122628",
"output": "49568064"
},
{
"input": "999463 261665 255021 255007",
"output": "1596941495"
},
{
"input": "43496 179847 327622 327621",
"output": "14250356892"
},
{
"input": "105126 379125 440715 440713",
"output": "46330970625"
},
{
"input": "1000000 1 1000000 999999",
"output": "1000000000000"
},
{
"input": "1000000 16 999952 999951",
"output": "999952000000"
},
{
"input": "1000000 49 999983 999982",
"output": "999983000023"
},
{
"input": "1000000 3 999997 999996",
"output": "999997000002"
},
{
"input": "1000000 11 999989 999988",
"output": "999989000010"
},
{
"input": "1000000 64 999956 999955",
"output": "999956000000"
},
{
"input": "1000000 531 999106 999105",
"output": "999106000236"
},
{
"input": "1000000 337 999956 999955",
"output": "999956000119"
},
{
"input": "1 1 2 1",
"output": "2"
},
{
"input": "1 1000000 5 3",
"output": "1000000"
},
{
"input": "1000000 1000000 1000000 999999",
"output": "1000000000000"
}
] | 1,619,245,598 | 4,298 | PyPy 3 | WRONG_ANSWER | TESTS | 9 | 342 | 6,041,600 | import sys
# from itertools import product
import logging
logging.root.setLevel(level=logging.INFO)
n,d,m,l = map(int,sys.stdin.readline().strip().split())
loc = 0
right = n*m+l
while loc % m <= l and loc <= right:
loc += d
print(loc) | Title: Platforms
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
In one one-dimensional world there are *n* platforms. Platform with index *k* (platforms are numbered from 1) is a segment with coordinates [(*k*<=-<=1)*m*,<=(*k*<=-<=1)*m*<=+<=*l*], and *l*<=<<=*m*. Grasshopper Bob starts to jump along the platforms from point 0, with each jump he moves exactly *d* units right. Find out the coordinate of the point, where Bob will fall down. The grasshopper falls down, if he finds himself not on the platform, but if he finds himself on the edge of the platform, he doesn't fall down.
Input Specification:
The first input line contains 4 integer numbers *n*, *d*, *m*, *l* (1<=≤<=*n*,<=*d*,<=*m*,<=*l*<=≤<=106,<=*l*<=<<=*m*) — respectively: amount of platforms, length of the grasshopper Bob's jump, and numbers *m* and *l* needed to find coordinates of the *k*-th platform: [(*k*<=-<=1)*m*,<=(*k*<=-<=1)*m*<=+<=*l*].
Output Specification:
Output the coordinates of the point, where the grosshopper will fall down. Don't forget that if Bob finds himself on the platform edge, he doesn't fall down.
Demo Input:
['2 2 5 3\n', '5 4 11 8\n']
Demo Output:
['4\n', '20\n']
Note:
none | ```python
import sys
# from itertools import product
import logging
logging.root.setLevel(level=logging.INFO)
n,d,m,l = map(int,sys.stdin.readline().strip().split())
loc = 0
right = n*m+l
while loc % m <= l and loc <= right:
loc += d
print(loc)
``` | 0 |
17 | A | Noldbach problem | PROGRAMMING | 1,000 | [
"brute force",
"math",
"number theory"
] | A. Noldbach problem | 2 | 64 | Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong. | The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000). | Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO. | [
"27 2\n",
"45 7\n"
] | [
"YES",
"NO"
] | In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form. | 0 | [
{
"input": "27 2",
"output": "YES"
},
{
"input": "45 7",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
},
{
"input": "15 1",
"output": "YES"
},
{
"input": "17 1",
"output": "YES"
},
{
"input": "34 5",
"output": "NO"
},
{
"input": "37 4",
"output": "YES"
},
{
"input": "43 5",
"output": "YES"
},
{
"input": "47 7",
"output": "NO"
},
{
"input": "50 5",
"output": "YES"
},
{
"input": "57 6",
"output": "YES"
},
{
"input": "60 8",
"output": "NO"
},
{
"input": "62 7",
"output": "YES"
},
{
"input": "76 9",
"output": "NO"
},
{
"input": "69 7",
"output": "YES"
},
{
"input": "113 10",
"output": "YES"
},
{
"input": "141 11",
"output": "YES"
},
{
"input": "207 16",
"output": "NO"
},
{
"input": "231 18",
"output": "NO"
},
{
"input": "296 19",
"output": "YES"
},
{
"input": "332 20",
"output": "YES"
},
{
"input": "378 24",
"output": "NO"
},
{
"input": "428 23",
"output": "YES"
},
{
"input": "497 27",
"output": "NO"
},
{
"input": "640 32",
"output": "YES"
},
{
"input": "798 35",
"output": "YES"
},
{
"input": "802 35",
"output": "YES"
},
{
"input": "864 40",
"output": "NO"
},
{
"input": "953 44",
"output": "NO"
},
{
"input": "995 44",
"output": "YES"
},
{
"input": "1000 44",
"output": "YES"
},
{
"input": "1000 0",
"output": "YES"
},
{
"input": "1000 1000",
"output": "NO"
},
{
"input": "2 1000",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
}
] | 1,646,433,007 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 31 | 154 | 2,048,000 | from math import sqrt
arr = [int(item) for item in input().split()]
parr = []
def isPrime(n):
if (n <= 0):
return
else:
for i in range(2, int(sqrt(n)) +1):
if (n % i == 0):
return False
else:
return True
for i in range(2, arr[0]+1): #making a list with all prime nums
if(isPrime(i)):
parr.append(i)
#creating permutations of all possibilites of combinations
def combinations():
sum = 0
if(arr[0] == 2):
return sum
for i in range(1, len(parr)):
summ = parr[i] + parr[i-1] + 1
if(isPrime(summ) and summ in parr):
sum += 1
return sum
var = combinations()
if(var == arr[1]):
print("YES")
else:
print("NO") | Title: Noldbach problem
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong.
Input Specification:
The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000).
Output Specification:
Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO.
Demo Input:
['27 2\n', '45 7\n']
Demo Output:
['YES', 'NO']
Note:
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form. | ```python
from math import sqrt
arr = [int(item) for item in input().split()]
parr = []
def isPrime(n):
if (n <= 0):
return
else:
for i in range(2, int(sqrt(n)) +1):
if (n % i == 0):
return False
else:
return True
for i in range(2, arr[0]+1): #making a list with all prime nums
if(isPrime(i)):
parr.append(i)
#creating permutations of all possibilites of combinations
def combinations():
sum = 0
if(arr[0] == 2):
return sum
for i in range(1, len(parr)):
summ = parr[i] + parr[i-1] + 1
if(isPrime(summ) and summ in parr):
sum += 1
return sum
var = combinations()
if(var == arr[1]):
print("YES")
else:
print("NO")
``` | 0 |
0 | none | none | none | 0 | [
"none"
] | null | null | Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot? | First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players. | Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise. | [
"4\n75 150 75 50\n",
"3\n100 150 250\n"
] | [
"Yes\n",
"No\n"
] | In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | 0 | [
{
"input": "4\n75 150 75 50",
"output": "Yes"
},
{
"input": "3\n100 150 250",
"output": "No"
},
{
"input": "7\n34 34 68 34 34 68 34",
"output": "Yes"
},
{
"input": "10\n72 96 12 18 81 20 6 2 54 1",
"output": "No"
},
{
"input": "20\n958692492 954966768 77387000 724664764 101294996 614007760 202904092 555293973 707655552 108023967 73123445 612562357 552908390 914853758 915004122 466129205 122853497 814592742 373389439 818473058",
"output": "No"
},
{
"input": "2\n1 1",
"output": "Yes"
},
{
"input": "2\n72 72",
"output": "Yes"
},
{
"input": "2\n49 42",
"output": "No"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "Yes"
},
{
"input": "6\n162000 96000 648000 1000 864000 432000",
"output": "Yes"
},
{
"input": "8\n600000 100000 100000 100000 900000 600000 900000 600000",
"output": "Yes"
},
{
"input": "12\n2048 1024 6144 1024 3072 3072 6144 1024 4096 2048 6144 3072",
"output": "Yes"
},
{
"input": "20\n246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246",
"output": "Yes"
},
{
"input": "50\n840868705 387420489 387420489 795385082 634350497 206851546 536870912 536870912 414927754 387420489 387420489 536870912 387420489 149011306 373106005 536870912 700746206 387420489 777952883 847215247 176645254 576664386 387420489 230876513 536870912 536870912 536870912 387420489 387420489 536870912 460495524 528643722 387420489 536870912 470369206 899619085 387420489 631148352 387420489 387420489 536870912 414666674 521349938 776784669 387420489 102428009 536870912 387420489 536870912 718311009",
"output": "No"
},
{
"input": "2\n5 6",
"output": "No"
},
{
"input": "3\n536870912 387420489 257407169",
"output": "No"
},
{
"input": "4\n2 2 5 2",
"output": "No"
},
{
"input": "2\n33554432 59049",
"output": "Yes"
},
{
"input": "3\n536870912 387420489 387420489",
"output": "Yes"
},
{
"input": "2\n1 5",
"output": "No"
},
{
"input": "18\n2 3 5 7 11 13 17 19 23 29 31 37 43 47 53 59 67 71",
"output": "No"
},
{
"input": "2\n1 30",
"output": "No"
},
{
"input": "3\n335544320 71744535 71744535",
"output": "Yes"
},
{
"input": "5\n1000000000 999999999 999999998 999999997 999999996",
"output": "No"
},
{
"input": "2\n25 5",
"output": "No"
},
{
"input": "4\n75 150 75 5",
"output": "No"
},
{
"input": "3\n536870912 387420489 362797056",
"output": "Yes"
},
{
"input": "3\n536870912 387420489 89",
"output": "No"
},
{
"input": "4\n547 2606459 222763549 143466789",
"output": "No"
},
{
"input": "3\n129140163 33554432 1",
"output": "Yes"
},
{
"input": "10\n244140625 244140625 244140625 244140625 244140625 244140625 244140625 244140625 536870912 387420489",
"output": "No"
},
{
"input": "3\n5 5 1",
"output": "No"
},
{
"input": "5\n3 7 29 36760123 823996703",
"output": "No"
}
] | 1,490,126,358 | 2,147,483,647 | Python 3 | OK | TESTS | 70 | 920 | 13,414,400 | n=int(input())
A=[]
A = [int(x) for x in input().split()]
for i in range(n):
while A[i]%2==0:
A[i]=A[i]//2
while A[i]%3==0:
A[i]=A[i]//3
#print (A)
for i in A:
if i!=A[0]:
print("No")
exit()
print("Yes")
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
Input Specification:
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
Output Specification:
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
Demo Input:
['4\n75 150 75 50\n', '3\n100 150 250\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | ```python
n=int(input())
A=[]
A = [int(x) for x in input().split()]
for i in range(n):
while A[i]%2==0:
A[i]=A[i]//2
while A[i]%3==0:
A[i]=A[i]//3
#print (A)
for i in A:
if i!=A[0]:
print("No")
exit()
print("Yes")
``` | 3 |
|
540 | C | Ice Cave | PROGRAMMING | 2,000 | [
"dfs and similar"
] | null | null | You play a computer game. Your character stands on some level of a multilevel ice cave. In order to move on forward, you need to descend one level lower and the only way to do this is to fall through the ice.
The level of the cave where you are is a rectangular square grid of *n* rows and *m* columns. Each cell consists either from intact or from cracked ice. From each cell you can move to cells that are side-adjacent with yours (due to some limitations of the game engine you cannot make jumps on the same place, i.e. jump from a cell to itself). If you move to the cell with cracked ice, then your character falls down through it and if you move to the cell with intact ice, then the ice on this cell becomes cracked.
Let's number the rows with integers from 1 to *n* from top to bottom and the columns with integers from 1 to *m* from left to right. Let's denote a cell on the intersection of the *r*-th row and the *c*-th column as (*r*,<=*c*).
You are staying in the cell (*r*1,<=*c*1) and this cell is cracked because you've just fallen here from a higher level. You need to fall down through the cell (*r*2,<=*c*2) since the exit to the next level is there. Can you do this? | The first line contains two integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=500) — the number of rows and columns in the cave description.
Each of the next *n* lines describes the initial state of the level of the cave, each line consists of *m* characters "." (that is, intact ice) and "X" (cracked ice).
The next line contains two integers, *r*1 and *c*1 (1<=≤<=*r*1<=≤<=*n*,<=1<=≤<=*c*1<=≤<=*m*) — your initial coordinates. It is guaranteed that the description of the cave contains character 'X' in cell (*r*1,<=*c*1), that is, the ice on the starting cell is initially cracked.
The next line contains two integers *r*2 and *c*2 (1<=≤<=*r*2<=≤<=*n*,<=1<=≤<=*c*2<=≤<=*m*) — the coordinates of the cell through which you need to fall. The final cell may coincide with the starting one. | If you can reach the destination, print 'YES', otherwise print 'NO'. | [
"4 6\nX...XX\n...XX.\n.X..X.\n......\n1 6\n2 2\n",
"5 4\n.X..\n...X\nX.X.\n....\n.XX.\n5 3\n1 1\n",
"4 7\n..X.XX.\n.XX..X.\nX...X..\nX......\n2 2\n1 6\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first sample test one possible path is:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c61f56de718beea14935ccdc85ae2c4ad45c1454.png" style="max-width: 100.0%;max-height: 100.0%;"/>
After the first visit of cell (2, 2) the ice on it cracks and when you step there for the second time, your character falls through the ice as intended. | 1,500 | [
{
"input": "4 6\nX...XX\n...XX.\n.X..X.\n......\n1 6\n2 2",
"output": "YES"
},
{
"input": "5 4\n.X..\n...X\nX.X.\n....\n.XX.\n5 3\n1 1",
"output": "NO"
},
{
"input": "4 7\n..X.XX.\n.XX..X.\nX...X..\nX......\n2 2\n1 6",
"output": "YES"
},
{
"input": "5 3\n.XX\n...\n.X.\n.X.\n...\n1 3\n4 1",
"output": "YES"
},
{
"input": "1 1\nX\n1 1\n1 1",
"output": "NO"
},
{
"input": "1 6\n.X...X\n1 2\n1 5",
"output": "NO"
},
{
"input": "7 1\nX\n.\n.\n.\nX\n.\n.\n5 1\n3 1",
"output": "YES"
},
{
"input": "1 2\nXX\n1 1\n1 1",
"output": "NO"
},
{
"input": "2 1\n.\nX\n2 1\n2 1",
"output": "YES"
},
{
"input": "3 4\n.X..\n..XX\n..X.\n1 2\n3 4",
"output": "NO"
},
{
"input": "3 5\n.X.XX\nX...X\nX.X..\n2 1\n1 5",
"output": "NO"
},
{
"input": "3 2\n..\nX.\n.X\n3 2\n3 1",
"output": "NO"
},
{
"input": "3 4\nXX.X\nX...\n.X.X\n1 2\n1 1",
"output": "YES"
},
{
"input": "1 2\nX.\n1 1\n1 2",
"output": "NO"
},
{
"input": "2 1\nX\nX\n2 1\n1 1",
"output": "YES"
},
{
"input": "2 2\nXX\nXX\n1 1\n2 2",
"output": "NO"
},
{
"input": "2 2\n..\n.X\n2 2\n1 1",
"output": "YES"
},
{
"input": "2 2\n.X\n.X\n1 2\n2 2",
"output": "YES"
},
{
"input": "2 2\n..\nXX\n2 1\n1 1",
"output": "YES"
},
{
"input": "4 2\nX.\n.X\n.X\nXX\n2 2\n3 1",
"output": "NO"
},
{
"input": "2 4\nX.XX\n.X..\n2 2\n2 3",
"output": "YES"
},
{
"input": "6 4\nX..X\n..X.\n.X..\n..X.\n.X..\nX..X\n1 1\n6 4",
"output": "NO"
},
{
"input": "5 4\nX...\n..X.\n.X..\nX..X\n....\n4 4\n3 1",
"output": "NO"
},
{
"input": "3 4\nX..X\n..XX\n.X..\n2 3\n3 1",
"output": "NO"
},
{
"input": "20 20\n....................\n.......X...........X\n............X......X\n.X...XX..X....X.....\n....X.....X.........\nX..........X........\n......X........X....\n....................\n...................X\n......X.............\n..............X.....\n......X.X...........\n.X.........X.X......\n.........X......X..X\n..................X.\n...X........X.......\n....................\n....................\n..X.....X...........\n........X......X.X..\n20 16\n5 20",
"output": "YES"
},
{
"input": "21 21\n.....X...X.........X.\n...X...XX......X.....\n..X........X.X...XX..\n.........X....X...X..\nX...X...........XX...\n...X...X....XX...XXX.\n.X............X......\n......X.X............\n.X...X.........X.X...\n......XX......X.X....\n....X.......X.XXX....\n.X.......X..XXX.X..X.\n..X........X....X...X\n.........X..........X\n.....XX.....X........\n...XX......X.........\n.....X...XX...X......\n..X.X....XX..XX.X....\nX........X.X..XX..X..\nX..X......X...X.X....\nX.....X.....X.X......\n20 4\n21 5",
"output": "YES"
},
{
"input": "2 1\nX\nX\n2 1\n2 1",
"output": "NO"
},
{
"input": "2 2\nXX\nX.\n1 1\n2 2",
"output": "NO"
},
{
"input": "2 1\nX\nX\n1 1\n1 1",
"output": "NO"
},
{
"input": "1 2\nXX\n1 2\n1 2",
"output": "NO"
},
{
"input": "1 2\nXX\n1 1\n1 2",
"output": "YES"
},
{
"input": "1 2\nXX\n1 2\n1 1",
"output": "YES"
},
{
"input": "2 1\nX\nX\n1 1\n2 1",
"output": "YES"
},
{
"input": "2 1\n.\nX\n2 1\n1 1",
"output": "NO"
},
{
"input": "2 1\nX\n.\n1 1\n2 1",
"output": "NO"
},
{
"input": "1 2\n.X\n1 2\n1 1",
"output": "NO"
},
{
"input": "2 1\nX\n.\n1 1\n1 1",
"output": "YES"
},
{
"input": "1 2\nX.\n1 1\n1 1",
"output": "YES"
},
{
"input": "1 2\n.X\n1 2\n1 2",
"output": "YES"
},
{
"input": "2 2\nX.\n..\n1 1\n2 2",
"output": "YES"
},
{
"input": "2 2\n..\nX.\n2 1\n1 1",
"output": "YES"
},
{
"input": "4 3\n..X\n..X\n.XX\n.XX\n4 2\n2 2",
"output": "YES"
},
{
"input": "3 3\nXXX\nX..\nXXX\n2 1\n2 2",
"output": "YES"
},
{
"input": "5 4\nXXXX\nX..X\nX..X\nXXXX\nXXXX\n4 2\n3 3",
"output": "YES"
}
] | 1,597,966,977 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 124 | 0 | def is_valid_tile(r, c, n, m):
return r >= 0 and c >= 0 and r < n and c < m
def gen_adjacents(r, c, n, m):
adjs = []
possibles = [(r + 1, c), (r - 1, c), (r, c + 1), (r, c - 1)]
for possible in possibles:
if is_valid_tile(*possible, n, m):
adjs.append(possible)
return adjs
def solve(table, r1, c1, r2, c2, n, m):
if r1 == r2 and c1 == c2:
adjacents = gen_adjacents(r1, c1, n, m)
for adjacent in adjacents:
if table[adjacent[0]][adjacent[1]] == "X":
return False
return True
if table[r2][c2] == ".":
adjacents = gen_adjacents(r2, c2, n, m)
solid = 0
for adj in adjacents:
if table[adj[0]][adj[1]] == ".":
solid += 1
if solid < 2:
return False
forward_visited = set([(r1, c1)]); backward_visited = set([(r2, c2)])
forward_nodes = [(r1, c1)]; backward_nodes = [(r2, c2)]
while forward_nodes and backward_nodes:
for i in range(len(forward_nodes)):
node = forward_nodes.pop(0)
adjacents = gen_adjacents(*node, n, m)
for adjacent in adjacents:
if adjacent in backward_visited:
return True
if adjacent not in forward_visited and table[adjacent[0]][adjacent[1]] == ".":
forward_visited.add(adjacent)
forward_nodes.append(adjacent)
for j in range(len(backward_nodes)):
node = backward_nodes.pop(0)
adjacents = gen_adjacents(*node, n, m)
for adjacent in adjacents:
if adjacent in forward_visited:
return True
if adjacent not in backward_visited and table[adjacent[0]][adjacent[1]] == ".":
backward_visited.add(adjacent)
backward_nodes.append(adjacent)
return False
def main():
n, m = map(int, input().split())
table = []
for _ in range(n):
row = input()
table.append(row)
r1, c1 = map(int, input().split())
r2, c2 = map(int, input().split())
r1 -= 1; c1 -= 1; r2 -= 1; c2 -= 1
if solve(table, r1, c1, r2, c2, n, m):
print("YES")
else:
print("NO")
main() | Title: Ice Cave
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You play a computer game. Your character stands on some level of a multilevel ice cave. In order to move on forward, you need to descend one level lower and the only way to do this is to fall through the ice.
The level of the cave where you are is a rectangular square grid of *n* rows and *m* columns. Each cell consists either from intact or from cracked ice. From each cell you can move to cells that are side-adjacent with yours (due to some limitations of the game engine you cannot make jumps on the same place, i.e. jump from a cell to itself). If you move to the cell with cracked ice, then your character falls down through it and if you move to the cell with intact ice, then the ice on this cell becomes cracked.
Let's number the rows with integers from 1 to *n* from top to bottom and the columns with integers from 1 to *m* from left to right. Let's denote a cell on the intersection of the *r*-th row and the *c*-th column as (*r*,<=*c*).
You are staying in the cell (*r*1,<=*c*1) and this cell is cracked because you've just fallen here from a higher level. You need to fall down through the cell (*r*2,<=*c*2) since the exit to the next level is there. Can you do this?
Input Specification:
The first line contains two integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=500) — the number of rows and columns in the cave description.
Each of the next *n* lines describes the initial state of the level of the cave, each line consists of *m* characters "." (that is, intact ice) and "X" (cracked ice).
The next line contains two integers, *r*1 and *c*1 (1<=≤<=*r*1<=≤<=*n*,<=1<=≤<=*c*1<=≤<=*m*) — your initial coordinates. It is guaranteed that the description of the cave contains character 'X' in cell (*r*1,<=*c*1), that is, the ice on the starting cell is initially cracked.
The next line contains two integers *r*2 and *c*2 (1<=≤<=*r*2<=≤<=*n*,<=1<=≤<=*c*2<=≤<=*m*) — the coordinates of the cell through which you need to fall. The final cell may coincide with the starting one.
Output Specification:
If you can reach the destination, print 'YES', otherwise print 'NO'.
Demo Input:
['4 6\nX...XX\n...XX.\n.X..X.\n......\n1 6\n2 2\n', '5 4\n.X..\n...X\nX.X.\n....\n.XX.\n5 3\n1 1\n', '4 7\n..X.XX.\n.XX..X.\nX...X..\nX......\n2 2\n1 6\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample test one possible path is:
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c61f56de718beea14935ccdc85ae2c4ad45c1454.png" style="max-width: 100.0%;max-height: 100.0%;"/>
After the first visit of cell (2, 2) the ice on it cracks and when you step there for the second time, your character falls through the ice as intended. | ```python
def is_valid_tile(r, c, n, m):
return r >= 0 and c >= 0 and r < n and c < m
def gen_adjacents(r, c, n, m):
adjs = []
possibles = [(r + 1, c), (r - 1, c), (r, c + 1), (r, c - 1)]
for possible in possibles:
if is_valid_tile(*possible, n, m):
adjs.append(possible)
return adjs
def solve(table, r1, c1, r2, c2, n, m):
if r1 == r2 and c1 == c2:
adjacents = gen_adjacents(r1, c1, n, m)
for adjacent in adjacents:
if table[adjacent[0]][adjacent[1]] == "X":
return False
return True
if table[r2][c2] == ".":
adjacents = gen_adjacents(r2, c2, n, m)
solid = 0
for adj in adjacents:
if table[adj[0]][adj[1]] == ".":
solid += 1
if solid < 2:
return False
forward_visited = set([(r1, c1)]); backward_visited = set([(r2, c2)])
forward_nodes = [(r1, c1)]; backward_nodes = [(r2, c2)]
while forward_nodes and backward_nodes:
for i in range(len(forward_nodes)):
node = forward_nodes.pop(0)
adjacents = gen_adjacents(*node, n, m)
for adjacent in adjacents:
if adjacent in backward_visited:
return True
if adjacent not in forward_visited and table[adjacent[0]][adjacent[1]] == ".":
forward_visited.add(adjacent)
forward_nodes.append(adjacent)
for j in range(len(backward_nodes)):
node = backward_nodes.pop(0)
adjacents = gen_adjacents(*node, n, m)
for adjacent in adjacents:
if adjacent in forward_visited:
return True
if adjacent not in backward_visited and table[adjacent[0]][adjacent[1]] == ".":
backward_visited.add(adjacent)
backward_nodes.append(adjacent)
return False
def main():
n, m = map(int, input().split())
table = []
for _ in range(n):
row = input()
table.append(row)
r1, c1 = map(int, input().split())
r2, c2 = map(int, input().split())
r1 -= 1; c1 -= 1; r2 -= 1; c2 -= 1
if solve(table, r1, c1, r2, c2, n, m):
print("YES")
else:
print("NO")
main()
``` | 0 |
|
743 | B | Chloe and the sequence | PROGRAMMING | 1,200 | [
"binary search",
"bitmasks",
"constructive algorithms",
"implementation"
] | null | null | Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step.
The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps.
Please help Chloe to solve the problem! | The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1). | Print single integer — the integer at the *k*-th position in the obtained sequence. | [
"3 2\n",
"4 8\n"
] | [
"2",
"4"
] | In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4. | 1,000 | [
{
"input": "3 2",
"output": "2"
},
{
"input": "4 8",
"output": "4"
},
{
"input": "5 27",
"output": "1"
},
{
"input": "7 44",
"output": "3"
},
{
"input": "15 18432",
"output": "12"
},
{
"input": "20 259676",
"output": "3"
},
{
"input": "30 671088640",
"output": "28"
},
{
"input": "38 137438953472",
"output": "38"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "3 3",
"output": "1"
},
{
"input": "4 12",
"output": "3"
},
{
"input": "8 224",
"output": "6"
},
{
"input": "13 1368",
"output": "4"
},
{
"input": "16 49152",
"output": "15"
},
{
"input": "19 1024",
"output": "11"
},
{
"input": "24 15204352",
"output": "20"
},
{
"input": "27 6586544",
"output": "5"
},
{
"input": "31 536870912",
"output": "30"
},
{
"input": "38 94489280512",
"output": "34"
},
{
"input": "42 215268308020",
"output": "3"
},
{
"input": "42 3452074721280",
"output": "21"
},
{
"input": "46 34394312982528",
"output": "15"
},
{
"input": "48 133865540681728",
"output": "39"
},
{
"input": "49 76541041985542",
"output": "2"
},
{
"input": "49 104822971826176",
"output": "34"
},
{
"input": "49 351843720888320",
"output": "47"
},
{
"input": "50 743042492421629",
"output": "1"
},
{
"input": "50 666266740436818",
"output": "2"
},
{
"input": "50 704239287953456",
"output": "5"
},
{
"input": "50 116938486513664",
"output": "17"
},
{
"input": "50 806423059496960",
"output": "37"
},
{
"input": "50 985162418487296",
"output": "48"
},
{
"input": "50 844424930131968",
"output": "49"
},
{
"input": "50 562949953421312",
"output": "50"
},
{
"input": "50 1",
"output": "1"
},
{
"input": "50 1125899906842623",
"output": "1"
},
{
"input": "50 1125899906842620",
"output": "3"
},
{
"input": "39 549755813887",
"output": "1"
},
{
"input": "50 100000000000",
"output": "12"
}
] | 1,632,881,408 | 2,147,483,647 | PyPy 3 | OK | TESTS | 39 | 108 | 20,172,800 | n, k = map(int, input().split())
cnt = 1
for i in range(n-1):
cnt = cnt*2 + 1
l = 1
r = cnt
p = n
while l <= r:
mid = (l + r) // 2
if mid == k:
print(n)
break
elif mid < k:
l = mid + 1
n -= 1
else:
r = mid - 1
n -=1
| Title: Chloe and the sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step.
The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps.
Please help Chloe to solve the problem!
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).
Output Specification:
Print single integer — the integer at the *k*-th position in the obtained sequence.
Demo Input:
['3 2\n', '4 8\n']
Demo Output:
['2', '4']
Note:
In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4. | ```python
n, k = map(int, input().split())
cnt = 1
for i in range(n-1):
cnt = cnt*2 + 1
l = 1
r = cnt
p = n
while l <= r:
mid = (l + r) // 2
if mid == k:
print(n)
break
elif mid < k:
l = mid + 1
n -= 1
else:
r = mid - 1
n -=1
``` | 3 |
|
622 | A | Infinite Sequence | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Consider the infinite sequence of integers: 1,<=1,<=2,<=1,<=2,<=3,<=1,<=2,<=3,<=4,<=1,<=2,<=3,<=4,<=5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the *n*-th position of the sequence. | The only line contains integer *n* (1<=≤<=*n*<=≤<=1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type. | Print the element in the *n*-th position of the sequence (the elements are numerated from one). | [
"3\n",
"5\n",
"10\n",
"55\n",
"56\n"
] | [
"2\n",
"2\n",
"4\n",
"10\n",
"1\n"
] | none | 0 | [
{
"input": "3",
"output": "2"
},
{
"input": "5",
"output": "2"
},
{
"input": "10",
"output": "4"
},
{
"input": "55",
"output": "10"
},
{
"input": "56",
"output": "1"
},
{
"input": "1000000000000",
"output": "88209"
},
{
"input": "847194127849",
"output": "255708"
},
{
"input": "294719472984",
"output": "593358"
},
{
"input": "999999911791",
"output": "1414213"
},
{
"input": "999999911792",
"output": "1"
},
{
"input": "100000000000000",
"output": "1749820"
},
{
"input": "1",
"output": "1"
},
{
"input": "99993",
"output": "312"
},
{
"input": "99994",
"output": "313"
},
{
"input": "99995",
"output": "314"
},
{
"input": "99990",
"output": "309"
},
{
"input": "2",
"output": "1"
},
{
"input": "99991",
"output": "310"
},
{
"input": "99992",
"output": "311"
},
{
"input": "99996",
"output": "315"
},
{
"input": "99997",
"output": "316"
},
{
"input": "99998",
"output": "317"
},
{
"input": "99999",
"output": "318"
},
{
"input": "1021",
"output": "31"
},
{
"input": "4",
"output": "1"
},
{
"input": "23",
"output": "2"
},
{
"input": "9994",
"output": "124"
},
{
"input": "99939",
"output": "258"
},
{
"input": "99999998250180",
"output": "14142135"
},
{
"input": "6",
"output": "3"
},
{
"input": "8",
"output": "2"
},
{
"input": "35",
"output": "7"
},
{
"input": "100",
"output": "9"
},
{
"input": "10101010",
"output": "745"
},
{
"input": "103",
"output": "12"
},
{
"input": "102",
"output": "11"
},
{
"input": "101",
"output": "10"
}
] | 1,683,031,698 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 10 | 46 | 0 | # LUOGU_RID: 109426988
l=1;r=10**7
end=int(input())
while l<r:
mid=(l+r)//2
if (mid**2+mid)//2 <= end:
l=mid+1
else:
r=mid
maxi=(l**2+l)//2
mini=((l-1)**2+l-1)//2
deta=end-mini
if deta==0:
print(l-1)
else:
print(deta)
| Title: Infinite Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider the infinite sequence of integers: 1,<=1,<=2,<=1,<=2,<=3,<=1,<=2,<=3,<=4,<=1,<=2,<=3,<=4,<=5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the *n*-th position of the sequence.
Input Specification:
The only line contains integer *n* (1<=≤<=*n*<=≤<=1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output Specification:
Print the element in the *n*-th position of the sequence (the elements are numerated from one).
Demo Input:
['3\n', '5\n', '10\n', '55\n', '56\n']
Demo Output:
['2\n', '2\n', '4\n', '10\n', '1\n']
Note:
none | ```python
# LUOGU_RID: 109426988
l=1;r=10**7
end=int(input())
while l<r:
mid=(l+r)//2
if (mid**2+mid)//2 <= end:
l=mid+1
else:
r=mid
maxi=(l**2+l)//2
mini=((l-1)**2+l-1)//2
deta=end-mini
if deta==0:
print(l-1)
else:
print(deta)
``` | 0 |
|
784 | F | Crunching Numbers Just for You | PROGRAMMING | 1,900 | [
"*special",
"implementation"
] | null | null | You are developing a new feature for the website which sells airline tickets: being able to sort tickets by price! You have already extracted the tickets' prices, so there's just the last step to be done...
You are given an array of integers. Sort it in non-descending order. | The input consists of a single line of space-separated integers. The first number is *n* (1<=≤<=*n*<=≤<=10) — the size of the array. The following *n* numbers are the elements of the array (1<=≤<=*a**i*<=≤<=100). | Output space-separated elements of the sorted array. | [
"3 3 1 2\n"
] | [
"1 2 3 \n"
] | Remember, this is a very important feature, and you have to make sure the customers appreciate it! | 0 | [
{
"input": "3 3 1 2",
"output": "1 2 3 "
},
{
"input": "10 54 100 27 1 33 27 80 49 27 6",
"output": "1 6 27 27 27 33 49 54 80 100 "
}
] | 1,691,719,893 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 31 | 0 |
print(*sorted(list(set(list(map(int,input().split())))))) | Title: Crunching Numbers Just for You
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are developing a new feature for the website which sells airline tickets: being able to sort tickets by price! You have already extracted the tickets' prices, so there's just the last step to be done...
You are given an array of integers. Sort it in non-descending order.
Input Specification:
The input consists of a single line of space-separated integers. The first number is *n* (1<=≤<=*n*<=≤<=10) — the size of the array. The following *n* numbers are the elements of the array (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output space-separated elements of the sorted array.
Demo Input:
['3 3 1 2\n']
Demo Output:
['1 2 3 \n']
Note:
Remember, this is a very important feature, and you have to make sure the customers appreciate it! | ```python
print(*sorted(list(set(list(map(int,input().split()))))))
``` | 0 |
|
133 | A | HQ9+ | PROGRAMMING | 900 | [
"implementation"
] | null | null | HQ9+ is a joke programming language which has only four one-character instructions:
- "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output. | The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive. | Output "YES", if executing the program will produce any output, and "NO" otherwise. | [
"Hi!\n",
"Codeforces\n"
] | [
"YES\n",
"NO\n"
] | In the first case the program contains only one instruction — "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions. | 500 | [
{
"input": "Hi!",
"output": "YES"
},
{
"input": "Codeforces",
"output": "NO"
},
{
"input": "a+b=c",
"output": "NO"
},
{
"input": "hq-lowercase",
"output": "NO"
},
{
"input": "Q",
"output": "YES"
},
{
"input": "9",
"output": "YES"
},
{
"input": "H",
"output": "YES"
},
{
"input": "+",
"output": "NO"
},
{
"input": "~",
"output": "NO"
},
{
"input": "dEHsbM'gS[\\brZ_dpjXw8f?L[4E\"s4Zc9*(,j:>p$}m7HD[_9nOWQ\\uvq2mHWR",
"output": "YES"
},
{
"input": "tt6l=RHOfStm.;Qd$-}zDes*E,.F7qn5-b%HC",
"output": "YES"
},
{
"input": "@F%K2=%RyL/",
"output": "NO"
},
{
"input": "juq)k(FT.^G=G\\zcqnO\"uJIE1_]KFH9S=1c\"mJ;F9F)%>&.WOdp09+k`Yc6}\"6xw,Aos:M\\_^^:xBb[CcsHm?J",
"output": "YES"
},
{
"input": "6G_\"Fq#<AWyHG=Rci1t%#Jc#x<Fpg'N@t%F=``YO7\\Zd;6PkMe<#91YgzTC)",
"output": "YES"
},
{
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},
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"output": "YES"
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{
"input": "O.&=qt(`z(",
"output": "NO"
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{
"input": "_^r6fyIc/~~;>l%9?aVEi7-{=,[<aMiB'-scSg$$|\"jAzY0N>QkHHGBZj2c\"=fhRlWd5;5K|GgU?7h]!;wl@",
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"output": "NO"
},
{
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"output": "YES"
},
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"output": "YES"
},
{
"input": "e{}2vQ+/r@p0}cLKNe4MCk",
"output": "YES"
},
{
"input": "mzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwyndmhtqvkgkbhtytszotwflegsjzzszfwtzfpnscguemwrczqxycivdqnkH",
"output": "YES"
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"output": "YES"
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"output": "YES"
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"input": "1H1",
"output": "YES"
},
{
"input": "+Q",
"output": "YES"
},
{
"input": "1ab",
"output": "NO"
},
{
"input": "!",
"output": "NO"
},
{
"input": "0+",
"output": "NO"
},
{
"input": "+H",
"output": "YES"
},
{
"input": "cH",
"output": "YES"
},
{
"input": "+8",
"output": "NO"
},
{
"input": "8+",
"output": "NO"
},
{
"input": "++++++++++++++++++++++++++",
"output": "NO"
},
{
"input": "(+)",
"output": "NO"
},
{
"input": "H+",
"output": "YES"
},
{
"input": "a!",
"output": "NO"
},
{
"input": "++++++++++++++",
"output": "NO"
},
{
"input": "+++++++++++++++++++++++++++++++++++++++++++++++++++++++++",
"output": "NO"
},
{
"input": "8",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "HQ9",
"output": "YES"
},
{
"input": "+++H",
"output": "YES"
},
{
"input": "++++++++++",
"output": "NO"
},
{
"input": "HHHHHHH",
"output": "YES"
},
{
"input": "abacabaH",
"output": "YES"
},
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}
] | 1,684,323,727 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 92 | 0 | l = ["H", "Q", "9", "+"]
s = input()
check = False
for i in range(len(s)):
if(s[i] in l):
check = True
break
if(check == False):
print("NO")
else:
print("YES") | Title: HQ9+
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
HQ9+ is a joke programming language which has only four one-character instructions:
- "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input Specification:
The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output Specification:
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Demo Input:
['Hi!\n', 'Codeforces\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first case the program contains only one instruction — "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions. | ```python
l = ["H", "Q", "9", "+"]
s = input()
check = False
for i in range(len(s)):
if(s[i] in l):
check = True
break
if(check == False):
print("NO")
else:
print("YES")
``` | 0 |
|
698 | A | Vacations | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out. | Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days. | [
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | 500 | [
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 3 2 2 3 2 3 3 3 3 3 3 1 2 2 3 1 3 3 2 2 2 3 1 0 3 3 3 2 3 3 1 1 3 1 3 3 3 1 3 1 3 0 1 3 2 3 2 1 1 3 2 3 3 3 2 3 1 3 3 3 3 2 2 2 1 3 1 3 3 3 3 1 3 2 3 3 0 3 3 3 3 3 1 0 2 1 3 3 0 2 3 3",
"output": "16"
},
{
"input": "10\n2 3 0 1 3 1 2 2 1 0",
"output": "3"
},
{
"input": "45\n3 3 2 3 2 3 3 3 0 3 3 3 3 3 3 3 1 3 2 3 2 3 2 2 2 3 2 3 3 3 3 3 1 2 3 3 2 2 2 3 3 3 3 1 3",
"output": "6"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n1 3",
"output": "0"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n0 0",
"output": "2"
},
{
"input": "2\n3 3",
"output": "0"
},
{
"input": "3\n3 3 3",
"output": "0"
},
{
"input": "2\n3 2",
"output": "0"
},
{
"input": "2\n0 2",
"output": "1"
},
{
"input": "10\n2 2 3 3 3 3 2 1 3 2",
"output": "2"
},
{
"input": "15\n0 1 0 0 0 2 0 1 0 0 0 2 0 0 0",
"output": "11"
},
{
"input": "15\n1 3 2 2 2 3 3 3 3 2 3 2 2 1 1",
"output": "4"
},
{
"input": "15\n3 1 3 2 3 2 2 2 3 3 3 3 2 3 2",
"output": "3"
},
{
"input": "20\n0 2 0 1 0 0 0 1 2 0 1 1 1 0 1 1 0 1 1 0",
"output": "12"
},
{
"input": "20\n2 3 2 3 3 3 3 2 0 3 1 1 2 3 0 3 2 3 0 3",
"output": "5"
},
{
"input": "20\n3 3 3 3 2 3 3 2 1 3 3 2 2 2 3 2 2 2 2 2",
"output": "4"
},
{
"input": "25\n0 0 1 0 0 1 0 0 1 0 0 1 0 2 0 0 2 0 0 1 0 2 0 1 1",
"output": "16"
},
{
"input": "25\n1 3 3 2 2 3 3 3 3 3 1 2 2 3 2 0 2 1 0 1 3 2 2 3 3",
"output": "5"
},
{
"input": "25\n2 3 1 3 3 2 1 3 3 3 1 3 3 1 3 2 3 3 1 3 3 3 2 3 3",
"output": "3"
},
{
"input": "30\n0 0 1 0 1 0 1 1 0 0 0 0 0 0 1 0 0 1 1 0 0 2 0 0 1 1 2 0 0 0",
"output": "22"
},
{
"input": "30\n1 1 3 2 2 0 3 2 3 3 1 2 0 1 1 2 3 3 2 3 1 3 2 3 0 2 0 3 3 2",
"output": "9"
},
{
"input": "30\n1 2 3 2 2 3 3 3 3 3 3 3 3 3 3 1 2 2 3 2 3 3 3 2 1 3 3 3 1 3",
"output": "2"
},
{
"input": "35\n0 1 1 0 0 2 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 2 1 0 2 2 1 0 1 0 1 1 1 0 0",
"output": "21"
},
{
"input": "35\n2 2 0 3 2 2 0 3 3 1 1 3 3 1 2 2 0 2 2 2 2 3 1 0 2 1 3 2 2 3 2 3 3 1 2",
"output": "11"
},
{
"input": "35\n1 2 2 3 3 3 3 3 2 2 3 3 2 3 3 2 3 2 3 3 2 2 2 3 3 2 3 3 3 1 3 3 2 2 2",
"output": "7"
},
{
"input": "40\n2 0 1 1 0 0 0 0 2 0 1 1 1 0 0 1 0 0 0 0 0 2 0 0 0 2 1 1 1 3 0 0 0 0 0 0 0 1 1 0",
"output": "28"
},
{
"input": "40\n2 2 3 2 0 2 3 2 1 2 3 0 2 3 2 1 1 3 1 1 0 2 3 1 3 3 1 1 3 3 2 2 1 3 3 3 2 3 3 1",
"output": "10"
},
{
"input": "40\n1 3 2 3 3 2 3 3 2 2 3 1 2 1 2 2 3 1 2 2 1 2 2 2 1 2 2 3 2 3 2 3 2 3 3 3 1 3 2 3",
"output": "8"
},
{
"input": "45\n2 1 0 0 0 2 1 0 1 0 0 2 2 1 1 0 0 2 0 0 0 0 0 0 1 0 0 2 0 0 1 1 0 0 1 0 0 1 1 2 0 0 2 0 2",
"output": "29"
},
{
"input": "45\n3 3 2 3 3 3 2 2 3 2 3 1 3 2 3 2 2 1 1 3 2 3 2 1 3 1 2 3 2 2 0 3 3 2 3 2 3 2 3 2 0 3 1 1 3",
"output": "8"
},
{
"input": "50\n3 0 0 0 2 0 0 0 0 0 0 0 2 1 0 2 0 1 0 1 3 0 2 1 1 0 0 1 1 0 0 1 2 1 1 2 1 1 0 0 0 0 0 0 0 1 2 2 0 0",
"output": "32"
},
{
"input": "50\n3 3 3 3 1 0 3 3 0 2 3 1 1 1 3 2 3 3 3 3 3 1 0 1 2 2 3 3 2 3 0 0 0 2 1 0 1 2 2 2 2 0 2 2 2 1 2 3 3 2",
"output": "16"
},
{
"input": "50\n3 2 3 1 2 1 2 3 3 2 3 3 2 1 3 3 3 3 3 3 2 3 2 3 2 2 3 3 3 2 3 3 3 3 2 3 1 2 3 3 2 3 3 1 2 2 1 1 3 3",
"output": "7"
},
{
"input": "55\n0 0 1 1 0 1 0 0 1 0 1 0 0 0 2 0 0 1 0 0 0 1 0 0 0 0 3 1 0 0 0 1 0 0 0 0 2 0 0 0 2 0 2 1 0 0 0 0 0 0 0 0 2 0 0",
"output": "40"
},
{
"input": "55\n3 0 3 3 3 2 0 2 3 0 3 2 3 3 0 3 3 1 3 3 1 2 3 2 0 3 3 2 1 2 3 2 3 0 3 2 2 1 2 3 2 2 1 3 2 2 3 1 3 2 2 3 3 2 2",
"output": "13"
},
{
"input": "55\n3 3 1 3 2 3 2 3 2 2 3 3 3 3 3 1 1 3 3 2 3 2 3 2 0 1 3 3 3 3 2 3 2 3 1 1 2 2 2 3 3 3 3 3 2 2 2 3 2 3 3 3 3 1 3",
"output": "7"
},
{
"input": "60\n0 1 0 0 0 0 0 0 0 2 1 1 3 0 0 0 0 0 1 0 1 1 0 0 0 3 0 1 0 1 0 2 0 0 0 0 0 1 0 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0",
"output": "44"
},
{
"input": "60\n3 2 1 3 2 2 3 3 3 1 1 3 2 2 3 3 1 3 2 2 3 3 2 2 2 2 0 2 2 3 2 3 0 3 3 3 2 3 3 0 1 3 2 1 3 1 1 2 1 3 1 1 2 2 1 3 3 3 2 2",
"output": "15"
},
{
"input": "60\n3 2 2 3 2 3 2 3 3 2 3 2 3 3 2 3 3 3 3 3 3 2 3 3 1 2 3 3 3 2 1 3 3 1 3 1 3 0 3 3 3 2 3 2 3 2 3 3 1 1 2 3 3 3 3 2 1 3 2 3",
"output": "8"
},
{
"input": "65\n1 0 2 1 1 0 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 1 2 0 2 1 0 2 1 0 1 0 1 1 0 1 1 1 2 1 0 1 0 0 0 0 1 2 2 1 0 0 1 2 1 2 0 2 0 0 0 1 1",
"output": "35"
},
{
"input": "65\n2 2 2 3 0 2 1 2 3 3 1 3 1 2 1 3 2 3 2 2 2 1 2 0 3 1 3 1 1 3 1 3 3 3 3 3 1 3 0 3 1 3 1 2 2 3 2 0 3 1 3 2 1 2 2 2 3 3 2 3 3 3 2 2 3",
"output": "13"
},
{
"input": "65\n3 2 3 3 3 2 3 2 3 3 3 3 3 3 3 3 3 2 3 2 3 2 2 3 3 3 3 3 2 2 2 3 3 2 3 3 2 3 3 3 3 2 3 3 3 2 2 3 3 3 3 3 3 2 2 3 3 2 3 3 1 3 3 3 3",
"output": "6"
},
{
"input": "70\n1 0 0 0 1 0 1 0 0 0 1 1 0 1 0 0 1 1 1 0 1 1 0 0 1 1 1 3 1 1 0 1 2 0 2 1 0 0 0 1 1 1 1 1 0 0 1 0 0 0 1 1 1 3 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1",
"output": "43"
},
{
"input": "70\n2 3 3 3 1 3 3 1 2 1 1 2 2 3 0 2 3 3 1 3 3 2 2 3 3 3 2 2 2 2 1 3 3 0 2 1 1 3 2 3 3 2 2 3 1 3 1 2 3 2 3 3 2 2 2 3 1 1 2 1 3 3 2 2 3 3 3 1 1 1",
"output": "16"
},
{
"input": "70\n3 3 2 2 1 2 1 2 2 2 2 2 3 3 2 3 3 3 3 2 2 2 2 3 3 3 1 3 3 3 2 3 3 3 3 2 3 3 1 3 1 3 2 3 3 2 3 3 3 2 3 2 3 3 1 2 3 3 2 2 2 3 2 3 3 3 3 3 3 1",
"output": "10"
},
{
"input": "75\n1 0 0 1 1 0 0 1 0 1 2 0 0 2 1 1 0 0 0 0 0 0 2 1 1 0 0 0 0 1 0 1 0 1 1 1 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 1 2 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1 1 0 1 0",
"output": "51"
},
{
"input": "75\n1 3 3 3 1 1 3 2 3 3 1 3 3 3 2 1 3 2 2 3 1 1 1 1 1 1 2 3 3 3 3 3 3 2 3 3 3 3 3 2 3 3 2 2 2 1 2 3 3 2 2 3 0 1 1 3 3 0 0 1 1 3 2 3 3 3 3 1 2 2 3 3 3 3 1",
"output": "16"
},
{
"input": "75\n3 3 3 3 2 2 3 2 2 3 2 2 1 2 3 3 2 2 3 3 1 2 2 2 1 3 3 3 1 2 2 3 3 3 2 3 2 2 2 3 3 1 3 2 2 3 3 3 0 3 2 1 3 3 2 3 3 3 3 1 2 3 3 3 2 2 3 3 3 3 2 2 3 3 1",
"output": "11"
},
{
"input": "80\n0 0 0 0 2 0 1 1 1 1 1 0 0 0 0 2 0 0 1 0 0 0 0 1 1 0 2 2 1 1 0 1 0 1 0 1 1 1 0 1 2 1 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 2 2 0 1 1 0 0 0 0 0 0 0 0 1",
"output": "56"
},
{
"input": "80\n2 2 3 3 2 1 0 1 0 3 2 2 3 2 1 3 1 3 3 2 3 3 3 2 3 3 3 2 1 3 3 1 3 3 3 3 3 3 2 2 2 1 3 2 1 3 2 1 1 0 1 1 2 1 3 0 1 2 3 2 2 3 2 3 1 3 3 2 1 1 0 3 3 3 3 1 2 1 2 0",
"output": "17"
},
{
"input": "80\n2 3 3 2 2 2 3 3 2 3 3 3 3 3 2 3 2 3 2 3 3 3 3 3 3 3 3 3 2 3 1 3 2 3 3 0 3 1 2 3 3 1 2 3 2 3 3 2 3 3 3 3 3 2 2 3 0 3 3 3 3 3 2 2 3 2 3 3 3 3 3 2 3 2 3 3 3 3 2 3",
"output": "9"
},
{
"input": "85\n0 1 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 2 0 1 0 0 2 0 1 1 0 0 0 0 2 2 0 0 0 1 0 0 0 1 2 0 1 0 0 0 2 1 1 2 0 3 1 0 2 2 1 0 0 1 1 0 0 0 0 1 0 2 1 1 2 1 0 0 1 2 1 2 0 0 1 0 1 0",
"output": "54"
},
{
"input": "85\n2 3 1 3 2 3 1 3 3 2 1 2 1 2 2 3 2 2 3 2 0 3 3 2 1 2 2 2 3 3 2 3 3 3 2 1 1 3 1 3 2 2 2 3 3 2 3 2 3 1 1 3 2 3 1 3 3 2 3 3 2 2 3 0 1 1 2 2 2 2 1 2 3 1 3 3 1 3 2 2 3 2 3 3 3",
"output": "19"
},
{
"input": "85\n1 2 1 2 3 2 3 3 3 3 3 3 3 2 1 3 2 3 3 3 3 2 3 3 3 1 3 3 3 3 2 3 3 3 3 3 3 2 2 1 3 3 3 3 2 2 3 1 1 2 3 3 3 2 3 3 3 3 3 2 3 3 3 2 2 3 3 1 1 1 3 3 3 3 1 3 3 3 1 3 3 1 3 2 3",
"output": "9"
},
{
"input": "90\n2 0 1 0 0 0 0 0 0 1 1 2 0 0 0 0 0 0 0 2 2 0 2 0 0 2 1 0 2 0 1 0 1 0 0 1 2 2 0 0 1 0 0 1 0 1 0 2 0 1 1 1 0 1 1 0 1 0 2 0 1 0 1 0 0 0 1 0 0 1 2 0 0 0 1 0 0 2 2 0 0 0 0 0 1 3 1 1 0 1",
"output": "57"
},
{
"input": "90\n2 3 3 3 2 3 2 1 3 0 3 2 3 3 2 1 3 3 2 3 2 3 3 2 1 3 1 3 3 1 2 2 3 3 2 1 2 3 2 3 0 3 3 2 2 3 1 0 3 3 1 3 3 3 3 2 1 2 2 1 3 2 1 3 3 1 2 0 2 2 3 2 2 3 3 3 1 3 2 1 2 3 3 2 3 2 3 3 2 1",
"output": "17"
},
{
"input": "90\n2 3 2 3 2 2 3 3 2 3 2 1 2 3 3 3 2 3 2 3 3 2 3 3 3 1 3 3 1 3 2 3 2 2 1 3 3 3 3 3 3 3 3 3 3 2 3 2 3 2 1 3 3 3 3 2 2 3 3 3 3 3 3 3 3 3 3 3 3 2 2 3 3 3 3 1 3 2 3 3 3 2 2 3 2 3 2 1 3 2",
"output": "9"
},
{
"input": "95\n0 0 3 0 2 0 1 0 0 2 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 0 1 0 0 2 1 0 0 1 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 1 2 0 1 2 2 0 0 1 0 2 0 0 0 1 0 2 1 2 1 0 1 0 0 0 1 0 0 1 1 2 1 1 1 1 2 0 0 0 0 0 1 1 0 1",
"output": "61"
},
{
"input": "95\n2 3 3 2 1 1 3 3 3 2 3 3 3 2 3 2 3 3 3 2 3 2 2 3 3 2 1 2 3 3 3 1 3 0 3 3 1 3 3 1 0 1 3 3 3 0 2 1 3 3 3 3 0 1 3 2 3 3 2 1 3 1 2 1 1 2 3 0 3 3 2 1 3 2 1 3 3 3 2 2 3 2 3 3 3 2 1 3 3 3 2 3 3 1 2",
"output": "15"
},
{
"input": "95\n2 3 3 2 3 2 2 1 3 1 2 1 2 3 1 2 3 3 1 3 3 3 1 2 3 2 2 2 2 3 3 3 2 2 3 3 3 3 3 1 2 2 3 3 3 3 2 3 2 2 2 3 3 2 3 3 3 3 3 3 3 0 3 2 0 3 3 1 3 3 3 2 3 2 3 2 3 3 3 3 2 2 1 1 3 3 3 3 3 1 3 3 3 3 2",
"output": "14"
},
{
"input": "100\n1 0 2 0 0 0 0 2 0 0 0 1 0 1 0 0 1 0 1 2 0 1 1 0 0 1 0 1 1 0 0 0 2 0 1 0 0 2 0 0 0 0 0 1 1 1 0 0 1 0 2 0 0 0 0 1 0 1 0 1 0 1 0 1 2 2 0 0 2 0 1 0 1 0 1 0 0 0 1 0 0 2 1 1 1 0 0 1 0 0 0 2 0 0 2 1 1 0 0 2",
"output": "63"
},
{
"input": "100\n3 2 1 3 2 3 2 3 2 2 3 1 3 3 3 3 3 2 2 3 2 2 3 2 3 3 3 2 3 1 2 1 3 3 3 3 1 3 3 3 3 3 2 3 2 1 3 3 1 2 2 3 1 3 3 1 2 2 1 3 1 3 2 2 3 3 1 3 2 3 1 2 1 2 3 3 2 2 1 2 3 3 3 3 3 1 3 3 3 3 2 1 3 0 3 3 3 2 3 3",
"output": "15"
},
{
"input": "100\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2",
"output": "0"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\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": "50"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "50"
},
{
"input": "99\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",
"output": "49"
},
{
"input": "100\n2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1",
"output": "0"
},
{
"input": "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 0 0 0",
"output": "100"
},
{
"input": "2\n0 3",
"output": "1"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "2\n2 0",
"output": "1"
},
{
"input": "2\n2 1",
"output": "0"
},
{
"input": "2\n2 3",
"output": "0"
},
{
"input": "2\n3 0",
"output": "1"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "100\n3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0",
"output": "50"
}
] | 1,686,921,382 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 5 | 46 | 0 | def f(v, prev):
c = 0
for tmp in range(len(v)):
if v[tmp] == 0:
c += 1
prev = 0
elif v[tmp] == 1 and prev != 1:
prev = 1
elif v[tmp] == 2 and prev != 2:
prev = 2
elif v[tmp] == 3 and prev != 0:
prev = 2 if prev == 1 else 1
elif v[tmp] == 3 and prev == 0:
c += min(f(v[tmp + 1:], 1), f(v[tmp + 1:], 2))
return c
else:
c += 1
return c
n = int(input())
v = list(map(int, input().split()))
print(f(v, 0))
| Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | ```python
def f(v, prev):
c = 0
for tmp in range(len(v)):
if v[tmp] == 0:
c += 1
prev = 0
elif v[tmp] == 1 and prev != 1:
prev = 1
elif v[tmp] == 2 and prev != 2:
prev = 2
elif v[tmp] == 3 and prev != 0:
prev = 2 if prev == 1 else 1
elif v[tmp] == 3 and prev == 0:
c += min(f(v[tmp + 1:], 1), f(v[tmp + 1:], 2))
return c
else:
c += 1
return c
n = int(input())
v = list(map(int, input().split()))
print(f(v, 0))
``` | 0 |
|
450 | A | Jzzhu and Children | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order? | The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100). | Output a single integer, representing the number of the last child. | [
"5 2\n1 3 1 4 2\n",
"6 4\n1 1 2 2 3 3\n"
] | [
"4\n",
"6\n"
] | Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home. | 500 | [
{
"input": "5 2\n1 3 1 4 2",
"output": "4"
},
{
"input": "6 4\n1 1 2 2 3 3",
"output": "6"
},
{
"input": "7 3\n6 1 5 4 2 3 1",
"output": "4"
},
{
"input": "10 5\n2 7 3 6 2 5 1 3 4 5",
"output": "4"
},
{
"input": "100 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 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": "100"
},
{
"input": "9 3\n9 5 2 3 7 1 8 4 6",
"output": "7"
},
{
"input": "20 10\n58 4 32 10 73 7 30 39 47 6 59 21 24 66 79 79 46 13 29 58",
"output": "16"
},
{
"input": "50 5\n89 56 3 2 40 37 56 52 83 59 43 83 43 59 29 74 22 58 53 41 53 67 78 30 57 32 58 29 95 46 45 85 60 49 41 82 8 71 52 40 45 26 6 71 84 91 4 93 40 54",
"output": "48"
},
{
"input": "50 1\n4 3 9 7 6 8 3 7 10 9 8 8 10 2 9 3 2 4 4 10 4 6 8 10 9 9 4 2 8 9 4 4 9 5 1 5 2 4 4 9 10 2 5 10 7 2 8 6 8 1",
"output": "44"
},
{
"input": "50 5\n3 9 10 8 3 3 4 6 8 2 9 9 3 1 2 10 6 8 7 2 7 4 2 7 5 10 2 2 2 5 10 5 6 6 8 7 10 4 3 2 10 8 6 6 8 6 4 4 1 3",
"output": "46"
},
{
"input": "50 2\n56 69 72 15 95 92 51 1 74 87 100 29 46 54 18 81 84 72 84 83 20 63 71 27 45 74 50 89 48 8 21 15 47 3 39 73 80 84 6 99 17 25 56 3 74 64 71 39 89 78",
"output": "40"
},
{
"input": "50 3\n31 39 64 16 86 3 1 9 25 54 98 42 20 3 49 41 73 37 55 62 33 77 64 22 33 82 26 13 10 13 7 40 48 18 46 79 94 72 19 12 11 61 16 37 10 49 14 94 48 69",
"output": "11"
},
{
"input": "50 100\n67 67 61 68 42 29 70 77 12 61 71 27 4 73 87 52 59 38 93 90 31 27 87 47 26 57 76 6 28 72 81 68 50 84 69 79 39 93 52 6 88 12 46 13 90 68 71 38 90 95",
"output": "50"
},
{
"input": "100 3\n4 14 20 11 19 11 14 20 5 7 6 12 11 17 5 11 7 6 2 10 13 5 12 8 5 17 20 18 7 19 11 7 7 20 20 8 10 17 17 19 20 5 15 16 19 7 11 16 4 17 2 10 1 20 20 16 19 9 9 11 5 7 12 9 9 6 20 18 13 19 8 4 8 1 2 4 10 11 15 14 1 7 17 12 13 19 12 2 3 14 15 15 5 17 14 12 17 14 16 9",
"output": "86"
},
{
"input": "100 5\n16 8 14 16 12 11 17 19 19 2 8 9 5 6 19 9 11 18 6 9 14 16 14 18 17 17 17 5 15 20 19 7 7 10 10 5 14 20 5 19 11 16 16 19 17 9 7 12 14 10 2 11 14 5 20 8 10 11 19 2 14 14 19 17 5 10 8 8 4 2 1 10 20 12 14 11 7 6 6 15 1 5 9 15 3 17 16 17 5 14 11 9 16 15 1 11 10 6 15 7",
"output": "93"
},
{
"input": "100 1\n58 94 18 50 17 14 96 62 83 80 75 5 9 22 25 41 3 96 74 45 66 37 2 37 13 85 68 54 77 11 85 19 25 21 52 59 90 61 72 89 82 22 10 16 3 68 61 29 55 76 28 85 65 76 27 3 14 10 56 37 86 18 35 38 56 68 23 88 33 38 52 87 55 83 94 34 100 41 83 56 91 77 32 74 97 13 67 31 57 81 53 39 5 88 46 1 79 4 49 42",
"output": "77"
},
{
"input": "100 2\n1 51 76 62 34 93 90 43 57 59 52 78 3 48 11 60 57 48 5 54 28 81 87 23 44 77 67 61 14 73 29 53 21 89 67 41 47 9 63 37 1 71 40 85 4 14 77 40 78 75 89 74 4 70 32 65 81 95 49 90 72 41 76 55 69 83 73 84 85 93 46 6 74 90 62 37 97 7 7 37 83 30 37 88 34 16 11 59 85 19 57 63 85 20 63 97 97 65 61 48",
"output": "97"
},
{
"input": "100 3\n30 83 14 55 61 66 34 98 90 62 89 74 45 93 33 31 75 35 82 100 63 69 48 18 99 2 36 71 14 30 70 76 96 85 97 90 49 36 6 76 37 94 70 3 63 73 75 48 39 29 13 2 46 26 9 56 1 18 54 53 85 34 2 12 1 93 75 67 77 77 14 26 33 25 55 9 57 70 75 6 87 66 18 3 41 69 73 24 49 2 20 72 39 58 91 54 74 56 66 78",
"output": "20"
},
{
"input": "100 4\n69 92 76 3 32 50 15 38 21 22 14 3 67 41 95 12 10 62 83 52 78 1 18 58 94 35 62 71 58 75 13 73 60 34 50 97 50 70 19 96 53 10 100 26 20 39 62 59 88 26 24 83 70 68 66 8 6 38 16 93 2 91 81 89 78 74 21 8 31 56 28 53 77 5 81 5 94 42 77 75 92 15 59 36 61 18 55 45 69 68 81 51 12 42 85 74 98 31 17 41",
"output": "97"
},
{
"input": "100 5\n2 72 10 60 6 50 72 34 97 77 35 43 80 64 40 53 46 6 90 22 29 70 26 68 52 19 72 88 83 18 55 32 99 81 11 21 39 42 41 63 60 97 30 23 55 78 89 35 24 50 99 52 27 76 24 8 20 27 51 37 17 82 69 18 46 19 26 77 52 83 76 65 43 66 84 84 13 30 66 88 84 23 37 1 17 26 11 50 73 56 54 37 40 29 35 8 1 39 50 82",
"output": "51"
},
{
"input": "100 7\n6 73 7 54 92 33 66 65 80 47 2 53 28 59 61 16 54 89 37 48 77 40 49 59 27 52 17 22 78 80 81 80 8 93 50 7 87 57 29 16 89 55 20 7 51 54 30 98 44 96 27 70 1 1 32 61 22 92 84 98 31 89 91 90 28 56 49 25 86 49 55 16 19 1 18 8 88 47 16 18 73 86 2 96 16 91 74 49 38 98 94 25 34 85 29 27 99 31 31 58",
"output": "97"
},
{
"input": "100 9\n36 4 45 16 19 6 10 87 44 82 71 49 70 35 83 19 40 76 45 94 44 96 10 54 82 77 86 63 11 37 21 3 15 89 80 88 89 16 72 23 25 9 51 25 10 45 96 5 6 18 51 31 42 57 41 51 42 15 89 61 45 82 16 48 61 67 19 40 9 33 90 36 78 36 79 79 16 10 83 87 9 22 84 12 23 76 36 14 2 81 56 33 56 23 57 84 76 55 35 88",
"output": "47"
},
{
"input": "100 10\n75 81 39 64 90 58 92 28 75 9 96 78 92 83 77 68 76 71 14 46 58 60 80 25 78 11 13 63 22 82 65 68 47 6 33 63 90 50 85 43 73 94 80 48 67 11 83 17 22 15 94 80 66 99 66 4 46 35 52 1 62 39 96 57 37 47 97 49 64 12 36 63 90 16 4 75 85 82 85 56 13 4 92 45 44 93 17 35 22 46 18 44 29 7 52 4 100 98 87 51",
"output": "98"
},
{
"input": "100 20\n21 19 61 70 54 97 98 14 61 72 25 94 24 56 55 25 12 80 76 11 35 17 80 26 11 94 52 47 84 61 10 2 74 25 10 21 2 79 55 50 30 75 10 64 44 5 60 96 52 16 74 41 20 77 20 44 8 86 74 36 49 61 99 13 54 64 19 99 50 43 12 73 48 48 83 55 72 73 63 81 30 27 95 9 97 82 24 3 89 90 33 14 47 88 22 78 12 75 58 67",
"output": "94"
},
{
"input": "100 30\n56 79 59 23 11 23 67 82 81 80 99 79 8 58 93 36 98 81 46 39 34 67 3 50 4 68 70 71 2 21 52 30 75 23 33 21 16 100 56 43 8 27 40 8 56 24 17 40 94 10 67 49 61 36 95 87 17 41 7 94 33 19 17 50 26 11 94 54 38 46 77 9 53 35 98 42 50 20 43 6 78 6 38 24 100 45 43 16 1 50 16 46 14 91 95 88 10 1 50 19",
"output": "95"
},
{
"input": "100 40\n86 11 97 17 38 95 11 5 13 83 67 75 50 2 46 39 84 68 22 85 70 23 64 46 59 93 39 80 35 78 93 21 83 19 64 1 49 59 99 83 44 81 70 58 15 82 83 47 55 65 91 10 2 92 4 77 37 32 12 57 78 11 42 8 59 21 96 69 61 30 44 29 12 70 91 14 10 83 11 75 14 10 19 39 8 98 5 81 66 66 79 55 36 29 22 45 19 24 55 49",
"output": "88"
},
{
"input": "100 50\n22 39 95 69 94 53 80 73 33 90 40 60 2 4 84 50 70 38 92 12 36 74 87 70 51 36 57 5 54 6 35 81 52 17 55 100 95 81 32 76 21 1 100 1 95 1 40 91 98 59 84 19 11 51 79 19 47 86 45 15 62 2 59 77 31 68 71 92 17 33 10 33 85 57 5 2 88 97 91 99 63 20 63 54 79 93 24 62 46 27 30 87 3 64 95 88 16 50 79 1",
"output": "99"
},
{
"input": "100 70\n61 48 89 17 97 6 93 13 64 50 66 88 24 52 46 99 6 65 93 64 82 37 57 41 47 1 84 5 97 83 79 46 16 35 40 7 64 15 44 96 37 17 30 92 51 67 26 3 14 56 27 68 66 93 36 39 51 6 40 55 79 26 71 54 8 48 18 2 71 12 55 60 29 37 31 97 26 37 25 68 67 70 3 87 100 41 5 82 65 92 24 66 76 48 89 8 40 93 31 95",
"output": "100"
},
{
"input": "100 90\n87 32 30 15 10 52 93 63 84 1 82 41 27 51 75 32 42 94 39 53 70 13 4 22 99 35 44 38 5 23 18 100 61 80 9 12 42 93 9 77 3 7 60 95 66 78 95 42 69 8 1 88 93 66 96 20 76 63 15 36 92 52 2 72 36 57 48 63 29 20 74 88 49 47 81 61 94 74 70 93 47 3 19 52 59 41 5 40 22 3 76 97 91 37 95 88 91 99 76 15",
"output": "98"
},
{
"input": "100 100\n79 75 7 28 6 96 38 35 57 95 41 74 24 96 32 78 81 13 63 84 24 95 3 23 66 1 60 6 96 49 41 5 14 18 31 97 66 19 49 89 49 70 51 28 20 99 18 1 28 77 24 46 69 21 40 32 31 66 28 6 66 97 9 16 70 90 91 30 34 82 93 41 65 11 39 52 1 88 63 43 80 50 60 49 28 56 18 76 24 57 74 1 28 99 36 35 79 54 18 16",
"output": "100"
},
{
"input": "1 3\n5",
"output": "1"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 3\n4 2",
"output": "1"
},
{
"input": "2 5\n99 97",
"output": "2"
},
{
"input": "3 4\n7 5 2",
"output": "2"
},
{
"input": "3 50\n47 86 51",
"output": "3"
},
{
"input": "5 100\n82 100 85 1 37",
"output": "5"
},
{
"input": "5 20\n40 39 21 5 20",
"output": "3"
},
{
"input": "1 27\n81",
"output": "1"
},
{
"input": "20 13\n7 8 29 83 74 28 93 85 7 8 3 9 8 70 49 50 39 41 57 1",
"output": "7"
},
{
"input": "2 1\n100 2",
"output": "1"
},
{
"input": "2 2\n6 4",
"output": "1"
},
{
"input": "5 2\n6 4 4 1 1",
"output": "1"
},
{
"input": "1 4\n3",
"output": "1"
},
{
"input": "3 2\n1 5 3",
"output": "2"
},
{
"input": "3 1\n3 2 2",
"output": "1"
},
{
"input": "3 1\n2 3 2",
"output": "2"
},
{
"input": "5 1\n5 1 1 1 1",
"output": "1"
},
{
"input": "2 3\n7 4",
"output": "1"
}
] | 1,639,861,157 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 78 | 0 | import sys
def main() :
n,m = map(int,input().split())
array = list(map(int,input().strip().split()))
max_ = 0
ind = -1
for i in range(len(array)):
a = array[i]//m
if a >= max_ :
max_ = a
ind = i+1
print(ind)
main() | Title: Jzzhu and Children
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output a single integer, representing the number of the last child.
Demo Input:
['5 2\n1 3 1 4 2\n', '6 4\n1 1 2 2 3 3\n']
Demo Output:
['4\n', '6\n']
Note:
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home. | ```python
import sys
def main() :
n,m = map(int,input().split())
array = list(map(int,input().strip().split()))
max_ = 0
ind = -1
for i in range(len(array)):
a = array[i]//m
if a >= max_ :
max_ = a
ind = i+1
print(ind)
main()
``` | 0 |
|
234 | A | Lefthanders and Righthanders | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | One fine October day a mathematics teacher Vasily Petrov went to a class and saw there *n* pupils who sat at the desks, two people at each desk. Vasily quickly realized that number *n* is even. Like all true mathematicians, Vasily has all students numbered from 1 to *n*.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists. | The first input line contains a single even integer *n* (4<=≤<=*n*<=≤<=100) — the number of students in the class. The second line contains exactly *n* capital English letters "L" and "R". If the *i*-th letter at the second line equals "L", then the student number *i* is a lefthander, otherwise he is a righthander. | Print integer pairs, one pair per line. In the *i*-th line print the numbers of students that will sit at the *i*-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them. | [
"6\nLLRLLL\n",
"4\nRRLL\n"
] | [
"1 4\n2 5\n6 3\n",
"3 1\n4 2\n"
] | none | 0 | [
{
"input": "6\nLLRLLL",
"output": "1 4\n2 5\n6 3"
},
{
"input": "4\nRRLL",
"output": "3 1\n4 2"
},
{
"input": "4\nLLRR",
"output": "1 3\n2 4"
},
{
"input": "6\nRLLRRL",
"output": "1 4\n2 5\n3 6"
},
{
"input": "8\nLRLRLLLR",
"output": "1 5\n6 2\n3 7\n4 8"
},
{
"input": "10\nRLLRLRRRLL",
"output": "1 6\n2 7\n3 8\n9 4\n5 10"
},
{
"input": "12\nLRRRRRLRRRRL",
"output": "1 7\n2 8\n3 9\n4 10\n5 11\n12 6"
},
{
"input": "14\nRLLRLLLLRLLLRL",
"output": "8 1\n2 9\n3 10\n11 4\n5 12\n6 13\n7 14"
},
{
"input": "16\nLLLRRRLRRLLRRLLL",
"output": "1 9\n2 10\n3 11\n4 12\n5 13\n14 6\n7 15\n16 8"
},
{
"input": "18\nRRRLLLLRRRLRLRLLRL",
"output": "1 10\n11 2\n3 12\n4 13\n5 14\n6 15\n7 16\n8 17\n18 9"
},
{
"input": "20\nRLRLLRLRRLLRRRRRRLRL",
"output": "11 1\n2 12\n3 13\n4 14\n5 15\n6 16\n7 17\n18 8\n9 19\n10 20"
},
{
"input": "22\nRLLLRLLLRRLRRRLRLLLLLL",
"output": "1 12\n2 13\n3 14\n4 15\n5 16\n6 17\n7 18\n8 19\n20 9\n21 10\n11 22"
},
{
"input": "24\nLRRRLRLLRLRRRRLLLLRRLRLR",
"output": "1 13\n2 14\n15 3\n16 4\n5 17\n18 6\n7 19\n8 20\n21 9\n10 22\n23 11\n12 24"
},
{
"input": "26\nRLRRLLRRLLRLRRLLRLLRRLRLRR",
"output": "1 14\n2 15\n16 3\n4 17\n5 18\n6 19\n7 20\n8 21\n9 22\n10 23\n24 11\n12 25\n13 26"
},
{
"input": "28\nLLLRRRRRLRRLRRRLRLRLRRLRLRRL",
"output": "1 15\n2 16\n3 17\n18 4\n5 19\n20 6\n7 21\n8 22\n9 23\n10 24\n25 11\n12 26\n13 27\n28 14"
},
{
"input": "30\nLRLLRLRRLLRLRLLRRRRRLRLRLRLLLL",
"output": "1 16\n2 17\n3 18\n4 19\n5 20\n6 21\n7 22\n23 8\n9 24\n10 25\n11 26\n12 27\n28 13\n14 29\n15 30"
},
{
"input": "32\nRLRLLRRLLRRLRLLRLRLRLLRLRRRLLRRR",
"output": "17 1\n2 18\n19 3\n4 20\n5 21\n22 6\n7 23\n8 24\n9 25\n10 26\n11 27\n12 28\n29 13\n14 30\n15 31\n16 32"
},
{
"input": "34\nLRRLRLRLLRRRRLLRLRRLRRLRLRRLRRRLLR",
"output": "1 18\n2 19\n20 3\n4 21\n5 22\n6 23\n7 24\n8 25\n9 26\n10 27\n28 11\n12 29\n13 30\n14 31\n15 32\n33 16\n17 34"
},
{
"input": "36\nRRLLLRRRLLLRRLLLRRLLRLLRLRLLRLRLRLLL",
"output": "19 1\n20 2\n3 21\n4 22\n5 23\n6 24\n25 7\n8 26\n9 27\n10 28\n11 29\n30 12\n13 31\n14 32\n15 33\n16 34\n35 17\n36 18"
},
{
"input": "38\nLLRRRLLRRRLRRLRLRRLRRLRLRLLRRRRLLLLRLL",
"output": "1 20\n2 21\n22 3\n4 23\n24 5\n6 25\n7 26\n27 8\n9 28\n10 29\n11 30\n12 31\n32 13\n14 33\n34 15\n16 35\n17 36\n37 18\n19 38"
},
{
"input": "40\nLRRRRRLRLLRRRLLRRLRLLRLRRLRRLLLRRLRRRLLL",
"output": "1 21\n2 22\n23 3\n4 24\n5 25\n26 6\n7 27\n8 28\n9 29\n10 30\n31 11\n12 32\n13 33\n14 34\n15 35\n16 36\n17 37\n18 38\n39 19\n20 40"
},
{
"input": "42\nRLRRLLLLLLLRRRLRLLLRRRLRLLLRLRLRLLLRLRLRRR",
"output": "1 22\n2 23\n3 24\n25 4\n5 26\n6 27\n7 28\n8 29\n9 30\n10 31\n11 32\n33 12\n34 13\n35 14\n15 36\n37 16\n17 38\n18 39\n19 40\n20 41\n21 42"
},
{
"input": "44\nLLLLRRLLRRLLRRLRLLRRRLRLRLLRLRLRRLLRLRRLLLRR",
"output": "1 23\n2 24\n3 25\n4 26\n27 5\n6 28\n7 29\n8 30\n31 9\n10 32\n11 33\n12 34\n35 13\n14 36\n15 37\n16 38\n17 39\n18 40\n41 19\n42 20\n21 43\n22 44"
},
{
"input": "46\nRRRLLLLRRLRLRRRRRLRLLRLRRLRLLLLLLLLRRLRLRLRLLL",
"output": "1 24\n2 25\n26 3\n4 27\n5 28\n6 29\n7 30\n31 8\n32 9\n10 33\n34 11\n12 35\n13 36\n14 37\n38 15\n16 39\n40 17\n18 41\n42 19\n20 43\n21 44\n45 22\n23 46"
},
{
"input": "48\nLLLLRRLRRRRLRRRLRLLLLLRRLLRLLRLLRRLRRLLRLRLRRRRL",
"output": "1 25\n2 26\n3 27\n4 28\n29 5\n6 30\n7 31\n32 8\n9 33\n10 34\n35 11\n12 36\n13 37\n38 14\n39 15\n16 40\n41 17\n18 42\n19 43\n20 44\n21 45\n22 46\n23 47\n48 24"
},
{
"input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 26\n2 27\n3 28\n4 29\n5 30\n6 31\n7 32\n8 33\n9 34\n10 35\n11 36\n12 37\n13 38\n14 39\n15 40\n16 41\n17 42\n18 43\n19 44\n20 45\n21 46\n22 47\n23 48\n24 49\n25 50"
},
{
"input": "52\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 27\n2 28\n3 29\n4 30\n5 31\n6 32\n7 33\n8 34\n9 35\n10 36\n11 37\n12 38\n13 39\n14 40\n15 41\n16 42\n17 43\n18 44\n19 45\n20 46\n21 47\n22 48\n23 49\n24 50\n25 51\n26 52"
},
{
"input": "54\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 28\n2 29\n3 30\n4 31\n5 32\n6 33\n7 34\n8 35\n9 36\n10 37\n11 38\n12 39\n13 40\n14 41\n15 42\n16 43\n17 44\n18 45\n19 46\n20 47\n21 48\n22 49\n23 50\n24 51\n25 52\n26 53\n27 54"
},
{
"input": "56\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 29\n2 30\n3 31\n4 32\n5 33\n6 34\n7 35\n8 36\n9 37\n10 38\n11 39\n12 40\n13 41\n14 42\n15 43\n16 44\n17 45\n18 46\n19 47\n20 48\n21 49\n22 50\n23 51\n24 52\n25 53\n26 54\n27 55\n28 56"
},
{
"input": "58\nRRRLLLRLLLLRRLRRRLLRLLRLRLLRLRRRRLLLLLLRLRRLRLRRRLRLRRLRRL",
"output": "1 30\n2 31\n3 32\n4 33\n5 34\n6 35\n36 7\n8 37\n9 38\n10 39\n11 40\n41 12\n13 42\n14 43\n44 15\n16 45\n46 17\n18 47\n19 48\n20 49\n21 50\n22 51\n52 23\n24 53\n25 54\n26 55\n27 56\n28 57\n29 58"
},
{
"input": "60\nRLLLLRRLLRRRLLLLRRRRRLRRRLRRRLLLRLLLRLRRRLRLLLRLLRRLLRRRRRLL",
"output": "31 1\n2 32\n3 33\n4 34\n5 35\n36 6\n7 37\n8 38\n9 39\n10 40\n11 41\n42 12\n13 43\n14 44\n15 45\n16 46\n17 47\n48 18\n49 19\n20 50\n21 51\n22 52\n53 23\n24 54\n25 55\n26 56\n27 57\n28 58\n59 29\n30 60"
},
{
"input": "62\nLRRLRLRLLLLRRLLLLRRRLRLLLLRRRLLLLLLRRRLLLLRRLRRLRLLLLLLLLRRLRR",
"output": "1 32\n33 2\n34 3\n4 35\n5 36\n6 37\n7 38\n8 39\n9 40\n10 41\n11 42\n12 43\n13 44\n14 45\n15 46\n16 47\n17 48\n18 49\n50 19\n51 20\n21 52\n53 22\n23 54\n24 55\n25 56\n26 57\n27 58\n28 59\n60 29\n30 61\n31 62"
},
{
"input": "64\nRLLLLRRRLRLLRRRRLRLLLRRRLLLRRRLLRLLRLRLRRRLLRRRRLRLRRRLLLLRRLLLL",
"output": "1 33\n2 34\n3 35\n4 36\n5 37\n6 38\n39 7\n8 40\n9 41\n10 42\n11 43\n12 44\n13 45\n14 46\n15 47\n16 48\n17 49\n18 50\n19 51\n20 52\n21 53\n22 54\n55 23\n56 24\n25 57\n26 58\n27 59\n28 60\n61 29\n62 30\n31 63\n32 64"
},
{
"input": "66\nLLRRRLLRLRLLRRRRRRRLLLLRRLLLLLLRLLLRLLLLLLRRRLRRLLRRRRRLRLLRLLLLRR",
"output": "1 34\n2 35\n3 36\n37 4\n38 5\n6 39\n7 40\n41 8\n9 42\n10 43\n11 44\n12 45\n46 13\n14 47\n15 48\n49 16\n50 17\n18 51\n19 52\n20 53\n21 54\n22 55\n23 56\n24 57\n58 25\n26 59\n27 60\n28 61\n29 62\n30 63\n31 64\n32 65\n33 66"
},
{
"input": "68\nRRLRLRLLRLRLRRRRRRLRRRLLLLRLLRLRLRLRRRRLRLRLLRRRRLRRLLRLRRLLRLRRLRRL",
"output": "35 1\n2 36\n3 37\n4 38\n5 39\n40 6\n7 41\n8 42\n9 43\n10 44\n45 11\n12 46\n13 47\n14 48\n15 49\n50 16\n17 51\n18 52\n19 53\n54 20\n21 55\n56 22\n23 57\n24 58\n25 59\n26 60\n27 61\n28 62\n29 63\n30 64\n31 65\n32 66\n33 67\n68 34"
},
{
"input": "70\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 36\n2 37\n3 38\n4 39\n5 40\n6 41\n7 42\n8 43\n9 44\n10 45\n11 46\n12 47\n13 48\n14 49\n15 50\n16 51\n17 52\n18 53\n19 54\n20 55\n21 56\n22 57\n23 58\n24 59\n25 60\n26 61\n27 62\n28 63\n29 64\n30 65\n31 66\n32 67\n33 68\n34 69\n35 70"
},
{
"input": "72\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 37\n2 38\n3 39\n4 40\n5 41\n6 42\n7 43\n8 44\n9 45\n10 46\n11 47\n12 48\n13 49\n14 50\n15 51\n16 52\n17 53\n18 54\n19 55\n20 56\n21 57\n22 58\n23 59\n24 60\n25 61\n26 62\n27 63\n28 64\n29 65\n30 66\n31 67\n32 68\n33 69\n34 70\n35 71\n36 72"
},
{
"input": "74\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 38\n2 39\n3 40\n4 41\n5 42\n6 43\n7 44\n8 45\n9 46\n10 47\n11 48\n12 49\n13 50\n14 51\n15 52\n16 53\n17 54\n18 55\n19 56\n20 57\n21 58\n22 59\n23 60\n24 61\n25 62\n26 63\n27 64\n28 65\n29 66\n30 67\n31 68\n32 69\n33 70\n34 71\n35 72\n36 73\n37 74"
},
{
"input": "76\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 39\n2 40\n3 41\n4 42\n5 43\n6 44\n7 45\n8 46\n9 47\n10 48\n11 49\n12 50\n13 51\n14 52\n15 53\n16 54\n17 55\n18 56\n19 57\n20 58\n21 59\n22 60\n23 61\n24 62\n25 63\n26 64\n27 65\n28 66\n29 67\n30 68\n31 69\n32 70\n33 71\n34 72\n35 73\n36 74\n37 75\n38 76"
},
{
"input": "78\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "1 40\n2 41\n3 42\n4 43\n5 44\n6 45\n7 46\n8 47\n9 48\n10 49\n11 50\n12 51\n13 52\n14 53\n15 54\n16 55\n17 56\n18 57\n19 58\n20 59\n21 60\n22 61\n23 62\n24 63\n25 64\n26 65\n27 66\n28 67\n29 68\n30 69\n31 70\n32 71\n33 72\n34 73\n35 74\n36 75\n37 76\n38 77\n39 78"
},
{
"input": "80\nLRLRRRRLRRRRLLLLRLLRLRLLRRLRLLLRRLLLLRLLLRLRLLRRRLRRRLRLRRRRRLRLLRLLRRLLLRLRRRLL",
"output": "1 41\n2 42\n3 43\n4 44\n45 5\n46 6\n7 47\n8 48\n9 49\n50 10\n11 51\n12 52\n13 53\n14 54\n15 55\n16 56\n17 57\n18 58\n19 59\n20 60\n21 61\n62 22\n23 63\n24 64\n65 25\n26 66\n27 67\n68 28\n29 69\n30 70\n31 71\n72 32\n73 33\n34 74\n35 75\n36 76\n37 77\n38 78\n39 79\n40 80"
},
{
"input": "82\nRLRRLLRLRLRLLLRLLLRRLLRRLRRRRLLRLLLLRRRRRLLLRRRLLLLRLRRLRRRLRLLLLRRRLRLRLLLRLLLLLR",
"output": "42 1\n2 43\n44 3\n4 45\n5 46\n6 47\n48 7\n8 49\n50 9\n10 51\n11 52\n12 53\n13 54\n14 55\n56 15\n16 57\n17 58\n18 59\n60 19\n20 61\n21 62\n22 63\n64 23\n65 24\n25 66\n26 67\n27 68\n69 28\n29 70\n30 71\n31 72\n73 32\n33 74\n34 75\n35 76\n36 77\n78 37\n79 38\n80 39\n81 40\n41 82"
},
{
"input": "84\nLRLRRRRRRLLLRLRLLLLLRRLRLRLRRRLLRLLLRLRLLLRRRLRLRRLRLRLLLLLLLLRRRRRRLLLRRLRLRLLLRLRR",
"output": "1 43\n2 44\n3 45\n46 4\n5 47\n48 6\n7 49\n8 50\n51 9\n10 52\n11 53\n12 54\n55 13\n14 56\n57 15\n16 58\n17 59\n18 60\n19 61\n20 62\n21 63\n22 64\n23 65\n24 66\n25 67\n26 68\n27 69\n70 28\n71 29\n30 72\n31 73\n32 74\n33 75\n34 76\n35 77\n36 78\n79 37\n38 80\n39 81\n40 82\n41 83\n42 84"
},
{
"input": "86\nRRRLLLRLLRLLRLRLRLLLRLRLRRLLRLLLRLLLLLLRRRLRLLRLLLRRRLRLLLLRLLRLRRLLRLLLRRRLLRLRLLRLLR",
"output": "1 44\n45 2\n46 3\n4 47\n5 48\n6 49\n50 7\n8 51\n9 52\n10 53\n11 54\n12 55\n56 13\n14 57\n58 15\n16 59\n17 60\n18 61\n19 62\n20 63\n64 21\n22 65\n23 66\n24 67\n68 25\n26 69\n27 70\n28 71\n72 29\n30 73\n31 74\n32 75\n76 33\n34 77\n35 78\n36 79\n37 80\n38 81\n39 82\n40 83\n84 41\n85 42\n43 86"
},
{
"input": "88\nLLRLRLRLLLLRRRRRRLRRLLLLLRRLRRLLLLLRLRLRLLLLLRLRLRRLRLRRLRLLRRLRLLLRLLLLRRLLRRLRLRLRRLRR",
"output": "1 45\n2 46\n47 3\n4 48\n49 5\n6 50\n7 51\n8 52\n9 53\n10 54\n11 55\n12 56\n57 13\n14 58\n59 15\n60 16\n17 61\n18 62\n63 19\n20 64\n21 65\n22 66\n23 67\n24 68\n25 69\n70 26\n71 27\n28 72\n29 73\n30 74\n31 75\n32 76\n33 77\n34 78\n35 79\n36 80\n37 81\n38 82\n39 83\n40 84\n41 85\n42 86\n43 87\n44 88"
},
{
"input": "90\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 46\n2 47\n3 48\n4 49\n5 50\n6 51\n7 52\n8 53\n9 54\n10 55\n11 56\n12 57\n13 58\n14 59\n15 60\n16 61\n17 62\n18 63\n19 64\n20 65\n21 66\n22 67\n23 68\n24 69\n25 70\n26 71\n27 72\n28 73\n29 74\n30 75\n31 76\n32 77\n33 78\n34 79\n35 80\n36 81\n37 82\n38 83\n39 84\n40 85\n41 86\n42 87\n43 88\n44 89\n45 90"
},
{
"input": "92\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 47\n2 48\n3 49\n4 50\n5 51\n6 52\n7 53\n8 54\n9 55\n10 56\n11 57\n12 58\n13 59\n14 60\n15 61\n16 62\n17 63\n18 64\n19 65\n20 66\n21 67\n22 68\n23 69\n24 70\n25 71\n26 72\n27 73\n28 74\n29 75\n30 76\n31 77\n32 78\n33 79\n34 80\n35 81\n36 82\n37 83\n38 84\n39 85\n40 86\n41 87\n42 88\n43 89\n44 90\n45 91\n46 92"
},
{
"input": "94\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 48\n2 49\n3 50\n4 51\n5 52\n6 53\n7 54\n8 55\n9 56\n10 57\n11 58\n12 59\n13 60\n14 61\n15 62\n16 63\n17 64\n18 65\n19 66\n20 67\n21 68\n22 69\n23 70\n24 71\n25 72\n26 73\n27 74\n28 75\n29 76\n30 77\n31 78\n32 79\n33 80\n34 81\n35 82\n36 83\n37 84\n38 85\n39 86\n40 87\n41 88\n42 89\n43 90\n44 91\n45 92\n46 93\n47 94"
},
{
"input": "96\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 49\n2 50\n3 51\n4 52\n5 53\n6 54\n7 55\n8 56\n9 57\n10 58\n11 59\n12 60\n13 61\n14 62\n15 63\n16 64\n17 65\n18 66\n19 67\n20 68\n21 69\n22 70\n23 71\n24 72\n25 73\n26 74\n27 75\n28 76\n29 77\n30 78\n31 79\n32 80\n33 81\n34 82\n35 83\n36 84\n37 85\n38 86\n39 87\n40 88\n41 89\n42 90\n43 91\n44 92\n45 93\n46 94\n47 95\n48 96"
},
{
"input": "98\nLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL",
"output": "1 50\n2 51\n3 52\n4 53\n5 54\n6 55\n7 56\n8 57\n9 58\n10 59\n11 60\n12 61\n13 62\n14 63\n15 64\n16 65\n17 66\n18 67\n19 68\n20 69\n21 70\n22 71\n23 72\n24 73\n25 74\n26 75\n27 76\n28 77\n29 78\n30 79\n31 80\n32 81\n33 82\n34 83\n35 84\n36 85\n37 86\n38 87\n39 88\n40 89\n41 90\n42 91\n43 92\n44 93\n45 94\n46 95\n47 96\n48 97\n49 98"
},
{
"input": "100\nRLRRRRLLLLRRRRLRRRRRRRRLRLRRLLRRRRRRRRLRRRRLLLLRRRRLRRLRLRRRLLRRLRRLLLRLRRLLLLLLRLRLRLRRLRLRLRRRLLLR",
"output": "1 51\n2 52\n3 53\n4 54\n55 5\n6 56\n7 57\n8 58\n9 59\n10 60\n61 11\n62 12\n13 63\n14 64\n15 65\n16 66\n17 67\n68 18\n69 19\n70 20\n21 71\n72 22\n23 73\n24 74\n75 25\n26 76\n77 27\n78 28\n29 79\n30 80\n31 81\n82 32\n33 83\n84 34\n35 85\n86 36\n37 87\n38 88\n39 89\n40 90\n91 41\n42 92\n93 43\n44 94\n45 95\n46 96\n47 97\n98 48\n99 49\n50 100"
},
{
"input": "100\nLRLLLLRLLLLRRRRRLRRRRLRRLRRLRLLRRLRRRRLLRRRLLLRLLLRRRRLLRLRLRRLRLLRRLLRRLRRLRRRRRLRRLRLRLRLLLLLLLLRL",
"output": "1 51\n2 52\n3 53\n4 54\n5 55\n6 56\n7 57\n8 58\n9 59\n10 60\n11 61\n12 62\n63 13\n14 64\n65 15\n66 16\n17 67\n18 68\n69 19\n70 20\n21 71\n22 72\n73 23\n24 74\n25 75\n76 26\n27 77\n28 78\n29 79\n30 80\n31 81\n82 32\n33 83\n34 84\n85 35\n36 86\n87 37\n38 88\n39 89\n40 90\n91 41\n92 42\n93 43\n44 94\n45 95\n46 96\n97 47\n48 98\n49 99\n50 100"
},
{
"input": "100\nLLLRRLLRLRLLLRLLLRLRLLRRRLRRLLLRLRLRRLLRLRRRLLLRRLLRLLRRLLRRRRRLRLRRLRLRRLRLRRLLRLRLLRLLLRLLRLLLLRLL",
"output": "1 51\n2 52\n3 53\n54 4\n5 55\n6 56\n7 57\n58 8\n9 59\n10 60\n11 61\n12 62\n13 63\n64 14\n15 65\n16 66\n17 67\n18 68\n19 69\n20 70\n21 71\n22 72\n23 73\n74 24\n25 75\n26 76\n27 77\n28 78\n29 79\n30 80\n31 81\n82 32\n33 83\n84 34\n35 85\n36 86\n87 37\n38 88\n39 89\n40 90\n41 91\n92 42\n43 93\n94 44\n45 95\n46 96\n47 97\n48 98\n99 49\n50 100"
},
{
"input": "100\nRLLLLRRLLLLRRRRLLRLRRRLLLRLLRLLLLLRRLLLLLLRRLRRRRRLRLLRLRRRLLLRLRLRLLLRRRLLLLLRRRRRLRRLLLLRLLLRRLLLL",
"output": "51 1\n2 52\n3 53\n4 54\n5 55\n56 6\n7 57\n8 58\n9 59\n10 60\n11 61\n62 12\n13 63\n64 14\n15 65\n16 66\n17 67\n68 18\n19 69\n70 20\n21 71\n22 72\n23 73\n24 74\n25 75\n76 26\n27 77\n28 78\n29 79\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n40 90\n41 91\n42 92\n93 43\n94 44\n45 95\n46 96\n97 47\n98 48\n99 49\n100 50"
},
{
"input": "100\nRLRRLRLRRLRLLRLLRRRLRRLLLLLRLRLRRRRRRRLLRRRLLRLRLLLRRRLLRRRLLRLRLLLLRRLRLLRLLRLLLLRRLRLRRLRLLLLRLRRR",
"output": "51 1\n2 52\n3 53\n4 54\n5 55\n56 6\n7 57\n8 58\n9 59\n10 60\n61 11\n12 62\n13 63\n14 64\n15 65\n16 66\n67 17\n68 18\n19 69\n20 70\n71 21\n22 72\n23 73\n24 74\n25 75\n26 76\n27 77\n28 78\n29 79\n80 30\n31 81\n82 32\n33 83\n34 84\n85 35\n36 86\n87 37\n38 88\n39 89\n40 90\n41 91\n92 42\n93 43\n44 94\n45 95\n46 96\n47 97\n48 98\n49 99\n50 100"
},
{
"input": "100\nLRRLRLRRRRRRLRRLRRLLLLLLRRLLRRLLRLLLLLLRRRLLRLRRRLLRLLRRLRRRLLRLRLLRRLRRRLLLRRRRLLRRRLLLRRRRRLLLLLLR",
"output": "1 51\n2 52\n53 3\n4 54\n5 55\n6 56\n57 7\n8 58\n9 59\n10 60\n61 11\n62 12\n13 63\n64 14\n15 65\n16 66\n67 17\n18 68\n19 69\n20 70\n21 71\n22 72\n23 73\n24 74\n75 25\n76 26\n27 77\n28 78\n29 79\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n40 90\n41 91\n42 92\n43 93\n44 94\n95 45\n46 96\n97 47\n98 48\n99 49\n50 100"
},
{
"input": "100\nRRLRRLRLRLRRRRLLRRLLRLRRLLRRRLLRLRRLRLRRLLLRRLLRRRRRRLLLRRRLLRRLLLLLLRLLLLLLRLLLRRRLRLLRRRRRLLRLLRRR",
"output": "1 51\n2 52\n3 53\n54 4\n55 5\n6 56\n7 57\n8 58\n9 59\n10 60\n61 11\n12 62\n13 63\n64 14\n15 65\n16 66\n67 17\n68 18\n19 69\n20 70\n71 21\n22 72\n73 23\n74 24\n25 75\n26 76\n27 77\n78 28\n79 29\n30 80\n31 81\n32 82\n33 83\n84 34\n35 85\n36 86\n87 37\n38 88\n39 89\n40 90\n41 91\n42 92\n43 93\n94 44\n45 95\n46 96\n47 97\n48 98\n49 99\n50 100"
},
{
"input": "100\nRRLLLRLRRLRLLRRLRRRLLRRRLRRLLLLLLLLLRRRLLRLRRLRRLRRLRRLRLLLLRLLRRRLLLLRLRRRLLRRRRLRRLLRRRRLRRRLRLLLR",
"output": "1 51\n52 2\n3 53\n4 54\n5 55\n6 56\n7 57\n58 8\n59 9\n10 60\n11 61\n12 62\n13 63\n14 64\n15 65\n16 66\n67 17\n68 18\n69 19\n20 70\n21 71\n72 22\n23 73\n24 74\n25 75\n76 26\n77 27\n28 78\n29 79\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n40 90\n41 91\n42 92\n43 93\n44 94\n95 45\n46 96\n97 47\n98 48\n49 99\n50 100"
},
{
"input": "100\nLLLLLRRLRRRRRRRLLRRRRRLRRLRLRLLRLRRLLLRRRRLLRRLRLLRLLLRLRLLRRRRRRRRRLRLLLRLRLLLLLRLRRRRLRLLRLRLRLRRL",
"output": "1 51\n2 52\n3 53\n4 54\n5 55\n56 6\n7 57\n8 58\n59 9\n10 60\n11 61\n12 62\n13 63\n14 64\n15 65\n16 66\n17 67\n18 68\n69 19\n20 70\n71 21\n72 22\n23 73\n24 74\n75 25\n26 76\n77 27\n28 78\n79 29\n30 80\n31 81\n32 82\n33 83\n34 84\n35 85\n36 86\n37 87\n38 88\n39 89\n90 40\n91 41\n42 92\n43 93\n44 94\n95 45\n46 96\n47 97\n48 98\n49 99\n50 100"
},
{
"input": "100\nLLRRRLLLRLLLLRLRLRLRRRLLLRRRLRLLRLLLRRRRRLRRLRRLRRRLRRLRRLLLRLRLLRRRRLRLRRRRRLRRLRLLRRRRLLLRRRRRLLLL",
"output": "1 51\n2 52\n3 53\n4 54\n55 5\n6 56\n7 57\n8 58\n59 9\n10 60\n11 61\n12 62\n13 63\n64 14\n15 65\n16 66\n17 67\n18 68\n19 69\n70 20\n21 71\n72 22\n23 73\n24 74\n25 75\n26 76\n27 77\n78 28\n29 79\n30 80\n31 81\n32 82\n83 33\n34 84\n35 85\n36 86\n37 87\n38 88\n89 39\n90 40\n91 41\n42 92\n43 93\n44 94\n45 95\n46 96\n97 47\n48 98\n99 49\n100 50"
}
] | 1,675,268,095 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | n = int(input())
x = list(input())
output = ""
for i in range(n//2):
p1 = x[i]
p2 = x[i+(n//2)]
if p1 == p2 or p1 == 'L':
output += f"{i+1} {i+(n//2)+1}\n"
else:
output += f"{i+(n//2)+1} {i+1}\n"
print(output)
#DHER BARA
| Title: Lefthanders and Righthanders
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One fine October day a mathematics teacher Vasily Petrov went to a class and saw there *n* pupils who sat at the desks, two people at each desk. Vasily quickly realized that number *n* is even. Like all true mathematicians, Vasily has all students numbered from 1 to *n*.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists.
Input Specification:
The first input line contains a single even integer *n* (4<=≤<=*n*<=≤<=100) — the number of students in the class. The second line contains exactly *n* capital English letters "L" and "R". If the *i*-th letter at the second line equals "L", then the student number *i* is a lefthander, otherwise he is a righthander.
Output Specification:
Print integer pairs, one pair per line. In the *i*-th line print the numbers of students that will sit at the *i*-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them.
Demo Input:
['6\nLLRLLL\n', '4\nRRLL\n']
Demo Output:
['1 4\n2 5\n6 3\n', '3 1\n4 2\n']
Note:
none | ```python
n = int(input())
x = list(input())
output = ""
for i in range(n//2):
p1 = x[i]
p2 = x[i+(n//2)]
if p1 == p2 or p1 == 'L':
output += f"{i+1} {i+(n//2)+1}\n"
else:
output += f"{i+(n//2)+1} {i+1}\n"
print(output)
#DHER BARA
``` | -1 |
|
854 | A | Fraction | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"math"
] | null | null | Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem. | In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction. | Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum. | [
"3\n",
"4\n",
"12\n"
] | [
"1 2\n",
"1 3\n",
"5 7\n"
] | none | 500 | [
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"output": "11 13"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "10",
"output": "3 7"
},
{
"input": "69",
"output": "34 35"
},
{
"input": "100",
"output": "49 51"
},
{
"input": "57",
"output": "28 29"
},
{
"input": "1000",
"output": "499 501"
},
{
"input": "999",
"output": "499 500"
},
{
"input": "998",
"output": "497 501"
},
{
"input": "997",
"output": "498 499"
},
{
"input": "996",
"output": "497 499"
},
{
"input": "995",
"output": "497 498"
},
{
"input": "994",
"output": "495 499"
},
{
"input": "5",
"output": "2 3"
},
{
"input": "6",
"output": "1 5"
},
{
"input": "8",
"output": "3 5"
},
{
"input": "9",
"output": "4 5"
},
{
"input": "423",
"output": "211 212"
},
{
"input": "876",
"output": "437 439"
},
{
"input": "29",
"output": "14 15"
}
] | 1,630,526,714 | 2,147,483,647 | PyPy 3 | OK | TESTS | 26 | 109 | 20,172,800 | from math import gcd
n = int(input())
k = n // 2
for i in range(k, n):
if gcd(i, n - i) == 1:
print(*sorted([n - i, i]))
break | Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none | ```python
from math import gcd
n = int(input())
k = n // 2
for i in range(k, n):
if gcd(i, n - i) == 1:
print(*sorted([n - i, i]))
break
``` | 3 |
|
382 | A | Ksenia and Pan Scales | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | Ksenia has ordinary pan scales and several weights of an equal mass. Ksenia has already put some weights on the scales, while other weights are untouched. Ksenia is now wondering whether it is possible to put all the remaining weights on the scales so that the scales were in equilibrium.
The scales is in equilibrium if the total sum of weights on the left pan is equal to the total sum of weights on the right pan. | The first line has a non-empty sequence of characters describing the scales. In this sequence, an uppercase English letter indicates a weight, and the symbol "|" indicates the delimiter (the character occurs in the sequence exactly once). All weights that are recorded in the sequence before the delimiter are initially on the left pan of the scale. All weights that are recorded in the sequence after the delimiter are initially on the right pan of the scale.
The second line contains a non-empty sequence containing uppercase English letters. Each letter indicates a weight which is not used yet.
It is guaranteed that all the English letters in the input data are different. It is guaranteed that the input does not contain any extra characters. | If you cannot put all the weights on the scales so that the scales were in equilibrium, print string "Impossible". Otherwise, print the description of the resulting scales, copy the format of the input.
If there are multiple answers, print any of them. | [
"AC|T\nL\n",
"|ABC\nXYZ\n",
"W|T\nF\n",
"ABC|\nD\n"
] | [
"AC|TL\n",
"XYZ|ABC\n",
"Impossible\n",
"Impossible\n"
] | none | 500 | [
{
"input": "AC|T\nL",
"output": "AC|TL"
},
{
"input": "|ABC\nXYZ",
"output": "XYZ|ABC"
},
{
"input": "W|T\nF",
"output": "Impossible"
},
{
"input": "ABC|\nD",
"output": "Impossible"
},
{
"input": "A|BC\nDEF",
"output": "ADF|BCE"
},
{
"input": "|\nABC",
"output": "Impossible"
},
{
"input": "|\nZXCVBANMIO",
"output": "XVAMO|ZCBNI"
},
{
"input": "|C\nA",
"output": "A|C"
},
{
"input": "|\nAB",
"output": "B|A"
},
{
"input": "A|XYZ\nUIOPL",
"output": "Impossible"
},
{
"input": "K|B\nY",
"output": "Impossible"
},
{
"input": "EQJWDOHKZRBISPLXUYVCMNFGT|\nA",
"output": "Impossible"
},
{
"input": "|MACKERIGZPVHNDYXJBUFLWSO\nQT",
"output": "Impossible"
},
{
"input": "ERACGIZOVPT|WXUYMDLJNQS\nKB",
"output": "ERACGIZOVPTB|WXUYMDLJNQSK"
},
{
"input": "CKQHRUZMISGE|FBVWPXDLTJYN\nOA",
"output": "CKQHRUZMISGEA|FBVWPXDLTJYNO"
},
{
"input": "V|CMOEUTAXBFWSK\nDLRZJGIYNQHP",
"output": "VDLRZJGIYNQHP|CMOEUTAXBFWSK"
},
{
"input": "QWHNMALDGKTJ|\nPBRYVXZUESCOIF",
"output": "QWHNMALDGKTJF|PBRYVXZUESCOI"
},
{
"input": "|\nFXCVMUEWZAHNDOSITPRLKQJYBG",
"output": "XVUWANOIPLQYG|FCMEZHDSTRKJB"
},
{
"input": "IB|PCGHZ\nFXWTJQNEKAUM",
"output": "Impossible"
},
{
"input": "EC|IWAXQ\nJUHSRKGZTOMYN",
"output": "ECJUHRGTMN|IWAXQSKZOY"
},
{
"input": "VDINYMA|UQKWBCLRHZJ\nXEGOF",
"output": "Impossible"
},
{
"input": "ZLTPSIQUBAR|XFDEMYC\nHNOJWG",
"output": "ZLTPSIQUBARG|XFDEMYCHNOJW"
},
{
"input": "R|FLZOTJNU\nGIYHKVX",
"output": "RGIYHKVX|FLZOTJNU"
},
{
"input": "W|TL\nQROFSADYPKHEJNMXBZVUCIG",
"output": "WQOSDPHJMBVCG|TLRFAYKENXZUI"
},
{
"input": "NRDFQSEKLAYMOT|ZH\nGUXIBJCVPW",
"output": "Impossible"
},
{
"input": "FGRT|\nAC",
"output": "Impossible"
},
{
"input": "|FGRT\nAC",
"output": "Impossible"
},
{
"input": "A|\nB",
"output": "A|B"
},
{
"input": "|A\nB",
"output": "B|A"
},
{
"input": "|\nA",
"output": "Impossible"
},
{
"input": "|\nQWERTYUIOPASDFGHJKLZXCVBNM",
"output": "WRYIPSFHKZCBM|QETUOADGJLXVN"
},
{
"input": "QWERTYUIOPASDFGHJKLZXCVBN|\nM",
"output": "Impossible"
},
{
"input": "QWERTY|VBN\nUIOPASDFGHJKLZXC",
"output": "Impossible"
},
{
"input": "ABC|D\nKSL",
"output": "Impossible"
},
{
"input": "A|BCDEF\nGH",
"output": "Impossible"
},
{
"input": "|ABC\nD",
"output": "Impossible"
},
{
"input": "A|BC\nDE",
"output": "Impossible"
},
{
"input": "|ASD\nX",
"output": "Impossible"
},
{
"input": "AB|CDEF\nXYZRT",
"output": "Impossible"
}
] | 1,662,484,981 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | # Ksenia and Pan Scales
left, right = list(map(str, input().split("|")))
w_add = input()
if (len(left) == len(right) + len(w_add)):
print(left, "|", right + w_add, sep="")
elif(len(left) + len(w_add) == len(right)):
print(left + w_add, "|", right, sep="")
else:
print("Impossible") | Title: Ksenia and Pan Scales
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ksenia has ordinary pan scales and several weights of an equal mass. Ksenia has already put some weights on the scales, while other weights are untouched. Ksenia is now wondering whether it is possible to put all the remaining weights on the scales so that the scales were in equilibrium.
The scales is in equilibrium if the total sum of weights on the left pan is equal to the total sum of weights on the right pan.
Input Specification:
The first line has a non-empty sequence of characters describing the scales. In this sequence, an uppercase English letter indicates a weight, and the symbol "|" indicates the delimiter (the character occurs in the sequence exactly once). All weights that are recorded in the sequence before the delimiter are initially on the left pan of the scale. All weights that are recorded in the sequence after the delimiter are initially on the right pan of the scale.
The second line contains a non-empty sequence containing uppercase English letters. Each letter indicates a weight which is not used yet.
It is guaranteed that all the English letters in the input data are different. It is guaranteed that the input does not contain any extra characters.
Output Specification:
If you cannot put all the weights on the scales so that the scales were in equilibrium, print string "Impossible". Otherwise, print the description of the resulting scales, copy the format of the input.
If there are multiple answers, print any of them.
Demo Input:
['AC|T\nL\n', '|ABC\nXYZ\n', 'W|T\nF\n', 'ABC|\nD\n']
Demo Output:
['AC|TL\n', 'XYZ|ABC\n', 'Impossible\n', 'Impossible\n']
Note:
none | ```python
# Ksenia and Pan Scales
left, right = list(map(str, input().split("|")))
w_add = input()
if (len(left) == len(right) + len(w_add)):
print(left, "|", right + w_add, sep="")
elif(len(left) + len(w_add) == len(right)):
print(left + w_add, "|", right, sep="")
else:
print("Impossible")
``` | 0 |
|
821 | C | Okabe and Boxes | PROGRAMMING | 1,500 | [
"data structures",
"greedy",
"trees"
] | null | null | Okabe and Super Hacker Daru are stacking and removing boxes. There are *n* boxes numbered from 1 to *n*. Initially there are no boxes on the stack.
Okabe, being a control freak, gives Daru 2*n* commands: *n* of which are to add a box to the top of the stack, and *n* of which are to remove a box from the top of the stack and throw it in the trash. Okabe wants Daru to throw away the boxes in the order from 1 to *n*. Of course, this means that it might be impossible for Daru to perform some of Okabe's remove commands, because the required box is not on the top of the stack.
That's why Daru can decide to wait until Okabe looks away and then reorder the boxes in the stack in any way he wants. He can do it at any point of time between Okabe's commands, but he can't add or remove boxes while he does it.
Tell Daru the minimum number of times he needs to reorder the boxes so that he can successfully complete all of Okabe's commands. It is guaranteed that every box is added before it is required to be removed. | The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of boxes.
Each of the next 2*n* lines of input starts with a string "add" or "remove". If the line starts with the "add", an integer *x* (1<=≤<=*x*<=≤<=*n*) follows, indicating that Daru should add the box with number *x* to the top of the stack.
It is guaranteed that exactly *n* lines contain "add" operations, all the boxes added are distinct, and *n* lines contain "remove" operations. It is also guaranteed that a box is always added before it is required to be removed. | Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands. | [
"3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove\n",
"7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove\n"
] | [
"1\n",
"2\n"
] | In the first sample, Daru should reorder the boxes after adding box 3 to the stack.
In the second sample, Daru should reorder the boxes after adding box 4 and box 7 to the stack. | 1,500 | [
{
"input": "3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove",
"output": "1"
},
{
"input": "7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove",
"output": "2"
},
{
"input": "4\nadd 1\nadd 3\nremove\nadd 4\nadd 2\nremove\nremove\nremove",
"output": "2"
},
{
"input": "2\nadd 1\nremove\nadd 2\nremove",
"output": "0"
},
{
"input": "1\nadd 1\nremove",
"output": "0"
},
{
"input": "15\nadd 12\nadd 7\nadd 10\nadd 11\nadd 5\nadd 2\nadd 1\nadd 6\nadd 8\nremove\nremove\nadd 15\nadd 4\nadd 13\nadd 9\nadd 3\nadd 14\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "2"
},
{
"input": "14\nadd 7\nadd 2\nadd 13\nadd 5\nadd 12\nadd 6\nadd 4\nadd 1\nadd 14\nremove\nadd 10\nremove\nadd 9\nadd 8\nadd 11\nadd 3\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "3"
},
{
"input": "11\nadd 10\nadd 9\nadd 11\nadd 1\nadd 5\nadd 6\nremove\nadd 3\nadd 8\nadd 2\nadd 4\nremove\nremove\nremove\nremove\nremove\nadd 7\nremove\nremove\nremove\nremove\nremove",
"output": "2"
},
{
"input": "3\nadd 3\nadd 2\nadd 1\nremove\nremove\nremove",
"output": "0"
},
{
"input": "4\nadd 1\nadd 3\nadd 4\nremove\nadd 2\nremove\nremove\nremove",
"output": "1"
},
{
"input": "6\nadd 3\nadd 4\nadd 5\nadd 1\nadd 6\nremove\nadd 2\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "16\nadd 1\nadd 2\nadd 3\nadd 4\nadd 5\nadd 6\nadd 7\nadd 8\nadd 9\nadd 10\nadd 11\nadd 12\nadd 13\nadd 14\nadd 15\nadd 16\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "2\nadd 2\nadd 1\nremove\nremove",
"output": "0"
},
{
"input": "17\nadd 1\nadd 2\nadd 3\nadd 4\nadd 5\nadd 6\nadd 7\nadd 8\nadd 9\nadd 10\nadd 11\nadd 12\nadd 13\nadd 14\nadd 15\nadd 16\nadd 17\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "18\nadd 1\nadd 2\nadd 3\nadd 4\nadd 5\nadd 6\nadd 7\nadd 8\nadd 9\nadd 10\nadd 11\nadd 12\nadd 13\nadd 14\nadd 15\nadd 16\nadd 17\nadd 18\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "4\nadd 1\nadd 2\nremove\nremove\nadd 4\nadd 3\nremove\nremove",
"output": "1"
},
{
"input": "19\nadd 1\nadd 2\nadd 3\nadd 4\nadd 5\nadd 6\nadd 7\nadd 8\nadd 9\nadd 10\nadd 11\nadd 12\nadd 13\nadd 14\nadd 15\nadd 16\nadd 17\nadd 18\nadd 19\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "5\nadd 4\nadd 3\nadd 1\nremove\nadd 2\nremove\nremove\nadd 5\nremove\nremove",
"output": "1"
},
{
"input": "7\nadd 4\nadd 6\nadd 1\nadd 5\nadd 7\nremove\nadd 2\nremove\nadd 3\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "8\nadd 1\nadd 2\nadd 3\nadd 7\nadd 8\nremove\nremove\nremove\nadd 6\nadd 5\nadd 4\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "4\nadd 1\nadd 4\nremove\nadd 3\nadd 2\nremove\nremove\nremove",
"output": "1"
},
{
"input": "7\nadd 1\nadd 2\nadd 3\nadd 5\nadd 7\nremove\nremove\nremove\nadd 4\nremove\nremove\nadd 6\nremove\nremove",
"output": "1"
},
{
"input": "4\nadd 4\nadd 1\nadd 2\nremove\nremove\nadd 3\nremove\nremove",
"output": "1"
},
{
"input": "5\nadd 1\nadd 3\nadd 4\nadd 5\nremove\nadd 2\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "5\nadd 2\nadd 1\nremove\nremove\nadd 5\nadd 3\nremove\nadd 4\nremove\nremove",
"output": "0"
},
{
"input": "9\nadd 3\nadd 2\nadd 1\nadd 4\nadd 6\nadd 9\nremove\nremove\nremove\nremove\nadd 5\nremove\nremove\nadd 8\nadd 7\nremove\nremove\nremove",
"output": "1"
},
{
"input": "10\nadd 9\nadd 10\nadd 4\nadd 3\nadd 2\nadd 1\nremove\nremove\nremove\nremove\nadd 8\nadd 7\nadd 5\nadd 6\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "1"
}
] | 1,607,503,567 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define all(c) (c).begin(), (c).end()
#define data ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define pb push_back
#define REP(i,a,b) for (int i = a; i < b; i++)
#define range(name, start,end,step) (int name = start; name < end; name+= step)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ar array;
int pow(int a, int b, int m){
int ans = 1;
while(b){
if(b&1) ans=(ans*a)%m;
b/=2;
a=(a*a)%m;
}
return ans;
}
int main(){
data;
int n, cur = 1, res = 0, x, len;
string s;
cin >> n;
vector<int> stack;
for (int i = 0; i< 2*n; i++){
cin >> s;
if (s == "add"){
cin >> x;
stack.push_back(x);
}else{
len = stack.size();
if (stack.at(len-1) != cur){
sort(stack.begin(), stack.end(), greater<int>());
res++;
}
stack.pop_back();
cur ++;
}
}
cout << res;
return 0;
} | Title: Okabe and Boxes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Okabe and Super Hacker Daru are stacking and removing boxes. There are *n* boxes numbered from 1 to *n*. Initially there are no boxes on the stack.
Okabe, being a control freak, gives Daru 2*n* commands: *n* of which are to add a box to the top of the stack, and *n* of which are to remove a box from the top of the stack and throw it in the trash. Okabe wants Daru to throw away the boxes in the order from 1 to *n*. Of course, this means that it might be impossible for Daru to perform some of Okabe's remove commands, because the required box is not on the top of the stack.
That's why Daru can decide to wait until Okabe looks away and then reorder the boxes in the stack in any way he wants. He can do it at any point of time between Okabe's commands, but he can't add or remove boxes while he does it.
Tell Daru the minimum number of times he needs to reorder the boxes so that he can successfully complete all of Okabe's commands. It is guaranteed that every box is added before it is required to be removed.
Input Specification:
The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of boxes.
Each of the next 2*n* lines of input starts with a string "add" or "remove". If the line starts with the "add", an integer *x* (1<=≤<=*x*<=≤<=*n*) follows, indicating that Daru should add the box with number *x* to the top of the stack.
It is guaranteed that exactly *n* lines contain "add" operations, all the boxes added are distinct, and *n* lines contain "remove" operations. It is also guaranteed that a box is always added before it is required to be removed.
Output Specification:
Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands.
Demo Input:
['3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove\n', '7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove\n']
Demo Output:
['1\n', '2\n']
Note:
In the first sample, Daru should reorder the boxes after adding box 3 to the stack.
In the second sample, Daru should reorder the boxes after adding box 4 and box 7 to the stack. | ```python
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define all(c) (c).begin(), (c).end()
#define data ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define pb push_back
#define REP(i,a,b) for (int i = a; i < b; i++)
#define range(name, start,end,step) (int name = start; name < end; name+= step)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ar array;
int pow(int a, int b, int m){
int ans = 1;
while(b){
if(b&1) ans=(ans*a)%m;
b/=2;
a=(a*a)%m;
}
return ans;
}
int main(){
data;
int n, cur = 1, res = 0, x, len;
string s;
cin >> n;
vector<int> stack;
for (int i = 0; i< 2*n; i++){
cin >> s;
if (s == "add"){
cin >> x;
stack.push_back(x);
}else{
len = stack.size();
if (stack.at(len-1) != cur){
sort(stack.begin(), stack.end(), greater<int>());
res++;
}
stack.pop_back();
cur ++;
}
}
cout << res;
return 0;
}
``` | -1 |
|
583 | A | Asphalting Roads | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted. | The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct. | In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1. | [
"2\n1 1\n1 2\n2 1\n2 2\n",
"1\n1 1\n"
] | [
"1 4 \n",
"1 \n"
] | In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | 500 | [
{
"input": "2\n1 1\n1 2\n2 1\n2 2",
"output": "1 4 "
},
{
"input": "1\n1 1",
"output": "1 "
},
{
"input": "2\n1 1\n2 2\n1 2\n2 1",
"output": "1 2 "
},
{
"input": "2\n1 2\n2 2\n2 1\n1 1",
"output": "1 3 "
},
{
"input": "3\n2 2\n1 2\n3 2\n3 3\n1 1\n2 3\n1 3\n3 1\n2 1",
"output": "1 4 5 "
},
{
"input": "3\n1 3\n3 1\n2 1\n1 1\n1 2\n2 2\n3 2\n3 3\n2 3",
"output": "1 2 6 "
},
{
"input": "4\n1 3\n2 3\n2 4\n4 4\n3 1\n1 1\n3 4\n2 1\n1 4\n4 3\n4 1\n3 2\n1 2\n4 2\n2 2\n3 3",
"output": "1 3 5 14 "
},
{
"input": "4\n3 3\n4 2\n2 3\n3 4\n4 4\n1 2\n3 2\n2 2\n1 4\n3 1\n4 1\n2 1\n1 3\n1 1\n4 3\n2 4",
"output": "1 2 9 12 "
},
{
"input": "9\n4 5\n2 3\n8 3\n5 6\n9 3\n4 4\n5 4\n4 7\n1 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n9 9\n8 7\n7 5\n3 7\n6 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n1 1\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 2\n3 9\n2 9\n7 7\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n3 3\n3 8\n5 5\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n8 8\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8",
"output": "1 2 4 9 10 14 16 32 56 "
},
{
"input": "8\n1 1\n1 2\n1 3\n1 4\n1 5\n8 6\n1 7\n1 8\n2 1\n8 5\n2 3\n2 4\n2 5\n2 6\n4 3\n2 2\n3 1\n3 2\n3 3\n3 4\n3 5\n3 6\n5 6\n3 8\n4 1\n4 2\n2 7\n4 4\n8 8\n4 6\n4 7\n4 8\n5 1\n5 2\n5 3\n6 5\n5 5\n3 7\n5 7\n5 8\n6 1\n6 2\n6 3\n6 4\n5 4\n6 6\n6 7\n6 8\n7 1\n7 2\n7 3\n7 4\n7 5\n7 6\n7 7\n7 8\n8 1\n8 2\n8 3\n8 4\n2 8\n1 6\n8 7\n4 5",
"output": "1 6 11 18 28 36 39 56 "
},
{
"input": "9\n9 9\n5 5\n8 8\n3 3\n2 2\n6 6\n4 4\n1 1\n7 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n1 7\n8 7\n7 5\n3 7\n5 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n4 5\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 3\n3 9\n2 9\n5 4\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n8 3\n3 8\n9 3\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n4 7\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8",
"output": "1 2 3 4 5 6 7 8 9 "
}
] | 1,451,023,186 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | n = int(input())
inter = [[False]*n]*2
for i in range(n*n):
h,v = input().split()
hnum = int(h)-1
vnum = int(v)-1
if not inter[0][hnum] and not inter[1][vnum]:
inter[0][hnum] = True
inter[1][vnum] = True
print(i+1,end=' ') | Title: Asphalting Roads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
Output Specification:
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
Demo Input:
['2\n1 1\n1 2\n2 1\n2 2\n', '1\n1 1\n']
Demo Output:
['1 4 \n', '1 \n']
Note:
In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | ```python
n = int(input())
inter = [[False]*n]*2
for i in range(n*n):
h,v = input().split()
hnum = int(h)-1
vnum = int(v)-1
if not inter[0][hnum] and not inter[1][vnum]:
inter[0][hnum] = True
inter[1][vnum] = True
print(i+1,end=' ')
``` | 0 |
|
768 | A | Oath of the Night's Watch | PROGRAMMING | 900 | [
"constructive algorithms",
"sortings"
] | null | null | "Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support? | First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow.
Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards. | Output a single integer representing the number of stewards which Jon will feed. | [
"2\n1 5\n",
"3\n1 2 5\n"
] | [
"0",
"1"
] | In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | 500 | [
{
"input": "2\n1 5",
"output": "0"
},
{
"input": "3\n1 2 5",
"output": "1"
},
{
"input": "4\n1 2 3 4",
"output": "2"
},
{
"input": "8\n7 8 9 4 5 6 1 2",
"output": "6"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n100",
"output": "0"
},
{
"input": "205\n5 5 3 3 6 2 9 3 8 9 6 6 10 8 1 5 3 3 1 2 9 9 9 3 9 10 3 9 8 3 5 6 6 4 6 9 2 9 10 9 5 6 6 7 4 2 6 3 4 1 10 1 7 2 7 7 3 2 6 5 5 2 9 3 8 8 7 6 6 4 2 2 6 2 3 5 7 2 2 10 1 4 6 9 2 3 7 2 2 7 4 4 9 10 7 5 8 6 5 3 6 10 2 7 5 6 6 8 3 3 9 4 3 5 7 9 3 2 1 1 3 2 1 9 3 1 4 4 10 2 5 5 8 1 4 8 5 3 1 10 8 6 5 8 3 5 4 5 4 4 6 7 2 8 10 8 7 6 6 9 6 7 1 10 3 2 5 10 4 4 5 4 3 4 8 5 3 8 10 3 10 9 7 2 1 8 6 4 6 5 8 10 2 6 7 4 9 4 5 1 8 7 10 3 1",
"output": "174"
},
{
"input": "4\n1000000000 99999999 1000000000 1000000000",
"output": "0"
},
{
"input": "3\n2 2 2",
"output": "0"
},
{
"input": "5\n1 1 1 1 1",
"output": "0"
},
{
"input": "3\n1 1 1",
"output": "0"
},
{
"input": "6\n1 1 3 3 2 2",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "4\n1 1 2 5",
"output": "1"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n1 1 1 1 5",
"output": "0"
},
{
"input": "5\n1 1 2 3 3",
"output": "1"
},
{
"input": "3\n1 1 3",
"output": "0"
},
{
"input": "3\n2 2 3",
"output": "0"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "5\n1 5 3 5 1",
"output": "1"
},
{
"input": "7\n1 2 2 2 2 2 3",
"output": "5"
},
{
"input": "4\n2 2 2 2",
"output": "0"
},
{
"input": "9\n2 2 2 3 4 5 6 6 6",
"output": "3"
},
{
"input": "10\n1 1 1 2 3 3 3 3 3 3",
"output": "1"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "0"
},
{
"input": "3\n0 0 1",
"output": "0"
},
{
"input": "9\n1 1 1 2 2 2 3 3 3",
"output": "3"
},
{
"input": "3\n1 2 2",
"output": "0"
},
{
"input": "6\n2 2 2 2 2 2",
"output": "0"
},
{
"input": "5\n2 2 2 2 2",
"output": "0"
},
{
"input": "5\n5 5 5 5 5",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "6\n1 2 5 5 5 5",
"output": "1"
},
{
"input": "5\n1 2 3 3 3",
"output": "1"
},
{
"input": "3\n1 1 2",
"output": "0"
},
{
"input": "6\n1 1 1 1 1 2",
"output": "0"
},
{
"input": "5\n1 1 2 4 4",
"output": "1"
},
{
"input": "3\n999999 5999999 9999999",
"output": "1"
},
{
"input": "4\n1 1 5 5",
"output": "0"
},
{
"input": "9\n1 1 1 2 2 2 4 4 4",
"output": "3"
},
{
"input": "5\n1 3 4 5 1",
"output": "2"
},
{
"input": "5\n3 3 3 3 3",
"output": "0"
},
{
"input": "5\n1 1 2 2 2",
"output": "0"
},
{
"input": "5\n2 1 1 1 3",
"output": "1"
},
{
"input": "5\n0 0 0 1 2",
"output": "1"
},
{
"input": "4\n2 2 2 3",
"output": "0"
},
{
"input": "7\n1 1 1 1 5 5 5",
"output": "0"
},
{
"input": "5\n1 2 3 4 4",
"output": "2"
},
{
"input": "2\n5 4",
"output": "0"
},
{
"input": "4\n5 5 5 5",
"output": "0"
},
{
"input": "5\n1 1 1 5 5",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "0"
},
{
"input": "4\n1 2 2 2",
"output": "0"
},
{
"input": "8\n1000000000 1000000000 1000000000 999999999 999999999 999999999 999999998 999999998",
"output": "3"
},
{
"input": "5\n1 1 3 4 4",
"output": "1"
},
{
"input": "6\n1 1 2 2 3 3",
"output": "2"
},
{
"input": "4\n1 1 1 1",
"output": "0"
},
{
"input": "9\n1 2 3 4 1 5 6 7 8",
"output": "6"
},
{
"input": "8\n5 4 4 6 6 4 4 3",
"output": "5"
},
{
"input": "8\n4 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "7\n4 3 3 3 3 3 3",
"output": "0"
},
{
"input": "6\n4 3 3 3 3 3",
"output": "0"
},
{
"input": "5\n4 3 3 3 3",
"output": "0"
}
] | 1,684,410,238 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 15 | 0 | a=int(input())
b=list(map(int,input().split()))
s=0
k=sorted(b)
for i in b:
if i<k[-1] and i > k[0]:
s+=1
if s ==0:
print(0)
else:
print(s-1)
| Title: Oath of the Night's Watch
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I pledge my life and honor to the Night's Watch, for this night and all the nights to come." — The Night's Watch oath.
With that begins the watch of Jon Snow. He is assigned the task to support the stewards.
This time he has *n* stewards with him whom he has to provide support. Each steward has his own strength. Jon Snow likes to support a steward only if there exists at least one steward who has strength strictly less than him and at least one steward who has strength strictly greater than him.
Can you find how many stewards will Jon support?
Input Specification:
First line consists of a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of stewards with Jon Snow.
Second line consists of *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) representing the values assigned to the stewards.
Output Specification:
Output a single integer representing the number of stewards which Jon will feed.
Demo Input:
['2\n1 5\n', '3\n1 2 5\n']
Demo Output:
['0', '1']
Note:
In the first sample, Jon Snow cannot support steward with strength 1 because there is no steward with strength less than 1 and he cannot support steward with strength 5 because there is no steward with strength greater than 5.
In the second sample, Jon Snow can support steward with strength 2 because there are stewards with strength less than 2 and greater than 2. | ```python
a=int(input())
b=list(map(int,input().split()))
s=0
k=sorted(b)
for i in b:
if i<k[-1] and i > k[0]:
s+=1
if s ==0:
print(0)
else:
print(s-1)
``` | 0 |
|
149 | A | Business trip | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters. | The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100). | Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1. | [
"5\n1 1 1 1 2 2 3 2 2 1 1 1\n",
"0\n0 0 0 0 0 0 0 1 1 2 3 0\n",
"11\n1 1 4 1 1 5 1 1 4 1 1 1\n"
] | [
"2\n",
"0\n",
"3\n"
] | Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all. | 500 | [
{
"input": "5\n1 1 1 1 2 2 3 2 2 1 1 1",
"output": "2"
},
{
"input": "0\n0 0 0 0 0 0 0 1 1 2 3 0",
"output": "0"
},
{
"input": "11\n1 1 4 1 1 5 1 1 4 1 1 1",
"output": "3"
},
{
"input": "15\n20 1 1 1 1 2 2 1 2 2 1 1",
"output": "1"
},
{
"input": "7\n8 9 100 12 14 17 21 10 11 100 23 10",
"output": "1"
},
{
"input": "52\n1 12 3 11 4 5 10 6 9 7 8 2",
"output": "6"
},
{
"input": "50\n2 2 3 4 5 4 4 5 7 3 2 7",
"output": "-1"
},
{
"input": "0\n55 81 28 48 99 20 67 95 6 19 10 93",
"output": "0"
},
{
"input": "93\n85 40 93 66 92 43 61 3 64 51 90 21",
"output": "1"
},
{
"input": "99\n36 34 22 0 0 0 52 12 0 0 33 47",
"output": "2"
},
{
"input": "99\n28 32 31 0 10 35 11 18 0 0 32 28",
"output": "3"
},
{
"input": "99\n19 17 0 1 18 11 29 9 29 22 0 8",
"output": "4"
},
{
"input": "76\n2 16 11 10 12 0 20 4 4 14 11 14",
"output": "5"
},
{
"input": "41\n2 1 7 7 4 2 4 4 9 3 10 0",
"output": "6"
},
{
"input": "47\n8 2 2 4 3 1 9 4 2 7 7 8",
"output": "7"
},
{
"input": "58\n6 11 7 0 5 6 3 9 4 9 5 1",
"output": "8"
},
{
"input": "32\n5 2 4 1 5 0 5 1 4 3 0 3",
"output": "9"
},
{
"input": "31\n6 1 0 4 4 5 1 0 5 3 2 0",
"output": "9"
},
{
"input": "35\n2 3 0 0 6 3 3 4 3 5 0 6",
"output": "9"
},
{
"input": "41\n3 1 3 4 3 6 6 1 4 4 0 6",
"output": "11"
},
{
"input": "97\n0 5 3 12 10 16 22 8 21 17 21 10",
"output": "5"
},
{
"input": "100\n21 21 0 0 4 13 0 26 0 0 0 15",
"output": "6"
},
{
"input": "100\n0 0 16 5 22 0 5 0 25 0 14 13",
"output": "7"
},
{
"input": "97\n17 0 10 0 0 0 18 0 14 23 15 0",
"output": "6"
},
{
"input": "100\n0 9 0 18 7 0 0 14 33 3 0 16",
"output": "7"
},
{
"input": "95\n5 2 13 0 15 18 17 0 6 11 0 8",
"output": "9"
},
{
"input": "94\n11 13 0 9 15 8 8 16 3 7 1 3",
"output": "11"
},
{
"input": "96\n8 4 12 15 8 0 4 10 6 6 12 11",
"output": "11"
},
{
"input": "100\n5 5 3 8 6 5 0 3 3 8 1 3",
"output": "-1"
},
{
"input": "100\n1 0 0 1 1 0 1 1 1 1 2 1",
"output": "-1"
},
{
"input": "100\n6 3 2 0 4 1 2 2 2 2 1 1",
"output": "-1"
},
{
"input": "0\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
},
{
"input": "0\n100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100",
"output": "1"
},
{
"input": "12\n1 1 1 1 1 1 1 1 1 1 1 1",
"output": "12"
},
{
"input": "13\n1 1 1 1 1 1 1 1 1 1 1 2",
"output": "12"
},
{
"input": "15\n10 1 1 1 1 1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1\n0 0 0 0 0 0 0 0 0 0 0 0",
"output": "-1"
}
] | 1,640,940,099 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 92 | 0 | k = int(input())
lis = sorted(list(map(int, input().split())))[::-1]
s =0
m =0
flag=0
if k==0:
print(0)
else:
for i in lis:
s+=i
m+=1
if s>=k:
flag=1
break
if flag==1:
print(m)
else:
print(-1)
| Title: Business trip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
Input Specification:
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
Output Specification:
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
Demo Input:
['5\n1 1 1 1 2 2 3 2 2 1 1 1\n', '0\n0 0 0 0 0 0 0 1 1 2 3 0\n', '11\n1 1 4 1 1 5 1 1 4 1 1 1\n']
Demo Output:
['2\n', '0\n', '3\n']
Note:
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all. | ```python
k = int(input())
lis = sorted(list(map(int, input().split())))[::-1]
s =0
m =0
flag=0
if k==0:
print(0)
else:
for i in lis:
s+=i
m+=1
if s>=k:
flag=1
break
if flag==1:
print(m)
else:
print(-1)
``` | 3 |
|
31 | C | Schedule | PROGRAMMING | 1,700 | [
"implementation"
] | C. Schedule | 2 | 256 | At the beginning of the new semester there is new schedule in the Berland State University. According to this schedule, *n* groups have lessons at the room 31. For each group the starting time of the lesson and the finishing time of the lesson are known. It has turned out that it is impossible to hold all lessons, because for some groups periods of their lessons intersect. If at some moment of time one groups finishes it's lesson, and the other group starts the lesson, their lessons don't intersect.
The dean wants to cancel the lesson in one group so that no two time periods of lessons of the remaining groups intersect. You are to find all ways to do that. | The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — amount of groups, which have lessons in the room 31. Then *n* lines follow, each of them contains two integers *l**i* *r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=106) — starting and finishing times of lesson of the *i*-th group. It is possible that initially no two lessons intersect (see sample 1). | Output integer *k* — amount of ways to cancel the lesson in exactly one group so that no two time periods of lessons of the remaining groups intersect. In the second line output *k* numbers — indexes of groups, where it is possible to cancel the lesson. Groups are numbered starting from 1 in the order that they were given in the input. Output the numbers in increasing order. | [
"3\n3 10\n20 30\n1 3\n",
"4\n3 10\n20 30\n1 3\n1 39\n",
"3\n1 5\n2 6\n3 7\n"
] | [
"3\n1 2 3 ",
"1\n4 ",
"0\n"
] | none | 1,500 | [
{
"input": "3\n3 10\n20 30\n1 3",
"output": "3\n1 2 3 "
},
{
"input": "4\n3 10\n20 30\n1 3\n1 39",
"output": "1\n4 "
},
{
"input": "3\n1 5\n2 6\n3 7",
"output": "0"
},
{
"input": "4\n1 5\n5 7\n6 9\n9 10",
"output": "2\n2 3 "
},
{
"input": "11\n717170 795210\n866429 970764\n163324 322182\n677099 717170\n241684 393937\n50433 114594\n970764 997956\n393937 664883\n235698 241684\n795210 832346\n114594 232438",
"output": "1\n3 "
},
{
"input": "16\n203671 381501\n58867 59732\n817520 962123\n125391 163027\n601766 617692\n381501 444610\n761937 817520\n16 10551\n21096 38291\n718073 761937\n583868 601766\n554859 731755\n678098 718073\n962123 992003\n163027 203671\n87917 96397",
"output": "1\n12 "
}
] | 1,522,078,581 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 122 | 6,963,200 | groups = []
overlapping = set()
for notused in range(int(input())):
groups.append(list(map(int, input().split(" "))))
for i in range(len(groups)-1):
if groups[-1][0] > groups[i][0] and groups[-1][0] < groups[i][1] or groups[-1][1] > groups[i][0] and groups[-1][1] < groups[i][1]:
overlapping.add(i)
overlapping.add(len(groups)-1)
if groups[i][0] > groups[-1][0] and groups[i][0] < groups[-1][1] or groups[i][1] > groups[-1][0] and groups[i][1] < groups[-1][1]:
overlapping.add(i)
overlapping.add(len(groups)-1)
if len(overlapping) == 0:
print(len(groups))
for i in range(len(groups)): print(i, end = " ")
print()
else:
# we now have a non-empty set of overlapping elements in the list.
possible = []
for item in overlapping:
newoverlaping = overlapping - {item}
ranges = []
overlap = False
for z in newoverlaping:
x = groups[z]
for y in ranges:
if x[0] > y[0] and x[0] < y[1] or x[1] > y[0] and x[1] < y[1]:
overlap = True
break;
if y[0] > x[0] and y[0] < x[1] or y[1] > x[0] and y[1] < x[1]:
overlap = True
break;
ranges.append(x)
if overlap: break;
if not overlap: possible.append(item+1)
print(len(possible))
for item in possible: print(item, end=" ")
if len(possible) != 0:print() | Title: Schedule
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
At the beginning of the new semester there is new schedule in the Berland State University. According to this schedule, *n* groups have lessons at the room 31. For each group the starting time of the lesson and the finishing time of the lesson are known. It has turned out that it is impossible to hold all lessons, because for some groups periods of their lessons intersect. If at some moment of time one groups finishes it's lesson, and the other group starts the lesson, their lessons don't intersect.
The dean wants to cancel the lesson in one group so that no two time periods of lessons of the remaining groups intersect. You are to find all ways to do that.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — amount of groups, which have lessons in the room 31. Then *n* lines follow, each of them contains two integers *l**i* *r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=106) — starting and finishing times of lesson of the *i*-th group. It is possible that initially no two lessons intersect (see sample 1).
Output Specification:
Output integer *k* — amount of ways to cancel the lesson in exactly one group so that no two time periods of lessons of the remaining groups intersect. In the second line output *k* numbers — indexes of groups, where it is possible to cancel the lesson. Groups are numbered starting from 1 in the order that they were given in the input. Output the numbers in increasing order.
Demo Input:
['3\n3 10\n20 30\n1 3\n', '4\n3 10\n20 30\n1 3\n1 39\n', '3\n1 5\n2 6\n3 7\n']
Demo Output:
['3\n1 2 3 ', '1\n4 ', '0\n']
Note:
none | ```python
groups = []
overlapping = set()
for notused in range(int(input())):
groups.append(list(map(int, input().split(" "))))
for i in range(len(groups)-1):
if groups[-1][0] > groups[i][0] and groups[-1][0] < groups[i][1] or groups[-1][1] > groups[i][0] and groups[-1][1] < groups[i][1]:
overlapping.add(i)
overlapping.add(len(groups)-1)
if groups[i][0] > groups[-1][0] and groups[i][0] < groups[-1][1] or groups[i][1] > groups[-1][0] and groups[i][1] < groups[-1][1]:
overlapping.add(i)
overlapping.add(len(groups)-1)
if len(overlapping) == 0:
print(len(groups))
for i in range(len(groups)): print(i, end = " ")
print()
else:
# we now have a non-empty set of overlapping elements in the list.
possible = []
for item in overlapping:
newoverlaping = overlapping - {item}
ranges = []
overlap = False
for z in newoverlaping:
x = groups[z]
for y in ranges:
if x[0] > y[0] and x[0] < y[1] or x[1] > y[0] and x[1] < y[1]:
overlap = True
break;
if y[0] > x[0] and y[0] < x[1] or y[1] > x[0] and y[1] < x[1]:
overlap = True
break;
ranges.append(x)
if overlap: break;
if not overlap: possible.append(item+1)
print(len(possible))
for item in possible: print(item, end=" ")
if len(possible) != 0:print()
``` | 0 |
873 | A | Chores | PROGRAMMING | 800 | [
"implementation"
] | null | null | Luba has to do *n* chores today. *i*-th chore takes *a**i* units of time to complete. It is guaranteed that for every the condition *a**i*<=≥<=*a**i*<=-<=1 is met, so the sequence is sorted.
Also Luba can work really hard on some chores. She can choose not more than *k* any chores and do each of them in *x* units of time instead of *a**i* ().
Luba is very responsible, so she has to do all *n* chores, and now she wants to know the minimum time she needs to do everything. Luba cannot do two chores simultaneously. | The first line contains three integers *n*,<=*k*,<=*x* (1<=≤<=*k*<=≤<=*n*<=≤<=100,<=1<=≤<=*x*<=≤<=99) — the number of chores Luba has to do, the number of chores she can do in *x* units of time, and the number *x* itself.
The second line contains *n* integer numbers *a**i* (2<=≤<=*a**i*<=≤<=100) — the time Luba has to spend to do *i*-th chore.
It is guaranteed that , and for each *a**i*<=≥<=*a**i*<=-<=1. | Print one number — minimum time Luba needs to do all *n* chores. | [
"4 2 2\n3 6 7 10\n",
"5 2 1\n100 100 100 100 100\n"
] | [
"13\n",
"302\n"
] | In the first example the best option would be to do the third and the fourth chore, spending *x* = 2 time on each instead of *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, respectively. Then the answer is 3 + 6 + 2 + 2 = 13.
In the second example Luba can choose any two chores to spend *x* time on them instead of *a*<sub class="lower-index">*i*</sub>. So the answer is 100·3 + 2·1 = 302. | 0 | [
{
"input": "4 2 2\n3 6 7 10",
"output": "13"
},
{
"input": "5 2 1\n100 100 100 100 100",
"output": "302"
},
{
"input": "1 1 1\n100",
"output": "1"
},
{
"input": "100 1 99\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": "9999"
},
{
"input": "100 100 1\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": "100"
},
{
"input": "100 50 50\n51 51 52 53 55 55 55 55 56 56 56 57 57 58 58 59 59 59 60 60 61 61 62 62 63 64 64 64 64 65 65 65 65 66 66 66 67 68 68 68 69 69 70 70 70 70 71 71 71 71 71 71 72 72 76 76 76 76 77 79 79 81 81 81 81 82 82 82 82 83 84 85 86 87 87 88 88 88 89 89 89 90 90 90 91 91 91 92 92 93 95 95 96 96 96 97 97 98 99 100",
"output": "5618"
},
{
"input": "100 100 1\n2 4 4 4 5 5 5 6 10 10 11 11 12 12 13 13 13 14 17 18 20 20 21 21 22 22 23 24 24 25 26 29 29 32 32 34 34 35 38 39 39 40 40 42 42 43 45 47 48 49 51 52 52 54 57 59 59 60 61 61 62 63 63 64 65 65 68 70 70 72 74 75 75 76 76 77 77 78 78 78 79 80 81 82 82 83 83 83 84 89 90 92 92 93 94 96 96 97 98 99",
"output": "100"
},
{
"input": "100 1 1\n3 3 5 7 8 8 8 9 9 9 11 13 14 15 18 18 19 20 21 22 22 25 27 27 29 31 32 33 33 34 36 37 37 38 40 42 44 44 46 47 47 48 48 48 50 50 51 51 54 54 54 55 55 56 56 56 60 61 62 62 63 64 65 65 68 70 70 71 71 71 71 75 75 76 76 79 79 79 79 81 81 82 82 86 86 86 86 88 90 90 92 96 97 97 98 98 98 98 100 100",
"output": "5202"
},
{
"input": "100 50 49\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51",
"output": "4950"
},
{
"input": "100 50 1\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "151"
},
{
"input": "100 1 1\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "199"
}
] | 1,636,384,035 | 2,147,483,647 | Python 3 | OK | TESTS | 11 | 46 | 4,300,800 | n,k,x=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
print(sum(a)+k*x-sum(a[n-k:])) | Title: Chores
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba has to do *n* chores today. *i*-th chore takes *a**i* units of time to complete. It is guaranteed that for every the condition *a**i*<=≥<=*a**i*<=-<=1 is met, so the sequence is sorted.
Also Luba can work really hard on some chores. She can choose not more than *k* any chores and do each of them in *x* units of time instead of *a**i* ().
Luba is very responsible, so she has to do all *n* chores, and now she wants to know the minimum time she needs to do everything. Luba cannot do two chores simultaneously.
Input Specification:
The first line contains three integers *n*,<=*k*,<=*x* (1<=≤<=*k*<=≤<=*n*<=≤<=100,<=1<=≤<=*x*<=≤<=99) — the number of chores Luba has to do, the number of chores she can do in *x* units of time, and the number *x* itself.
The second line contains *n* integer numbers *a**i* (2<=≤<=*a**i*<=≤<=100) — the time Luba has to spend to do *i*-th chore.
It is guaranteed that , and for each *a**i*<=≥<=*a**i*<=-<=1.
Output Specification:
Print one number — minimum time Luba needs to do all *n* chores.
Demo Input:
['4 2 2\n3 6 7 10\n', '5 2 1\n100 100 100 100 100\n']
Demo Output:
['13\n', '302\n']
Note:
In the first example the best option would be to do the third and the fourth chore, spending *x* = 2 time on each instead of *a*<sub class="lower-index">3</sub> and *a*<sub class="lower-index">4</sub>, respectively. Then the answer is 3 + 6 + 2 + 2 = 13.
In the second example Luba can choose any two chores to spend *x* time on them instead of *a*<sub class="lower-index">*i*</sub>. So the answer is 100·3 + 2·1 = 302. | ```python
n,k,x=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
print(sum(a)+k*x-sum(a[n-k:]))
``` | 3 |
|
136 | A | Presents | PROGRAMMING | 800 | [
"implementation"
] | null | null | Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves. | Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*. | [
"4\n2 3 4 1\n",
"3\n1 3 2\n",
"2\n1 2\n"
] | [
"4 1 2 3\n",
"1 3 2\n",
"1 2\n"
] | none | 500 | [
{
"input": "4\n2 3 4 1",
"output": "4 1 2 3"
},
{
"input": "3\n1 3 2",
"output": "1 3 2"
},
{
"input": "2\n1 2",
"output": "1 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 3 2 6 4 5 7 9 8 10",
"output": "1 3 2 5 6 4 7 9 8 10"
},
{
"input": "5\n5 4 3 2 1",
"output": "5 4 3 2 1"
},
{
"input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19"
},
{
"input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19",
"output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19"
},
{
"input": "10\n3 4 5 6 7 8 9 10 1 2",
"output": "9 10 1 2 3 4 5 6 7 8"
},
{
"input": "8\n1 5 3 7 2 6 4 8",
"output": "1 5 3 7 2 6 4 8"
},
{
"input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47",
"output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19"
},
{
"input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17",
"output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25"
},
{
"input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85",
"output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30"
},
{
"input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49",
"output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49"
},
{
"input": "12\n3 8 7 4 6 5 2 1 11 9 10 12",
"output": "8 7 1 4 6 5 3 2 10 11 9 12"
},
{
"input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76",
"output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4"
},
{
"input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43",
"output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1"
},
{
"input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27",
"output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3"
},
{
"input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22",
"output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53"
},
{
"input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13",
"output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24"
},
{
"input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21",
"output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61"
},
{
"input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33",
"output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14"
},
{
"input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23",
"output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21"
},
{
"input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12",
"output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9"
},
{
"input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31",
"output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13"
},
{
"input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3",
"output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16"
},
{
"input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8",
"output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6"
},
{
"input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 100 51 6 20 23 81 15 17 3 60 71",
"output": "83 54 98 86 88 92 22 28 57 15 38 29 70 48 96 7 97 56 81 93 50 25 94 44 26 55 85 72 76 30 3 37 13 79 19 58 11 82 31 87 84 36 40 32 47 62 18 35 27 42 91 77 60 49 41 2 33 9 65 99 14 17 23 34 8 63 20 68 21 39 100 71 46 53 61 16 10 1 73 59 95 78 24 52 45 64 67 74 12 5 4 75 66 69 6 89 80 51 43 90"
},
{
"input": "22\n12 8 11 2 16 7 13 6 22 21 20 10 4 14 18 1 5 15 3 19 17 9",
"output": "16 4 19 13 17 8 6 2 22 12 3 1 7 14 18 5 21 15 20 11 10 9"
},
{
"input": "72\n16 11 49 51 3 27 60 55 23 40 66 7 53 70 13 5 15 32 18 72 33 30 8 31 46 12 28 67 25 38 50 22 69 34 71 52 58 39 24 35 42 9 41 26 62 1 63 65 36 64 68 61 37 14 45 47 6 57 54 20 17 2 56 59 29 10 4 48 21 43 19 44",
"output": "46 62 5 67 16 57 12 23 42 66 2 26 15 54 17 1 61 19 71 60 69 32 9 39 29 44 6 27 65 22 24 18 21 34 40 49 53 30 38 10 43 41 70 72 55 25 56 68 3 31 4 36 13 59 8 63 58 37 64 7 52 45 47 50 48 11 28 51 33 14 35 20"
},
{
"input": "63\n21 56 11 10 62 24 20 42 28 52 38 2 37 43 48 22 7 8 40 14 13 46 53 1 23 4 60 63 51 36 25 12 39 32 49 16 58 44 31 61 33 50 55 54 45 6 47 41 9 57 30 29 26 18 19 27 15 34 3 35 59 5 17",
"output": "24 12 59 26 62 46 17 18 49 4 3 32 21 20 57 36 63 54 55 7 1 16 25 6 31 53 56 9 52 51 39 34 41 58 60 30 13 11 33 19 48 8 14 38 45 22 47 15 35 42 29 10 23 44 43 2 50 37 61 27 40 5 28"
},
{
"input": "18\n2 16 8 4 18 12 3 6 5 9 10 15 11 17 14 13 1 7",
"output": "17 1 7 4 9 8 18 3 10 11 13 6 16 15 12 2 14 5"
},
{
"input": "47\n6 9 10 41 25 3 4 37 20 1 36 22 29 27 11 24 43 31 12 17 34 42 38 39 13 2 7 21 18 5 15 35 44 26 33 46 19 40 30 14 28 23 47 32 45 8 16",
"output": "10 26 6 7 30 1 27 46 2 3 15 19 25 40 31 47 20 29 37 9 28 12 42 16 5 34 14 41 13 39 18 44 35 21 32 11 8 23 24 38 4 22 17 33 45 36 43"
},
{
"input": "96\n41 91 48 88 29 57 1 19 44 43 37 5 10 75 25 63 30 78 76 53 8 92 18 70 39 17 49 60 9 16 3 34 86 59 23 79 55 45 72 51 28 33 96 40 26 54 6 32 89 61 85 74 7 82 52 31 64 66 94 95 11 22 2 73 35 13 42 71 14 47 84 69 50 67 58 12 77 46 38 68 15 36 20 93 27 90 83 56 87 4 21 24 81 62 80 65",
"output": "7 63 31 90 12 47 53 21 29 13 61 76 66 69 81 30 26 23 8 83 91 62 35 92 15 45 85 41 5 17 56 48 42 32 65 82 11 79 25 44 1 67 10 9 38 78 70 3 27 73 40 55 20 46 37 88 6 75 34 28 50 94 16 57 96 58 74 80 72 24 68 39 64 52 14 19 77 18 36 95 93 54 87 71 51 33 89 4 49 86 2 22 84 59 60 43"
},
{
"input": "73\n67 24 39 22 23 20 48 34 42 40 19 70 65 69 64 21 53 11 59 15 26 10 30 33 72 29 55 25 56 71 8 9 57 49 41 61 13 12 6 27 66 36 47 50 73 60 2 37 7 4 51 17 1 46 14 62 35 3 45 63 43 58 54 32 31 5 28 44 18 52 68 38 16",
"output": "53 47 58 50 66 39 49 31 32 22 18 38 37 55 20 73 52 69 11 6 16 4 5 2 28 21 40 67 26 23 65 64 24 8 57 42 48 72 3 10 35 9 61 68 59 54 43 7 34 44 51 70 17 63 27 29 33 62 19 46 36 56 60 15 13 41 1 71 14 12 30 25 45"
},
{
"input": "81\n25 2 78 40 12 80 69 13 49 43 17 33 23 54 32 61 77 66 27 71 24 26 42 55 60 9 5 30 7 37 45 63 53 11 38 44 68 34 28 52 67 22 57 46 47 50 8 16 79 62 4 36 20 14 73 64 6 76 35 74 58 10 29 81 59 31 19 1 75 39 70 18 41 21 72 65 3 48 15 56 51",
"output": "68 2 77 51 27 57 29 47 26 62 34 5 8 54 79 48 11 72 67 53 74 42 13 21 1 22 19 39 63 28 66 15 12 38 59 52 30 35 70 4 73 23 10 36 31 44 45 78 9 46 81 40 33 14 24 80 43 61 65 25 16 50 32 56 76 18 41 37 7 71 20 75 55 60 69 58 17 3 49 6 64"
},
{
"input": "12\n12 3 1 5 11 6 7 10 2 8 9 4",
"output": "3 9 2 12 4 6 7 10 11 8 5 1"
},
{
"input": "47\n7 21 41 18 40 31 12 28 24 14 43 23 33 10 19 38 26 8 34 15 29 44 5 13 39 25 3 27 20 42 35 9 2 1 30 46 36 32 4 22 37 45 6 47 11 16 17",
"output": "34 33 27 39 23 43 1 18 32 14 45 7 24 10 20 46 47 4 15 29 2 40 12 9 26 17 28 8 21 35 6 38 13 19 31 37 41 16 25 5 3 30 11 22 42 36 44"
},
{
"input": "8\n1 3 5 2 4 8 6 7",
"output": "1 4 2 5 3 7 8 6"
},
{
"input": "38\n28 8 2 33 20 32 26 29 23 31 15 38 11 37 18 21 22 19 4 34 1 35 16 7 17 6 27 30 36 12 9 24 25 13 5 3 10 14",
"output": "21 3 36 19 35 26 24 2 31 37 13 30 34 38 11 23 25 15 18 5 16 17 9 32 33 7 27 1 8 28 10 6 4 20 22 29 14 12"
},
{
"input": "10\n2 9 4 6 10 1 7 5 3 8",
"output": "6 1 9 3 8 4 7 10 2 5"
},
{
"input": "23\n20 11 15 1 5 12 23 9 2 22 13 19 16 14 7 4 8 21 6 17 18 10 3",
"output": "4 9 23 16 5 19 15 17 8 22 2 6 11 14 3 13 20 21 12 1 18 10 7"
},
{
"input": "10\n2 4 9 3 6 8 10 5 1 7",
"output": "9 1 4 2 8 5 10 6 3 7"
},
{
"input": "55\n9 48 23 49 11 24 4 22 34 32 17 45 39 13 14 21 19 25 2 31 37 7 55 36 20 51 5 12 54 10 35 40 43 1 46 18 53 41 38 26 29 50 3 42 52 27 8 28 47 33 6 16 30 44 15",
"output": "34 19 43 7 27 51 22 47 1 30 5 28 14 15 55 52 11 36 17 25 16 8 3 6 18 40 46 48 41 53 20 10 50 9 31 24 21 39 13 32 38 44 33 54 12 35 49 2 4 42 26 45 37 29 23"
},
{
"input": "58\n49 13 12 54 2 38 56 11 33 25 26 19 28 8 23 41 20 36 46 55 15 35 9 7 32 37 58 6 3 14 47 31 40 30 53 44 4 50 29 34 10 43 39 57 5 22 27 45 51 42 24 16 18 21 52 17 48 1",
"output": "58 5 29 37 45 28 24 14 23 41 8 3 2 30 21 52 56 53 12 17 54 46 15 51 10 11 47 13 39 34 32 25 9 40 22 18 26 6 43 33 16 50 42 36 48 19 31 57 1 38 49 55 35 4 20 7 44 27"
},
{
"input": "34\n20 25 2 3 33 29 1 16 14 7 21 9 32 31 6 26 22 4 27 23 24 10 34 12 19 15 5 18 28 17 13 8 11 30",
"output": "7 3 4 18 27 15 10 32 12 22 33 24 31 9 26 8 30 28 25 1 11 17 20 21 2 16 19 29 6 34 14 13 5 23"
},
{
"input": "53\n47 29 46 25 23 13 7 31 33 4 38 11 35 16 42 14 15 43 34 39 28 18 6 45 30 1 40 20 2 37 5 32 24 12 44 26 27 3 19 51 36 21 22 9 10 50 41 48 49 53 8 17 52",
"output": "26 29 38 10 31 23 7 51 44 45 12 34 6 16 17 14 52 22 39 28 42 43 5 33 4 36 37 21 2 25 8 32 9 19 13 41 30 11 20 27 47 15 18 35 24 3 1 48 49 46 40 53 50"
},
{
"input": "99\n77 87 90 48 53 38 68 6 28 57 35 82 63 71 60 41 3 12 86 65 10 59 22 67 33 74 93 27 24 1 61 43 25 4 51 52 15 88 9 31 30 42 89 49 23 21 29 32 46 73 37 16 5 69 56 26 92 64 20 54 75 14 98 13 94 2 95 7 36 66 58 8 50 78 84 45 11 96 76 62 97 80 40 39 47 85 34 79 83 17 91 72 19 44 70 81 55 99 18",
"output": "30 66 17 34 53 8 68 72 39 21 77 18 64 62 37 52 90 99 93 59 46 23 45 29 33 56 28 9 47 41 40 48 25 87 11 69 51 6 84 83 16 42 32 94 76 49 85 4 44 73 35 36 5 60 97 55 10 71 22 15 31 80 13 58 20 70 24 7 54 95 14 92 50 26 61 79 1 74 88 82 96 12 89 75 86 19 2 38 43 3 91 57 27 65 67 78 81 63 98"
},
{
"input": "32\n17 29 2 6 30 8 26 7 1 27 10 9 13 24 31 21 15 19 22 18 4 11 25 28 32 3 23 12 5 14 20 16",
"output": "9 3 26 21 29 4 8 6 12 11 22 28 13 30 17 32 1 20 18 31 16 19 27 14 23 7 10 24 2 5 15 25"
},
{
"input": "65\n18 40 1 60 17 19 4 6 12 49 28 58 2 25 13 14 64 56 61 34 62 30 59 51 26 8 33 63 36 48 46 7 43 21 31 27 11 44 29 5 32 23 35 9 53 57 52 50 15 38 42 3 54 65 55 41 20 24 22 47 45 10 39 16 37",
"output": "3 13 52 7 40 8 32 26 44 62 37 9 15 16 49 64 5 1 6 57 34 59 42 58 14 25 36 11 39 22 35 41 27 20 43 29 65 50 63 2 56 51 33 38 61 31 60 30 10 48 24 47 45 53 55 18 46 12 23 4 19 21 28 17 54"
},
{
"input": "71\n35 50 55 58 25 32 26 40 63 34 44 53 24 18 37 7 64 27 56 65 1 19 2 43 42 14 57 47 22 13 59 61 39 67 30 45 54 38 33 48 6 5 3 69 36 21 41 4 16 46 20 17 15 12 10 70 68 23 60 31 52 29 66 28 51 49 62 11 8 9 71",
"output": "21 23 43 48 42 41 16 69 70 55 68 54 30 26 53 49 52 14 22 51 46 29 58 13 5 7 18 64 62 35 60 6 39 10 1 45 15 38 33 8 47 25 24 11 36 50 28 40 66 2 65 61 12 37 3 19 27 4 31 59 32 67 9 17 20 63 34 57 44 56 71"
},
{
"input": "74\n33 8 42 63 64 61 31 74 11 50 68 14 36 25 57 30 7 44 21 15 6 9 23 59 46 3 73 16 62 51 40 60 41 54 5 39 35 28 48 4 58 12 66 69 13 26 71 1 24 19 29 52 37 2 20 43 18 72 17 56 34 38 65 67 27 10 47 70 53 32 45 55 49 22",
"output": "48 54 26 40 35 21 17 2 22 66 9 42 45 12 20 28 59 57 50 55 19 74 23 49 14 46 65 38 51 16 7 70 1 61 37 13 53 62 36 31 33 3 56 18 71 25 67 39 73 10 30 52 69 34 72 60 15 41 24 32 6 29 4 5 63 43 64 11 44 68 47 58 27 8"
},
{
"input": "96\n78 10 82 46 38 91 77 69 2 27 58 80 79 44 59 41 6 31 76 11 42 48 51 37 19 87 43 25 52 32 1 39 63 29 21 65 53 74 92 16 15 95 90 83 30 73 71 5 50 17 96 33 86 60 67 64 20 26 61 40 55 88 94 93 9 72 47 57 14 45 22 3 54 68 13 24 4 7 56 81 89 70 49 8 84 28 18 62 35 36 75 23 66 85 34 12",
"output": "31 9 72 77 48 17 78 84 65 2 20 96 75 69 41 40 50 87 25 57 35 71 92 76 28 58 10 86 34 45 18 30 52 95 89 90 24 5 32 60 16 21 27 14 70 4 67 22 83 49 23 29 37 73 61 79 68 11 15 54 59 88 33 56 36 93 55 74 8 82 47 66 46 38 91 19 7 1 13 12 80 3 44 85 94 53 26 62 81 43 6 39 64 63 42 51"
},
{
"input": "7\n2 1 5 7 3 4 6",
"output": "2 1 5 6 3 7 4"
},
{
"input": "51\n8 33 37 2 16 22 24 30 4 9 5 15 27 3 18 39 31 26 10 17 46 41 25 14 6 1 29 48 36 20 51 49 21 43 19 13 38 50 47 34 11 23 28 12 42 7 32 40 44 45 35",
"output": "26 4 14 9 11 25 46 1 10 19 41 44 36 24 12 5 20 15 35 30 33 6 42 7 23 18 13 43 27 8 17 47 2 40 51 29 3 37 16 48 22 45 34 49 50 21 39 28 32 38 31"
},
{
"input": "27\n12 14 7 3 20 21 25 13 22 15 23 4 2 24 10 17 19 8 26 11 27 18 9 5 6 1 16",
"output": "26 13 4 12 24 25 3 18 23 15 20 1 8 2 10 27 16 22 17 5 6 9 11 14 7 19 21"
},
{
"input": "71\n51 13 20 48 54 23 24 64 14 62 71 67 57 53 3 30 55 43 33 25 39 40 66 6 46 18 5 19 61 16 32 68 70 41 60 44 29 49 27 69 50 38 10 17 45 56 9 21 26 63 28 35 7 59 1 65 2 15 8 11 12 34 37 47 58 22 31 4 36 42 52",
"output": "55 57 15 68 27 24 53 59 47 43 60 61 2 9 58 30 44 26 28 3 48 66 6 7 20 49 39 51 37 16 67 31 19 62 52 69 63 42 21 22 34 70 18 36 45 25 64 4 38 41 1 71 14 5 17 46 13 65 54 35 29 10 50 8 56 23 12 32 40 33 11"
},
{
"input": "9\n8 5 2 6 1 9 4 7 3",
"output": "5 3 9 7 2 4 8 1 6"
},
{
"input": "29\n10 24 11 5 26 25 2 9 22 15 8 14 29 21 4 1 23 17 3 12 13 16 18 28 19 20 7 6 27",
"output": "16 7 19 15 4 28 27 11 8 1 3 20 21 12 10 22 18 23 25 26 14 9 17 2 6 5 29 24 13"
},
{
"input": "60\n39 25 42 4 55 60 16 18 47 1 11 40 7 50 19 35 49 54 12 3 30 38 2 58 17 26 45 6 33 43 37 32 52 36 15 23 27 59 24 20 28 14 8 9 13 29 44 46 41 21 5 48 51 22 31 56 57 53 10 34",
"output": "10 23 20 4 51 28 13 43 44 59 11 19 45 42 35 7 25 8 15 40 50 54 36 39 2 26 37 41 46 21 55 32 29 60 16 34 31 22 1 12 49 3 30 47 27 48 9 52 17 14 53 33 58 18 5 56 57 24 38 6"
},
{
"input": "50\n37 45 22 5 12 21 28 24 18 47 20 25 8 50 14 2 34 43 11 16 49 41 48 1 19 31 39 46 32 23 15 42 3 35 38 30 44 26 10 9 40 36 7 17 33 4 27 6 13 29",
"output": "24 16 33 46 4 48 43 13 40 39 19 5 49 15 31 20 44 9 25 11 6 3 30 8 12 38 47 7 50 36 26 29 45 17 34 42 1 35 27 41 22 32 18 37 2 28 10 23 21 14"
},
{
"input": "30\n8 29 28 16 17 25 27 15 21 11 6 20 2 13 1 30 5 4 24 10 14 3 23 18 26 9 12 22 19 7",
"output": "15 13 22 18 17 11 30 1 26 20 10 27 14 21 8 4 5 24 29 12 9 28 23 19 6 25 7 3 2 16"
},
{
"input": "46\n15 2 44 43 38 19 31 42 4 37 29 30 24 45 27 41 8 20 33 7 35 3 18 46 36 26 1 28 21 40 16 22 32 11 14 13 12 9 25 39 10 6 23 17 5 34",
"output": "27 2 22 9 45 42 20 17 38 41 34 37 36 35 1 31 44 23 6 18 29 32 43 13 39 26 15 28 11 12 7 33 19 46 21 25 10 5 40 30 16 8 4 3 14 24"
},
{
"input": "9\n4 8 6 5 3 9 2 7 1",
"output": "9 7 5 1 4 3 8 2 6"
},
{
"input": "46\n31 30 33 23 45 7 36 8 11 3 32 39 41 20 1 28 6 27 18 24 17 5 16 37 26 13 22 14 2 38 15 46 9 4 19 21 12 44 10 35 25 34 42 43 40 29",
"output": "15 29 10 34 22 17 6 8 33 39 9 37 26 28 31 23 21 19 35 14 36 27 4 20 41 25 18 16 46 2 1 11 3 42 40 7 24 30 12 45 13 43 44 38 5 32"
},
{
"input": "66\n27 12 37 48 46 21 34 58 38 28 66 2 64 32 44 31 13 36 40 15 19 11 22 5 30 29 6 7 61 39 20 42 23 54 51 33 50 9 60 8 57 45 49 10 62 41 59 3 55 63 52 24 25 26 43 56 65 4 16 14 1 35 18 17 53 47",
"output": "61 12 48 58 24 27 28 40 38 44 22 2 17 60 20 59 64 63 21 31 6 23 33 52 53 54 1 10 26 25 16 14 36 7 62 18 3 9 30 19 46 32 55 15 42 5 66 4 43 37 35 51 65 34 49 56 41 8 47 39 29 45 50 13 57 11"
},
{
"input": "13\n3 12 9 2 8 5 13 4 11 1 10 7 6",
"output": "10 4 1 8 6 13 12 5 3 11 9 2 7"
},
{
"input": "80\n21 25 56 50 20 61 7 74 51 69 8 2 46 57 45 71 14 52 17 43 9 30 70 78 31 10 38 13 23 15 37 79 6 16 77 73 80 4 49 48 18 28 26 58 33 41 64 22 54 72 59 60 40 63 53 27 1 5 75 67 62 34 19 39 68 65 44 55 3 32 11 42 76 12 35 47 66 36 24 29",
"output": "57 12 69 38 58 33 7 11 21 26 71 74 28 17 30 34 19 41 63 5 1 48 29 79 2 43 56 42 80 22 25 70 45 62 75 78 31 27 64 53 46 72 20 67 15 13 76 40 39 4 9 18 55 49 68 3 14 44 51 52 6 61 54 47 66 77 60 65 10 23 16 50 36 8 59 73 35 24 32 37"
},
{
"input": "63\n9 49 53 25 40 46 43 51 54 22 58 16 23 26 10 47 5 27 2 8 61 59 19 35 63 56 28 20 34 4 62 38 6 55 36 31 57 15 29 33 1 48 50 37 7 30 18 42 32 52 12 41 14 21 45 11 24 17 39 13 44 60 3",
"output": "41 19 63 30 17 33 45 20 1 15 56 51 60 53 38 12 58 47 23 28 54 10 13 57 4 14 18 27 39 46 36 49 40 29 24 35 44 32 59 5 52 48 7 61 55 6 16 42 2 43 8 50 3 9 34 26 37 11 22 62 21 31 25"
},
{
"input": "26\n11 4 19 13 17 9 2 24 6 5 22 23 14 15 3 25 16 8 18 10 21 1 12 26 7 20",
"output": "22 7 15 2 10 9 25 18 6 20 1 23 4 13 14 17 5 19 3 26 21 11 12 8 16 24"
},
{
"input": "69\n40 22 11 66 4 27 31 29 64 53 37 55 51 2 7 36 18 52 6 1 30 21 17 20 14 9 59 62 49 68 3 50 65 57 44 5 67 46 33 13 34 15 24 48 63 58 38 25 41 35 16 54 32 10 60 61 39 12 69 8 23 45 26 47 56 43 28 19 42",
"output": "20 14 31 5 36 19 15 60 26 54 3 58 40 25 42 51 23 17 68 24 22 2 61 43 48 63 6 67 8 21 7 53 39 41 50 16 11 47 57 1 49 69 66 35 62 38 64 44 29 32 13 18 10 52 12 65 34 46 27 55 56 28 45 9 33 4 37 30 59"
},
{
"input": "6\n4 3 6 5 1 2",
"output": "5 6 2 1 4 3"
},
{
"input": "9\n7 8 5 3 1 4 2 9 6",
"output": "5 7 4 6 3 9 1 2 8"
},
{
"input": "41\n27 24 16 30 25 8 32 2 26 20 39 33 41 22 40 14 36 9 28 4 34 11 31 23 19 18 17 35 3 10 6 13 5 15 29 38 7 21 1 12 37",
"output": "39 8 29 20 33 31 37 6 18 30 22 40 32 16 34 3 27 26 25 10 38 14 24 2 5 9 1 19 35 4 23 7 12 21 28 17 41 36 11 15 13"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "20\n2 6 4 18 7 10 17 13 16 8 14 9 20 5 19 12 1 3 15 11",
"output": "17 1 18 3 14 2 5 10 12 6 20 16 8 11 19 9 7 4 15 13"
},
{
"input": "2\n2 1",
"output": "2 1"
},
{
"input": "60\n2 4 31 51 11 7 34 20 3 14 18 23 48 54 15 36 38 60 49 40 5 33 41 26 55 58 10 8 13 9 27 30 37 1 21 59 44 57 35 19 46 43 42 45 12 22 39 32 24 16 6 56 53 52 25 17 47 29 50 28",
"output": "34 1 9 2 21 51 6 28 30 27 5 45 29 10 15 50 56 11 40 8 35 46 12 49 55 24 31 60 58 32 3 48 22 7 39 16 33 17 47 20 23 43 42 37 44 41 57 13 19 59 4 54 53 14 25 52 38 26 36 18"
},
{
"input": "14\n14 6 3 12 11 2 7 1 10 9 8 5 4 13",
"output": "8 6 3 13 12 2 7 11 10 9 5 4 14 1"
},
{
"input": "81\n13 43 79 8 7 21 73 46 63 4 62 78 56 11 70 68 61 53 60 49 16 27 59 47 69 5 22 44 77 57 52 48 1 9 72 81 28 55 58 33 51 18 31 17 41 20 42 3 32 54 19 2 75 34 64 10 65 50 30 29 67 12 71 66 74 15 26 23 6 38 25 35 37 24 80 76 40 45 39 36 14",
"output": "33 52 48 10 26 69 5 4 34 56 14 62 1 81 66 21 44 42 51 46 6 27 68 74 71 67 22 37 60 59 43 49 40 54 72 80 73 70 79 77 45 47 2 28 78 8 24 32 20 58 41 31 18 50 38 13 30 39 23 19 17 11 9 55 57 64 61 16 25 15 63 35 7 65 53 76 29 12 3 75 36"
},
{
"input": "42\n41 11 10 8 21 37 32 19 31 25 1 15 36 5 6 27 4 3 13 7 16 17 2 23 34 24 38 28 12 20 30 42 18 26 39 35 33 40 9 14 22 29",
"output": "11 23 18 17 14 15 20 4 39 3 2 29 19 40 12 21 22 33 8 30 5 41 24 26 10 34 16 28 42 31 9 7 37 25 36 13 6 27 35 38 1 32"
},
{
"input": "97\n20 6 76 42 4 18 35 59 39 63 27 7 66 47 61 52 15 36 88 93 19 33 10 92 1 34 46 86 78 57 51 94 77 29 26 73 41 2 58 97 43 65 17 74 21 49 25 3 91 82 95 12 96 13 84 90 69 24 72 37 16 55 54 71 64 62 48 89 11 70 80 67 30 40 44 85 53 83 79 9 56 45 75 87 22 14 81 68 8 38 60 50 28 23 31 32 5",
"output": "25 38 48 5 97 2 12 89 80 23 69 52 54 86 17 61 43 6 21 1 45 85 94 58 47 35 11 93 34 73 95 96 22 26 7 18 60 90 9 74 37 4 41 75 82 27 14 67 46 92 31 16 77 63 62 81 30 39 8 91 15 66 10 65 42 13 72 88 57 70 64 59 36 44 83 3 33 29 79 71 87 50 78 55 76 28 84 19 68 56 49 24 20 32 51 53 40"
},
{
"input": "62\n15 27 46 6 8 51 14 56 23 48 42 49 52 22 20 31 29 12 47 3 62 34 37 35 32 57 19 25 5 60 61 38 18 10 11 55 45 53 17 30 9 36 4 50 41 16 44 28 40 59 24 1 13 39 26 7 33 58 2 43 21 54",
"output": "52 59 20 43 29 4 56 5 41 34 35 18 53 7 1 46 39 33 27 15 61 14 9 51 28 55 2 48 17 40 16 25 57 22 24 42 23 32 54 49 45 11 60 47 37 3 19 10 12 44 6 13 38 62 36 8 26 58 50 30 31 21"
},
{
"input": "61\n35 27 4 61 52 32 41 46 14 37 17 54 55 31 11 26 44 49 15 30 9 50 45 39 7 38 53 3 58 40 13 56 18 19 28 6 43 5 21 42 20 34 2 25 36 12 33 57 16 60 1 8 59 10 22 23 24 48 51 47 29",
"output": "51 43 28 3 38 36 25 52 21 54 15 46 31 9 19 49 11 33 34 41 39 55 56 57 44 16 2 35 61 20 14 6 47 42 1 45 10 26 24 30 7 40 37 17 23 8 60 58 18 22 59 5 27 12 13 32 48 29 53 50 4"
},
{
"input": "59\n31 26 36 15 17 19 10 53 11 34 13 46 55 9 44 7 8 37 32 52 47 25 51 22 35 39 41 4 43 24 5 27 20 57 6 38 3 28 21 40 50 18 14 56 33 45 12 2 49 59 54 29 16 48 42 58 1 30 23",
"output": "57 48 37 28 31 35 16 17 14 7 9 47 11 43 4 53 5 42 6 33 39 24 59 30 22 2 32 38 52 58 1 19 45 10 25 3 18 36 26 40 27 55 29 15 46 12 21 54 49 41 23 20 8 51 13 44 34 56 50"
},
{
"input": "10\n2 10 7 4 1 5 8 6 3 9",
"output": "5 1 9 4 6 8 3 7 10 2"
},
{
"input": "14\n14 2 1 8 6 12 11 10 9 7 3 4 5 13",
"output": "3 2 11 12 13 5 10 4 9 8 7 6 14 1"
},
{
"input": "43\n28 38 15 14 31 42 27 30 19 33 43 26 22 29 18 32 3 13 1 8 35 34 4 12 11 17 41 21 5 25 39 37 20 23 7 24 16 10 40 9 6 36 2",
"output": "19 43 17 23 29 41 35 20 40 38 25 24 18 4 3 37 26 15 9 33 28 13 34 36 30 12 7 1 14 8 5 16 10 22 21 42 32 2 31 39 27 6 11"
},
{
"input": "86\n39 11 20 31 28 76 29 64 35 21 41 71 12 82 5 37 80 73 38 26 79 75 23 15 59 45 47 6 3 62 50 49 51 22 2 65 86 60 70 42 74 17 1 30 55 44 8 66 81 27 57 77 43 13 54 32 72 46 48 56 14 34 78 52 36 85 24 19 69 83 25 61 7 4 84 33 63 58 18 40 68 10 67 9 16 53",
"output": "43 35 29 74 15 28 73 47 84 82 2 13 54 61 24 85 42 79 68 3 10 34 23 67 71 20 50 5 7 44 4 56 76 62 9 65 16 19 1 80 11 40 53 46 26 58 27 59 32 31 33 64 86 55 45 60 51 78 25 38 72 30 77 8 36 48 83 81 69 39 12 57 18 41 22 6 52 63 21 17 49 14 70 75 66 37"
},
{
"input": "99\n65 78 56 98 33 24 61 40 29 93 1 64 57 22 25 52 67 95 50 3 31 15 90 68 71 83 38 36 6 46 89 26 4 87 14 88 72 37 23 43 63 12 80 96 5 34 73 86 9 48 92 62 99 10 16 20 66 27 28 2 82 70 30 94 49 8 84 69 18 60 58 59 44 39 21 7 91 76 54 19 75 85 74 47 55 32 97 77 51 13 35 79 45 42 11 41 17 81 53",
"output": "11 60 20 33 45 29 76 66 49 54 95 42 90 35 22 55 97 69 80 56 75 14 39 6 15 32 58 59 9 63 21 86 5 46 91 28 38 27 74 8 96 94 40 73 93 30 84 50 65 19 89 16 99 79 85 3 13 71 72 70 7 52 41 12 1 57 17 24 68 62 25 37 47 83 81 78 88 2 92 43 98 61 26 67 82 48 34 36 31 23 77 51 10 64 18 44 87 4 53"
},
{
"input": "100\n42 23 48 88 36 6 18 70 96 1 34 40 46 22 39 55 85 93 45 67 71 75 59 9 21 3 86 63 65 68 20 38 73 31 84 90 50 51 56 95 72 33 49 19 83 76 54 74 100 30 17 98 15 94 4 97 5 99 81 27 92 32 89 12 13 91 87 29 60 11 52 43 35 58 10 25 16 80 28 2 44 61 8 82 66 69 41 24 57 62 78 37 79 77 53 7 14 47 26 64",
"output": "10 80 26 55 57 6 96 83 24 75 70 64 65 97 53 77 51 7 44 31 25 14 2 88 76 99 60 79 68 50 34 62 42 11 73 5 92 32 15 12 87 1 72 81 19 13 98 3 43 37 38 71 95 47 16 39 89 74 23 69 82 90 28 100 29 85 20 30 86 8 21 41 33 48 22 46 94 91 93 78 59 84 45 35 17 27 67 4 63 36 66 61 18 54 40 9 56 52 58 49"
},
{
"input": "99\n8 68 94 75 71 60 57 58 6 11 5 48 65 41 49 12 46 72 95 59 13 70 74 7 84 62 17 36 55 76 38 79 2 85 23 10 32 99 87 50 83 28 54 91 53 51 1 3 97 81 21 89 93 78 61 26 82 96 4 98 25 40 31 44 24 47 30 52 14 16 39 27 9 29 45 18 67 63 37 43 90 66 19 69 88 22 92 77 34 42 73 80 56 64 20 35 15 33 86",
"output": "47 33 48 59 11 9 24 1 73 36 10 16 21 69 97 70 27 76 83 95 51 86 35 65 61 56 72 42 74 67 63 37 98 89 96 28 79 31 71 62 14 90 80 64 75 17 66 12 15 40 46 68 45 43 29 93 7 8 20 6 55 26 78 94 13 82 77 2 84 22 5 18 91 23 4 30 88 54 32 92 50 57 41 25 34 99 39 85 52 81 44 87 53 3 19 58 49 60 38"
},
{
"input": "99\n12 99 88 13 7 19 74 47 23 90 16 29 26 11 58 60 64 98 37 18 82 67 72 46 51 85 17 92 87 20 77 36 78 71 57 35 80 54 73 15 14 62 97 45 31 79 94 56 76 96 28 63 8 44 38 86 49 2 52 66 61 59 10 43 55 50 22 34 83 53 95 40 81 21 30 42 27 3 5 41 1 70 69 25 93 48 65 6 24 89 91 33 39 68 9 4 32 84 75",
"output": "81 58 78 96 79 88 5 53 95 63 14 1 4 41 40 11 27 20 6 30 74 67 9 89 84 13 77 51 12 75 45 97 92 68 36 32 19 55 93 72 80 76 64 54 44 24 8 86 57 66 25 59 70 38 65 48 35 15 62 16 61 42 52 17 87 60 22 94 83 82 34 23 39 7 99 49 31 33 46 37 73 21 69 98 26 56 29 3 90 10 91 28 85 47 71 50 43 18 2"
},
{
"input": "99\n20 79 26 75 99 69 98 47 93 62 18 42 43 38 90 66 67 8 13 84 76 58 81 60 64 46 56 23 78 17 86 36 19 52 85 39 48 27 96 49 37 95 5 31 10 24 12 1 80 35 92 33 16 68 57 54 32 29 45 88 72 77 4 87 97 89 59 3 21 22 61 94 83 15 44 34 70 91 55 9 51 50 73 11 14 6 40 7 63 25 2 82 41 65 28 74 71 30 53",
"output": "48 91 68 63 43 86 88 18 80 45 84 47 19 85 74 53 30 11 33 1 69 70 28 46 90 3 38 95 58 98 44 57 52 76 50 32 41 14 36 87 93 12 13 75 59 26 8 37 40 82 81 34 99 56 79 27 55 22 67 24 71 10 89 25 94 16 17 54 6 77 97 61 83 96 4 21 62 29 2 49 23 92 73 20 35 31 64 60 66 15 78 51 9 72 42 39 65 7 5"
},
{
"input": "99\n74 20 9 1 60 85 65 13 4 25 40 99 5 53 64 3 36 31 73 44 55 50 45 63 98 51 68 6 47 37 71 82 88 34 84 18 19 12 93 58 86 7 11 46 90 17 33 27 81 69 42 59 56 32 95 52 76 61 96 62 78 43 66 21 49 97 75 14 41 72 89 16 30 79 22 23 15 83 91 38 48 2 87 26 28 80 94 70 54 92 57 10 8 35 67 77 29 24 39",
"output": "4 82 16 9 13 28 42 93 3 92 43 38 8 68 77 72 46 36 37 2 64 75 76 98 10 84 48 85 97 73 18 54 47 34 94 17 30 80 99 11 69 51 62 20 23 44 29 81 65 22 26 56 14 89 21 53 91 40 52 5 58 60 24 15 7 63 95 27 50 88 31 70 19 1 67 57 96 61 74 86 49 32 78 35 6 41 83 33 71 45 79 90 39 87 55 59 66 25 12"
},
{
"input": "99\n50 94 2 18 69 90 59 83 75 68 77 97 39 78 25 7 16 9 49 4 42 89 44 48 17 96 61 70 3 10 5 81 56 57 88 6 98 1 46 67 92 37 11 30 85 41 8 36 51 29 20 71 19 79 74 93 43 34 55 40 38 21 64 63 32 24 72 14 12 86 82 15 65 23 66 22 28 53 13 26 95 99 91 52 76 27 60 45 47 33 73 84 31 35 54 80 58 62 87",
"output": "38 3 29 20 31 36 16 47 18 30 43 69 79 68 72 17 25 4 53 51 62 76 74 66 15 80 86 77 50 44 93 65 90 58 94 48 42 61 13 60 46 21 57 23 88 39 89 24 19 1 49 84 78 95 59 33 34 97 7 87 27 98 64 63 73 75 40 10 5 28 52 67 91 55 9 85 11 14 54 96 32 71 8 92 45 70 99 35 22 6 83 41 56 2 81 26 12 37 82"
},
{
"input": "99\n19 93 14 34 39 37 33 15 52 88 7 43 69 27 9 77 94 31 48 22 63 70 79 17 50 6 81 8 76 58 23 74 86 11 57 62 41 87 75 51 12 18 68 56 95 3 80 83 84 29 24 61 71 78 59 96 20 85 90 28 45 36 38 97 1 49 40 98 44 67 13 73 72 91 47 10 30 54 35 42 4 2 92 26 64 60 53 21 5 82 46 32 55 66 16 89 99 65 25",
"output": "65 82 46 81 89 26 11 28 15 76 34 41 71 3 8 95 24 42 1 57 88 20 31 51 99 84 14 60 50 77 18 92 7 4 79 62 6 63 5 67 37 80 12 69 61 91 75 19 66 25 40 9 87 78 93 44 35 30 55 86 52 36 21 85 98 94 70 43 13 22 53 73 72 32 39 29 16 54 23 47 27 90 48 49 58 33 38 10 96 59 74 83 2 17 45 56 64 68 97"
},
{
"input": "99\n86 25 50 51 62 39 41 67 44 20 45 14 80 88 66 7 36 59 13 84 78 58 96 75 2 43 48 47 69 12 19 98 22 38 28 55 11 76 68 46 53 70 85 34 16 33 91 30 8 40 74 60 94 82 87 32 37 4 5 10 89 73 90 29 35 26 23 57 27 65 24 3 9 83 77 72 6 31 15 92 93 79 64 18 63 42 56 1 52 97 17 81 71 21 49 99 54 95 61",
"output": "88 25 72 58 59 77 16 49 73 60 37 30 19 12 79 45 91 84 31 10 94 33 67 71 2 66 69 35 64 48 78 56 46 44 65 17 57 34 6 50 7 86 26 9 11 40 28 27 95 3 4 89 41 97 36 87 68 22 18 52 99 5 85 83 70 15 8 39 29 42 93 76 62 51 24 38 75 21 82 13 92 54 74 20 43 1 55 14 61 63 47 80 81 53 98 23 90 32 96"
},
{
"input": "100\n66 44 99 15 43 79 28 33 88 90 49 68 82 38 9 74 4 58 29 81 31 94 10 42 89 21 63 40 62 61 18 6 84 72 48 25 67 69 71 85 98 34 83 70 65 78 91 77 93 41 23 24 87 11 55 12 59 73 36 97 7 14 26 39 30 27 45 20 50 17 53 2 57 47 95 56 75 19 37 96 16 35 8 3 76 60 13 86 5 32 64 80 46 51 54 100 1 22 52 92",
"output": "97 72 84 17 89 32 61 83 15 23 54 56 87 62 4 81 70 31 78 68 26 98 51 52 36 63 66 7 19 65 21 90 8 42 82 59 79 14 64 28 50 24 5 2 67 93 74 35 11 69 94 99 71 95 55 76 73 18 57 86 30 29 27 91 45 1 37 12 38 44 39 34 58 16 77 85 48 46 6 92 20 13 43 33 40 88 53 9 25 10 47 100 49 22 75 80 60 41 3 96"
},
{
"input": "99\n3 73 32 37 25 15 93 63 85 8 91 78 80 5 39 48 46 7 83 70 23 96 9 29 77 53 30 20 56 50 13 45 21 76 87 99 65 31 16 18 14 72 51 28 43 2 81 34 38 40 66 54 74 26 71 4 61 17 58 24 22 33 49 36 42 11 12 55 60 27 62 90 79 92 94 68 1 52 84 41 86 35 69 75 47 10 64 88 97 98 67 19 89 95 59 82 57 44 6",
"output": "77 46 1 56 14 99 18 10 23 86 66 67 31 41 6 39 58 40 92 28 33 61 21 60 5 54 70 44 24 27 38 3 62 48 82 64 4 49 15 50 80 65 45 98 32 17 85 16 63 30 43 78 26 52 68 29 97 59 95 69 57 71 8 87 37 51 91 76 83 20 55 42 2 53 84 34 25 12 73 13 47 96 19 79 9 81 35 88 93 72 11 74 7 75 94 22 89 90 36"
},
{
"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": "100 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"
}
] | 1,695,300,617 | 2,147,483,647 | Python 3 | OK | TESTS | 101 | 122 | 0 | n=int(input())
my_str=input()
my_list=[]
a=my_str.split()
h=1
while h<=n:
index=a.index(str(h))
my_list.append(str(index+1))
h +=1
result=' '.join(my_list)
print(result) | Title: Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Output Specification:
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
Demo Input:
['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n']
Demo Output:
['4 1 2 3\n', '1 3 2\n', '1 2\n']
Note:
none | ```python
n=int(input())
my_str=input()
my_list=[]
a=my_str.split()
h=1
while h<=n:
index=a.index(str(h))
my_list.append(str(index+1))
h +=1
result=' '.join(my_list)
print(result)
``` | 3 |
|
20 | C | Dijkstra? | PROGRAMMING | 1,900 | [
"graphs",
"shortest paths"
] | C. Dijkstra? | 1 | 64 | You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*. | The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | [
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n",
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n"
] | [
"1 4 3 5 ",
"1 4 3 5 "
] | none | 1,500 | [
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "2 1\n1 2 1",
"output": "1 2 "
},
{
"input": "3 1\n1 2 1",
"output": "-1"
},
{
"input": "3 3\n1 2 1\n1 3 2\n2 3 1",
"output": "1 3 "
},
{
"input": "10 10\n1 5 12\n2 4 140\n2 10 149\n3 6 154\n3 7 9\n3 8 226\n3 10 132\n4 10 55\n5 8 33\n7 8 173",
"output": "1 5 8 7 3 10 "
},
{
"input": "10 10\n1 5 178\n1 8 221\n2 7 92\n2 8 159\n3 5 55\n3 6 179\n3 10 237\n4 8 205\n5 6 191\n8 10 157",
"output": "1 8 10 "
},
{
"input": "10 10\n1 4 200\n1 9 197\n3 4 79\n3 5 213\n3 6 149\n5 8 3\n5 9 189\n6 7 130\n6 9 51\n8 10 135",
"output": "1 9 5 8 10 "
},
{
"input": "10 10\n1 4 201\n2 3 238\n3 4 40\n3 6 231\n3 8 45\n4 5 227\n4 6 58\n4 9 55\n5 7 14\n6 10 242",
"output": "1 4 6 10 "
}
] | 1,599,946,878 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 27 | 1,000 | 26,521,600 | import heapq
from collections import defaultdict
def distances(graph, start,leng):
d = {vertex: float('infinity') for vertex in range(1,leng+1)}
prev = {vertex: 1 for vertex in graph}
d[start] = 0
pq = [(0, start)]
while len(pq) > 0:
dis, ver = heapq.heappop(pq)
if dis > d[ver]:
continue
for nei, weight in graph[ver].items():
alt = dis + weight
if alt < d[nei]:
d[nei] = alt
heapq.heappush(pq, (alt, nei))
prev[nei] = ver
return prev, d
def backtrace(parent, start, end):
path = [end]
while path[-1] != start:
path.append(parent[path[-1]])
path.reverse()
return path
g = defaultdict(dict)
n, m = map(int, input().split())
for i in range(m):
u,v,w = map(int,input().split())
g[u][v] = w
g[v][u] = w
prev,d = distances(g,1,n)
print((' '.join(str(i) for i in backtrace(prev,1,n))) if d[n] < float('infinity') else -1) | Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none | ```python
import heapq
from collections import defaultdict
def distances(graph, start,leng):
d = {vertex: float('infinity') for vertex in range(1,leng+1)}
prev = {vertex: 1 for vertex in graph}
d[start] = 0
pq = [(0, start)]
while len(pq) > 0:
dis, ver = heapq.heappop(pq)
if dis > d[ver]:
continue
for nei, weight in graph[ver].items():
alt = dis + weight
if alt < d[nei]:
d[nei] = alt
heapq.heappush(pq, (alt, nei))
prev[nei] = ver
return prev, d
def backtrace(parent, start, end):
path = [end]
while path[-1] != start:
path.append(parent[path[-1]])
path.reverse()
return path
g = defaultdict(dict)
n, m = map(int, input().split())
for i in range(m):
u,v,w = map(int,input().split())
g[u][v] = w
g[v][u] = w
prev,d = distances(g,1,n)
print((' '.join(str(i) for i in backtrace(prev,1,n))) if d[n] < float('infinity') else -1)
``` | 0 |
785 | A | Anton and Polyhedrons | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number! | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron. | Output one number — the total number of faces in all the polyhedrons in Anton's collection. | [
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] | [
"42\n",
"28\n"
] | In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | 500 | [
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosahedron\nTetrahedron\nOctahedron\nDodecahedron\nIcosahedron\nOctahedron\nIcosahedron\nTetrahedron\nDodecahedron\nTetrahedron\nOctahedron\nCube\nCube\nDodecahedron\nTetrahedron",
"output": "256"
},
{
"input": "1\nTetrahedron",
"output": "4"
},
{
"input": "1\nCube",
"output": "6"
},
{
"input": "1\nOctahedron",
"output": "8"
},
{
"input": "1\nDodecahedron",
"output": "12"
},
{
"input": "1\nIcosahedron",
"output": "20"
},
{
"input": "28\nOctahedron\nDodecahedron\nOctahedron\nOctahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nDodecahedron\nDodecahedron\nCube\nDodecahedron\nCube\nTetrahedron\nCube\nCube\nTetrahedron\nDodecahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nIcosahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron",
"output": "340"
}
] | 1,674,398,724 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 30 | 1,091 | 9,216,000 | d = {"Tetrahedron":4, "Cube":6, "Octahedron":8, "Dodecahedron":12, "Icosahedron":20}
n = int(input())
c=0
for i in range(n):
c += d[str(input())]
print(c) | Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | ```python
d = {"Tetrahedron":4, "Cube":6, "Octahedron":8, "Dodecahedron":12, "Icosahedron":20}
n = int(input())
c=0
for i in range(n):
c += d[str(input())]
print(c)
``` | 3 |
|
527 | C | Glass Carving | PROGRAMMING | 1,500 | [
"binary search",
"data structures",
"implementation"
] | null | null | Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree? | The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts. | After each cut print on a single line the area of the maximum available glass fragment in mm2. | [
"4 3 4\nH 2\nV 2\nV 3\nV 1\n",
"7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n"
] | [
"8\n4\n4\n2\n",
"28\n16\n12\n6\n4\n"
] | Picture for the first sample test: | 1,500 | [
{
"input": "4 3 4\nH 2\nV 2\nV 3\nV 1",
"output": "8\n4\n4\n2"
},
{
"input": "7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1",
"output": "28\n16\n12\n6\n4"
},
{
"input": "2 2 1\nV 1",
"output": "2"
},
{
"input": "2 2 1\nH 1",
"output": "2"
},
{
"input": "2 2 2\nV 1\nH 1",
"output": "2\n1"
},
{
"input": "2 2 2\nH 1\nV 1",
"output": "2\n1"
},
{
"input": "10 10 10\nV 6\nH 8\nV 4\nV 8\nH 2\nH 5\nV 9\nH 7\nH 3\nV 7",
"output": "60\n48\n32\n32\n24\n12\n12\n12\n8\n8"
},
{
"input": "5 15 10\nH 8\nH 9\nV 1\nH 2\nH 6\nH 4\nH 1\nV 2\nH 13\nV 3",
"output": "40\n40\n32\n24\n24\n24\n24\n18\n12\n8"
},
{
"input": "15 5 10\nV 13\nV 10\nV 3\nH 2\nV 9\nV 7\nV 2\nH 1\nV 4\nH 3",
"output": "65\n50\n35\n21\n18\n12\n12\n12\n9\n6"
},
{
"input": "2 3 1\nH 1",
"output": "4"
},
{
"input": "200000 200000 1\nH 1",
"output": "39999800000"
},
{
"input": "2 4 1\nH 2",
"output": "4"
}
] | 1,568,708,618 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 93 | 307,200 | __author__ = 'ruckus'
class Split:
def __init__(self, order, offset) -> None:
super().__init__()
self.order = order
self.offset = int(offset)
class Glass:
h_splits = [0]
w_splits = [0]
max_p = 0
max_p_h_offset = 0
max_p_w_offset = 0
def __init__(self, h, w) -> None:
super().__init__()
self.h_splits.append(h)
self.w_splits.append(w)
self.max_p = h * w
def split(self, split: Split):
if split.order == 'H':
self.h_splits.append(split.offset)
self.h_splits = self.h_splits.sort()
else:
self.w_splits.append(split.offset)
self.w_splits = self.w_splits.sort()
max_p = 0
max_p_h_offset = 0
max_p_w_offset = 0
for i in range(1,len(self.h_splits)):
for j in range(1,len(self.w_splits)):
p = (self.h_splits[i] - self.h_splits[i - 1]) * (self.w_splits[j] - self.w_splits[j - 1])
if p > max_p:
max_p = p
max_p_h_offset = self.h_splits[i - 1]
max_p_w_offset = self.w_splits[j - 1]
self.max_p = max_p
self.max_p_h_offset = max_p_h_offset
self.max_p_w_offset = max_p_w_offset
print(max_p)
w, h, n = map(int, input().split())
glass = Glass(h, w)
for i in range(n):
order, offset = input().split()
glass.split(Split(order, offset)) | Title: Glass Carving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?
Input Specification:
The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.
Output Specification:
After each cut print on a single line the area of the maximum available glass fragment in mm2.
Demo Input:
['4 3 4\nH 2\nV 2\nV 3\nV 1\n', '7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n']
Demo Output:
['8\n4\n4\n2\n', '28\n16\n12\n6\n4\n']
Note:
Picture for the first sample test: | ```python
__author__ = 'ruckus'
class Split:
def __init__(self, order, offset) -> None:
super().__init__()
self.order = order
self.offset = int(offset)
class Glass:
h_splits = [0]
w_splits = [0]
max_p = 0
max_p_h_offset = 0
max_p_w_offset = 0
def __init__(self, h, w) -> None:
super().__init__()
self.h_splits.append(h)
self.w_splits.append(w)
self.max_p = h * w
def split(self, split: Split):
if split.order == 'H':
self.h_splits.append(split.offset)
self.h_splits = self.h_splits.sort()
else:
self.w_splits.append(split.offset)
self.w_splits = self.w_splits.sort()
max_p = 0
max_p_h_offset = 0
max_p_w_offset = 0
for i in range(1,len(self.h_splits)):
for j in range(1,len(self.w_splits)):
p = (self.h_splits[i] - self.h_splits[i - 1]) * (self.w_splits[j] - self.w_splits[j - 1])
if p > max_p:
max_p = p
max_p_h_offset = self.h_splits[i - 1]
max_p_w_offset = self.w_splits[j - 1]
self.max_p = max_p
self.max_p_h_offset = max_p_h_offset
self.max_p_w_offset = max_p_w_offset
print(max_p)
w, h, n = map(int, input().split())
glass = Glass(h, w)
for i in range(n):
order, offset = input().split()
glass.split(Split(order, offset))
``` | -1 |
|
104 | A | Blackjack | PROGRAMMING | 800 | [
"implementation"
] | A. Blackjack | 2 | 256 | One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. | The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. | Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. | [
"12\n",
"20\n",
"10\n"
] | [
"4",
"15",
"0"
] | In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | 500 | [
{
"input": "12",
"output": "4"
},
{
"input": "20",
"output": "15"
},
{
"input": "10",
"output": "0"
},
{
"input": "11",
"output": "4"
},
{
"input": "15",
"output": "4"
},
{
"input": "18",
"output": "4"
},
{
"input": "25",
"output": "0"
},
{
"input": "22",
"output": "0"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "6",
"output": "0"
},
{
"input": "7",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "9",
"output": "0"
},
{
"input": "13",
"output": "4"
},
{
"input": "14",
"output": "4"
},
{
"input": "16",
"output": "4"
},
{
"input": "17",
"output": "4"
},
{
"input": "19",
"output": "4"
},
{
"input": "21",
"output": "4"
},
{
"input": "23",
"output": "0"
},
{
"input": "24",
"output": "0"
}
] | 1,671,963,750 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 8 | 92 | 0 | n = int(input())
req = abs(n - 10)
if req == 1 or req == 11:
print(4)
elif req >= 2 and req <= 9:
print(4)
elif req == 10:
print(15)
else:
print(0) | Title: Blackjack
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.
Input Specification:
The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.
Output Specification:
Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.
Demo Input:
['12\n', '20\n', '10\n']
Demo Output:
['4', '15', '0']
Note:
In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | ```python
n = int(input())
req = abs(n - 10)
if req == 1 or req == 11:
print(4)
elif req >= 2 and req <= 9:
print(4)
elif req == 10:
print(15)
else:
print(0)
``` | 0 |
886 | B | Vlad and Cafes | PROGRAMMING | 1,000 | [] | null | null | Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe. | In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted. | Print one integer — index of the cafe that Vlad hasn't visited for as long as possible. | [
"5\n1 3 2 1 2\n",
"6\n2 1 2 2 4 1\n"
] | [
"3\n",
"2\n"
] | In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes. | 1,000 | [
{
"input": "5\n1 3 2 1 2",
"output": "3"
},
{
"input": "6\n2 1 2 2 4 1",
"output": "2"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n200000",
"output": "200000"
},
{
"input": "2\n2018 2017",
"output": "2018"
},
{
"input": "5\n100 1000 1000 1000 1000",
"output": "100"
},
{
"input": "8\n200000 1 200000 1 200000 1 200000 2",
"output": "1"
},
{
"input": "5\n20000 1 2 3 4",
"output": "20000"
},
{
"input": "2\n200000 1",
"output": "200000"
},
{
"input": "3\n2 100 2",
"output": "100"
},
{
"input": "2\n10 1",
"output": "10"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "3\n5 5 5",
"output": "5"
},
{
"input": "2\n5 1",
"output": "5"
}
] | 1,510,507,621 | 4,921 | Python 3 | TIME_LIMIT_EXCEEDED | PRETESTS | 4 | 2,000 | 14,233,600 | a = int(input())
b = list(map(int, input().split()))
minnum = 0
k = a - 1
keys = []
while k >= 0:
if b[k] not in keys:
keys.append(b[k])
min = k
minnum = b[k]
else:
if b[k] == minnum:
if k - min > 1:
k = min - 1
min = k
minnum = b[k]
else:
min = k
minnum = b[k]
k -= 1
print(minnum) | Title: Vlad and Cafes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
Input Specification:
In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
Output Specification:
Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.
Demo Input:
['5\n1 3 2 1 2\n', '6\n2 1 2 2 4 1\n']
Demo Output:
['3\n', '2\n']
Note:
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes. | ```python
a = int(input())
b = list(map(int, input().split()))
minnum = 0
k = a - 1
keys = []
while k >= 0:
if b[k] not in keys:
keys.append(b[k])
min = k
minnum = b[k]
else:
if b[k] == minnum:
if k - min > 1:
k = min - 1
min = k
minnum = b[k]
else:
min = k
minnum = b[k]
k -= 1
print(minnum)
``` | 0 |
|
617 | A | Elephant | PROGRAMMING | 800 | [
"math"
] | null | null | An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house. | The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house. | Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*. | [
"5\n",
"12\n"
] | [
"1\n",
"3\n"
] | In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves. | 500 | [
{
"input": "5",
"output": "1"
},
{
"input": "12",
"output": "3"
},
{
"input": "999999",
"output": "200000"
},
{
"input": "41",
"output": "9"
},
{
"input": "1000000",
"output": "200000"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "1"
},
{
"input": "534204",
"output": "106841"
},
{
"input": "469569",
"output": "93914"
},
{
"input": "502877",
"output": "100576"
},
{
"input": "942212",
"output": "188443"
},
{
"input": "97",
"output": "20"
},
{
"input": "53",
"output": "11"
},
{
"input": "89",
"output": "18"
},
{
"input": "574",
"output": "115"
},
{
"input": "716",
"output": "144"
},
{
"input": "729",
"output": "146"
},
{
"input": "8901",
"output": "1781"
},
{
"input": "3645",
"output": "729"
},
{
"input": "4426",
"output": "886"
},
{
"input": "46573",
"output": "9315"
},
{
"input": "86380",
"output": "17276"
},
{
"input": "94190",
"output": "18838"
},
{
"input": "999990",
"output": "199998"
},
{
"input": "999991",
"output": "199999"
},
{
"input": "999992",
"output": "199999"
},
{
"input": "999993",
"output": "199999"
},
{
"input": "999994",
"output": "199999"
},
{
"input": "999995",
"output": "199999"
},
{
"input": "999996",
"output": "200000"
},
{
"input": "999997",
"output": "200000"
},
{
"input": "999998",
"output": "200000"
}
] | 1,699,022,901 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 61 | 0 | x=int(input())
if x%5==1: print(x//5)
else : print(x//5+1) | Title: Elephant
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
Input Specification:
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
Output Specification:
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
Demo Input:
['5\n', '12\n']
Demo Output:
['1\n', '3\n']
Note:
In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves. | ```python
x=int(input())
if x%5==1: print(x//5)
else : print(x//5+1)
``` | 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"
}
] | 1,581,761,997 | 2,147,483,647 | Python 3 | OK | TESTS | 76 | 109 | 307,200 | a = int(input())
if a % 2 != 0:
print(0)
exit()
print((a // 2 - 1) // 2) | 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
a = int(input())
if a % 2 != 0:
print(0)
exit()
print((a // 2 - 1) // 2)
``` | 3 |
|
224 | B | Array | PROGRAMMING | 1,500 | [
"bitmasks",
"implementation",
"two pointers"
] | null | null | You've got an array *a*, consisting of *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. Your task is to find a minimal by inclusion segment [*l*,<=*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) such, that among numbers *a**l*,<= *a**l*<=+<=1,<= ...,<= *a**r* there are exactly *k* distinct numbers.
Segment [*l*,<=*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; *l*,<=*r* are integers) of length *m*<==<=*r*<=-<=*l*<=+<=1, satisfying the given property, is called minimal by inclusion, if there is no segment [*x*,<=*y*] satisfying the property and less then *m* in length, such that 1<=≤<=*l*<=≤<=*x*<=≤<=*y*<=≤<=*r*<=≤<=*n*. Note that the segment [*l*,<=*r*] doesn't have to be minimal in length among all segments, satisfying the given property. | The first line contains two space-separated integers: *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* — elements of the array *a* (1<=≤<=*a**i*<=≤<=105). | Print a space-separated pair of integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) such, that the segment [*l*,<=*r*] is the answer to the problem. If the sought segment does not exist, print "-1 -1" without the quotes. If there are multiple correct answers, print any of them. | [
"4 2\n1 2 2 3\n",
"8 3\n1 1 2 2 3 3 4 5\n",
"7 4\n4 7 7 4 7 4 7\n"
] | [
"1 2\n",
"2 5\n",
"-1 -1\n"
] | In the first sample among numbers *a*<sub class="lower-index">1</sub> and *a*<sub class="lower-index">2</sub> there are exactly two distinct numbers.
In the second sample segment [2, 5] is a minimal by inclusion segment with three distinct numbers, but it is not minimal in length among such segments.
In the third sample there is no segment with four distinct numbers. | 1,000 | [
{
"input": "4 2\n1 2 2 3",
"output": "1 2"
},
{
"input": "8 3\n1 1 2 2 3 3 4 5",
"output": "2 5"
},
{
"input": "7 4\n4 7 7 4 7 4 7",
"output": "-1 -1"
},
{
"input": "5 1\n1 7 2 3 2",
"output": "1 1"
},
{
"input": "1 2\n666",
"output": "-1 -1"
},
{
"input": "1 1\n5",
"output": "1 1"
},
{
"input": "10 4\n1 1 2 2 3 3 4 4 4 4",
"output": "2 7"
},
{
"input": "4 2\n3 3 4 3",
"output": "2 3"
},
{
"input": "4 3\n4 4 4 2",
"output": "-1 -1"
},
{
"input": "10 5\n15 17 2 13 3 16 4 5 9 12",
"output": "1 5"
},
{
"input": "17 13\n34 15 156 11 183 147 192 112 145 30 88 37 1 98 3 162 148",
"output": "1 13"
},
{
"input": "17 14\n271 158 573 88 792 767 392 646 392 392 271 549 402 767 573 925 796",
"output": "-1 -1"
},
{
"input": "8 5\n1 2 1 1 2 3 4 5",
"output": "4 8"
},
{
"input": "7 3\n2 1 2 2 1 2 3",
"output": "5 7"
},
{
"input": "6 3\n1 3 1 1 4 5",
"output": "2 5"
},
{
"input": "5 3\n1 2 1 1 3",
"output": "2 5"
},
{
"input": "9 3\n1 2 1 2 1 2 2 3 1",
"output": "5 8"
},
{
"input": "4 3\n1 2 1 3",
"output": "2 4"
},
{
"input": "5 3\n1 3 1 3 4",
"output": "3 5"
},
{
"input": "6 3\n1 3 3 1 4 4",
"output": "3 5"
},
{
"input": "5 3\n1 2 1 2 3",
"output": "3 5"
},
{
"input": "8 4\n1 2 3 2 1 2 3 4",
"output": "5 8"
},
{
"input": "10 4\n1 2 3 1 2 3 4 3 2 1",
"output": "4 7"
},
{
"input": "10 3\n1 1 1 2 1 2 3 3 3 4",
"output": "5 7"
},
{
"input": "10 3\n1 1 2 1 2 2 3 4 5 6",
"output": "4 7"
}
] | 1,590,317,378 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 342 | 7,680,000 | #####################
#1: Array,
#R -> ++Count[]
#L -> --Count[]
line1 = input().split(" ")
arr = list(map(int,input().split()))
n = int(line1[0])
k = int(line1[1])
counts = {}
l = 0
r = 0
count = 0
for i in range(len(arr)):
if arr[i] in counts:
counts[arr[i]] += 1
else:
count += 1
counts[arr[i]] = 1
if count >= k:
r = i
break
for i in range(len(arr)):
l = i
counts[arr[i]] -= 1
if counts[arr[i]] == 0:
l = i
break
if count < k:
print("-1 -1")
else:
print(l + 1, r + 1)
| Title: Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got an array *a*, consisting of *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. Your task is to find a minimal by inclusion segment [*l*,<=*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) such, that among numbers *a**l*,<= *a**l*<=+<=1,<= ...,<= *a**r* there are exactly *k* distinct numbers.
Segment [*l*,<=*r*] (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; *l*,<=*r* are integers) of length *m*<==<=*r*<=-<=*l*<=+<=1, satisfying the given property, is called minimal by inclusion, if there is no segment [*x*,<=*y*] satisfying the property and less then *m* in length, such that 1<=≤<=*l*<=≤<=*x*<=≤<=*y*<=≤<=*r*<=≤<=*n*. Note that the segment [*l*,<=*r*] doesn't have to be minimal in length among all segments, satisfying the given property.
Input Specification:
The first line contains two space-separated integers: *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* — elements of the array *a* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Print a space-separated pair of integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) such, that the segment [*l*,<=*r*] is the answer to the problem. If the sought segment does not exist, print "-1 -1" without the quotes. If there are multiple correct answers, print any of them.
Demo Input:
['4 2\n1 2 2 3\n', '8 3\n1 1 2 2 3 3 4 5\n', '7 4\n4 7 7 4 7 4 7\n']
Demo Output:
['1 2\n', '2 5\n', '-1 -1\n']
Note:
In the first sample among numbers *a*<sub class="lower-index">1</sub> and *a*<sub class="lower-index">2</sub> there are exactly two distinct numbers.
In the second sample segment [2, 5] is a minimal by inclusion segment with three distinct numbers, but it is not minimal in length among such segments.
In the third sample there is no segment with four distinct numbers. | ```python
#####################
#1: Array,
#R -> ++Count[]
#L -> --Count[]
line1 = input().split(" ")
arr = list(map(int,input().split()))
n = int(line1[0])
k = int(line1[1])
counts = {}
l = 0
r = 0
count = 0
for i in range(len(arr)):
if arr[i] in counts:
counts[arr[i]] += 1
else:
count += 1
counts[arr[i]] = 1
if count >= k:
r = i
break
for i in range(len(arr)):
l = i
counts[arr[i]] -= 1
if counts[arr[i]] == 0:
l = i
break
if count < k:
print("-1 -1")
else:
print(l + 1, r + 1)
``` | 3 |
|
675 | A | Infinite Sequence | PROGRAMMING | 1,100 | [
"math"
] | null | null | Vasya likes everything infinite. Now he is studying the properties of a sequence *s*, such that its first element is equal to *a* (*s*1<==<=*a*), and the difference between any two neighbouring elements is equal to *c* (*s**i*<=-<=*s**i*<=-<=1<==<=*c*). In particular, Vasya wonders if his favourite integer *b* appears in this sequence, that is, there exists a positive integer *i*, such that *s**i*<==<=*b*. Of course, you are the person he asks for a help. | The first line of the input contain three integers *a*, *b* and *c* (<=-<=109<=≤<=*a*,<=*b*,<=*c*<=≤<=109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively. | If *b* appears in the sequence *s* print "YES" (without quotes), otherwise print "NO" (without quotes). | [
"1 7 3\n",
"10 10 0\n",
"1 -4 5\n",
"0 60 50\n"
] | [
"YES\n",
"YES\n",
"NO\n",
"NO\n"
] | In the first sample, the sequence starts from integers 1, 4, 7, so 7 is its element.
In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.
In the third sample all elements of the sequence are greater than Vasya's favorite integer.
In the fourth sample, the sequence starts from 0, 50, 100, and all the following elements are greater than Vasya's favorite integer. | 500 | [
{
"input": "1 7 3",
"output": "YES"
},
{
"input": "10 10 0",
"output": "YES"
},
{
"input": "1 -4 5",
"output": "NO"
},
{
"input": "0 60 50",
"output": "NO"
},
{
"input": "1 -4 -5",
"output": "YES"
},
{
"input": "0 1 0",
"output": "NO"
},
{
"input": "10 10 42",
"output": "YES"
},
{
"input": "-1000000000 1000000000 -1",
"output": "NO"
},
{
"input": "10 16 4",
"output": "NO"
},
{
"input": "-1000000000 1000000000 5",
"output": "YES"
},
{
"input": "1000000000 -1000000000 5",
"output": "NO"
},
{
"input": "1000000000 -1000000000 0",
"output": "NO"
},
{
"input": "1000000000 1000000000 0",
"output": "YES"
},
{
"input": "115078364 -899474523 -1",
"output": "YES"
},
{
"input": "-245436499 416383245 992",
"output": "YES"
},
{
"input": "-719636354 536952440 2",
"output": "YES"
},
{
"input": "-198350539 963391024 68337739",
"output": "YES"
},
{
"input": "-652811055 875986516 1091",
"output": "YES"
},
{
"input": "119057893 -516914539 -39748277",
"output": "YES"
},
{
"input": "989140430 731276607 -36837689",
"output": "YES"
},
{
"input": "677168390 494583489 -985071853",
"output": "NO"
},
{
"input": "58090193 777423708 395693923",
"output": "NO"
},
{
"input": "479823846 -403424770 -653472589",
"output": "NO"
},
{
"input": "-52536829 -132023273 -736287999",
"output": "NO"
},
{
"input": "-198893776 740026818 -547885271",
"output": "NO"
},
{
"input": "-2 -2 -2",
"output": "YES"
},
{
"input": "-2 -2 -1",
"output": "YES"
},
{
"input": "-2 -2 0",
"output": "YES"
},
{
"input": "-2 -2 1",
"output": "YES"
},
{
"input": "-2 -2 2",
"output": "YES"
},
{
"input": "-2 -1 -2",
"output": "NO"
},
{
"input": "-2 -1 -1",
"output": "NO"
},
{
"input": "-2 -1 0",
"output": "NO"
},
{
"input": "-2 -1 1",
"output": "YES"
},
{
"input": "-2 -1 2",
"output": "NO"
},
{
"input": "-2 0 -2",
"output": "NO"
},
{
"input": "-2 0 -1",
"output": "NO"
},
{
"input": "-2 0 0",
"output": "NO"
},
{
"input": "-2 0 1",
"output": "YES"
},
{
"input": "-2 0 2",
"output": "YES"
},
{
"input": "-2 1 -2",
"output": "NO"
},
{
"input": "-2 1 -1",
"output": "NO"
},
{
"input": "-2 1 0",
"output": "NO"
},
{
"input": "-2 1 1",
"output": "YES"
},
{
"input": "-2 1 2",
"output": "NO"
},
{
"input": "-2 2 -2",
"output": "NO"
},
{
"input": "-2 2 -1",
"output": "NO"
},
{
"input": "-2 2 0",
"output": "NO"
},
{
"input": "-2 2 1",
"output": "YES"
},
{
"input": "-2 2 2",
"output": "YES"
},
{
"input": "-1 -2 -2",
"output": "NO"
},
{
"input": "-1 -2 -1",
"output": "YES"
},
{
"input": "-1 -2 0",
"output": "NO"
},
{
"input": "-1 -2 1",
"output": "NO"
},
{
"input": "-1 -2 2",
"output": "NO"
},
{
"input": "-1 -1 -2",
"output": "YES"
},
{
"input": "-1 -1 -1",
"output": "YES"
},
{
"input": "-1 -1 0",
"output": "YES"
},
{
"input": "-1 -1 1",
"output": "YES"
},
{
"input": "-1 -1 2",
"output": "YES"
},
{
"input": "-1 0 -2",
"output": "NO"
},
{
"input": "-1 0 -1",
"output": "NO"
},
{
"input": "-1 0 0",
"output": "NO"
},
{
"input": "-1 0 1",
"output": "YES"
},
{
"input": "-1 0 2",
"output": "NO"
},
{
"input": "-1 1 -2",
"output": "NO"
},
{
"input": "-1 1 -1",
"output": "NO"
},
{
"input": "-1 1 0",
"output": "NO"
},
{
"input": "-1 1 1",
"output": "YES"
},
{
"input": "-1 1 2",
"output": "YES"
},
{
"input": "-1 2 -2",
"output": "NO"
},
{
"input": "-1 2 -1",
"output": "NO"
},
{
"input": "-1 2 0",
"output": "NO"
},
{
"input": "-1 2 1",
"output": "YES"
},
{
"input": "-1 2 2",
"output": "NO"
},
{
"input": "0 -2 -2",
"output": "YES"
},
{
"input": "0 -2 -1",
"output": "YES"
},
{
"input": "0 -2 0",
"output": "NO"
},
{
"input": "0 -2 1",
"output": "NO"
},
{
"input": "0 -2 2",
"output": "NO"
},
{
"input": "0 -1 -2",
"output": "NO"
},
{
"input": "0 -1 -1",
"output": "YES"
},
{
"input": "0 -1 0",
"output": "NO"
},
{
"input": "0 -1 1",
"output": "NO"
},
{
"input": "0 -1 2",
"output": "NO"
},
{
"input": "0 0 -2",
"output": "YES"
},
{
"input": "0 0 -1",
"output": "YES"
},
{
"input": "0 0 0",
"output": "YES"
},
{
"input": "0 0 1",
"output": "YES"
},
{
"input": "0 0 2",
"output": "YES"
},
{
"input": "0 1 -2",
"output": "NO"
},
{
"input": "0 1 -1",
"output": "NO"
},
{
"input": "0 1 0",
"output": "NO"
},
{
"input": "0 1 1",
"output": "YES"
},
{
"input": "0 1 2",
"output": "NO"
},
{
"input": "0 2 -2",
"output": "NO"
},
{
"input": "0 2 -1",
"output": "NO"
},
{
"input": "0 2 0",
"output": "NO"
},
{
"input": "0 2 1",
"output": "YES"
},
{
"input": "0 2 2",
"output": "YES"
},
{
"input": "1 -2 -2",
"output": "NO"
},
{
"input": "1 -2 -1",
"output": "YES"
},
{
"input": "1 -2 0",
"output": "NO"
},
{
"input": "1 -2 1",
"output": "NO"
},
{
"input": "1 -2 2",
"output": "NO"
},
{
"input": "1 -1 -2",
"output": "YES"
},
{
"input": "1 -1 -1",
"output": "YES"
},
{
"input": "1 -1 0",
"output": "NO"
},
{
"input": "1 -1 1",
"output": "NO"
},
{
"input": "1 -1 2",
"output": "NO"
},
{
"input": "1 0 -2",
"output": "NO"
},
{
"input": "1 0 -1",
"output": "YES"
},
{
"input": "1 0 0",
"output": "NO"
},
{
"input": "1 0 1",
"output": "NO"
},
{
"input": "1 0 2",
"output": "NO"
},
{
"input": "1 1 -2",
"output": "YES"
},
{
"input": "1 1 -1",
"output": "YES"
},
{
"input": "1 1 0",
"output": "YES"
},
{
"input": "1 1 1",
"output": "YES"
},
{
"input": "1 1 2",
"output": "YES"
},
{
"input": "1 2 -2",
"output": "NO"
},
{
"input": "1 2 -1",
"output": "NO"
},
{
"input": "1 2 0",
"output": "NO"
},
{
"input": "1 2 1",
"output": "YES"
},
{
"input": "1 2 2",
"output": "NO"
},
{
"input": "2 -2 -2",
"output": "YES"
},
{
"input": "2 -2 -1",
"output": "YES"
},
{
"input": "2 -2 0",
"output": "NO"
},
{
"input": "2 -2 1",
"output": "NO"
},
{
"input": "2 -2 2",
"output": "NO"
},
{
"input": "2 -1 -2",
"output": "NO"
},
{
"input": "2 -1 -1",
"output": "YES"
},
{
"input": "2 -1 0",
"output": "NO"
},
{
"input": "2 -1 1",
"output": "NO"
},
{
"input": "2 -1 2",
"output": "NO"
},
{
"input": "2 0 -2",
"output": "YES"
},
{
"input": "2 0 -1",
"output": "YES"
},
{
"input": "2 0 0",
"output": "NO"
},
{
"input": "2 0 1",
"output": "NO"
},
{
"input": "2 0 2",
"output": "NO"
},
{
"input": "2 1 -2",
"output": "NO"
},
{
"input": "2 1 -1",
"output": "YES"
},
{
"input": "2 1 0",
"output": "NO"
},
{
"input": "2 1 1",
"output": "NO"
},
{
"input": "2 1 2",
"output": "NO"
},
{
"input": "2 2 -2",
"output": "YES"
},
{
"input": "2 2 -1",
"output": "YES"
},
{
"input": "2 2 0",
"output": "YES"
},
{
"input": "2 2 1",
"output": "YES"
},
{
"input": "2 2 2",
"output": "YES"
},
{
"input": "-1000000000 1000000000 1",
"output": "YES"
},
{
"input": "-1000000000 1000000000 2",
"output": "YES"
},
{
"input": "1000000000 -1000000000 -1",
"output": "YES"
},
{
"input": "5 2 3",
"output": "NO"
},
{
"input": "2 1 -1",
"output": "YES"
},
{
"input": "3 2 1",
"output": "NO"
},
{
"input": "0 -5 -3",
"output": "NO"
},
{
"input": "2 5 5",
"output": "NO"
},
{
"input": "0 10 1",
"output": "YES"
},
{
"input": "15 5 -5",
"output": "YES"
},
{
"input": "2 1 1",
"output": "NO"
},
{
"input": "20 10 0",
"output": "NO"
},
{
"input": "20 15 5",
"output": "NO"
},
{
"input": "1 6 1",
"output": "YES"
},
{
"input": "1000000000 0 -1000000000",
"output": "YES"
},
{
"input": "1 1 -5",
"output": "YES"
},
{
"input": "4 6 1",
"output": "YES"
},
{
"input": "-5 -10 -5",
"output": "YES"
},
{
"input": "2 0 0",
"output": "NO"
},
{
"input": "10 9 -1",
"output": "YES"
},
{
"input": "-2 -1 -1",
"output": "NO"
},
{
"input": "1 13 3",
"output": "YES"
},
{
"input": "2 3 0",
"output": "NO"
},
{
"input": "1 1 -1",
"output": "YES"
},
{
"input": "5 -10 -5",
"output": "YES"
},
{
"input": "5 3 1",
"output": "NO"
},
{
"input": "1 1000000000 1",
"output": "YES"
},
{
"input": "-1000000000 1000000000 1000000000",
"output": "YES"
}
] | 1,598,621,782 | 2,147,483,647 | PyPy 3 | OK | TESTS | 178 | 156 | 0 | def f(l):
a,b,c = l
if c==0:
return a==b
return (b-a)%c==0 and (b-a)//c>=0
l = list(map(int,input().split()))
print('YES' if f(l) else 'NO')
| Title: Infinite Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya likes everything infinite. Now he is studying the properties of a sequence *s*, such that its first element is equal to *a* (*s*1<==<=*a*), and the difference between any two neighbouring elements is equal to *c* (*s**i*<=-<=*s**i*<=-<=1<==<=*c*). In particular, Vasya wonders if his favourite integer *b* appears in this sequence, that is, there exists a positive integer *i*, such that *s**i*<==<=*b*. Of course, you are the person he asks for a help.
Input Specification:
The first line of the input contain three integers *a*, *b* and *c* (<=-<=109<=≤<=*a*,<=*b*,<=*c*<=≤<=109) — the first element of the sequence, Vasya's favorite number and the difference between any two neighbouring elements of the sequence, respectively.
Output Specification:
If *b* appears in the sequence *s* print "YES" (without quotes), otherwise print "NO" (without quotes).
Demo Input:
['1 7 3\n', '10 10 0\n', '1 -4 5\n', '0 60 50\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'NO\n']
Note:
In the first sample, the sequence starts from integers 1, 4, 7, so 7 is its element.
In the second sample, the favorite integer of Vasya is equal to the first element of the sequence.
In the third sample all elements of the sequence are greater than Vasya's favorite integer.
In the fourth sample, the sequence starts from 0, 50, 100, and all the following elements are greater than Vasya's favorite integer. | ```python
def f(l):
a,b,c = l
if c==0:
return a==b
return (b-a)%c==0 and (b-a)//c>=0
l = list(map(int,input().split()))
print('YES' if f(l) else 'NO')
``` | 3 |
|
34 | B | Sale | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | B. Sale | 2 | 256 | Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets. | Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets. | [
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] | [
"8\n",
"7\n"
] | none | 1,000 | [
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,623,589,393 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 186 | 0 | n, m = list(map(int, input().split()))
prices = list(map(int, input().split()))
negative_prices = []
for price in prices:
if price < 0:
negative_prices.append(price)
negative_prices.sort()
result = 0
for i in range(len(negative_prices)):
result -= negative_prices[i]
print(result) | Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none | ```python
n, m = list(map(int, input().split()))
prices = list(map(int, input().split()))
negative_prices = []
for price in prices:
if price < 0:
negative_prices.append(price)
negative_prices.sort()
result = 0
for i in range(len(negative_prices)):
result -= negative_prices[i]
print(result)
``` | 0 |
116 | A | Tram | PROGRAMMING | 800 | [
"implementation"
] | null | null | Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram. | The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0. | Print a single integer denoting the minimum possible capacity of the tram (0 is allowed). | [
"4\n0 3\n2 5\n4 2\n4 0\n"
] | [
"6\n"
] | For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | 500 | [
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
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"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,697,377,920 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 92 | 0 | cur = 0
ans = 0
for _ in range(int(input())):
a, b = list(map(int, input().split()))
cur -= a
cur += b
ans = max(ans, cur)
print(ans) | Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | ```python
cur = 0
ans = 0
for _ in range(int(input())):
a, b = list(map(int, input().split()))
cur -= a
cur += b
ans = max(ans, cur)
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"
}
] | 1,584,365,565 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 124 | 0 | # You
# Dont read my code
a,b = map(int,input().split())
div = [1]
if a % 2 != 0:
for i in range(3, a,2):
if a % i == 0:
div.append(i)
div.append(a)
else:
for i in range(2,a,3):
if a % i == 0:
div.append(i)
div.append(a)
if b <= len(div):
print(div[b-1])
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
# You
# Dont read my code
a,b = map(int,input().split())
div = [1]
if a % 2 != 0:
for i in range(3, a,2):
if a % i == 0:
div.append(i)
div.append(a)
else:
for i in range(2,a,3):
if a % i == 0:
div.append(i)
div.append(a)
if b <= len(div):
print(div[b-1])
else:
print(-1)
``` | 0 |
|
3 | B | Lorry | PROGRAMMING | 1,900 | [
"greedy",
"sortings"
] | B. Lorry | 2 | 64 | A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body. | The first line contains a pair of integer numbers *n* and *v* (1<=≤<=*n*<=≤<=105; 1<=≤<=*v*<=≤<=109), where *n* is the number of waterborne vehicles in the boat depot, and *v* is the truck body volume of the lorry in cubic metres. The following *n* lines contain the information about the waterborne vehicles, that is a pair of numbers *t**i*,<=*p**i* (1<=≤<=*t**i*<=≤<=2; 1<=≤<=*p**i*<=≤<=104), where *t**i* is the vehicle type (1 – a kayak, 2 – a catamaran), and *p**i* is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file. | In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them. | [
"3 2\n1 2\n2 7\n1 3\n"
] | [
"7\n2\n"
] | none | 0 | [
{
"input": "3 2\n1 2\n2 7\n1 3",
"output": "7\n2"
},
{
"input": "5 3\n1 9\n2 9\n1 9\n2 10\n1 6",
"output": "24\n3 1 5"
},
{
"input": "10 10\n1 14\n2 15\n2 11\n2 12\n2 9\n1 14\n2 15\n1 9\n2 11\n2 6",
"output": "81\n6 1 7 2 4 9"
},
{
"input": "20 19\n2 47\n1 37\n1 48\n2 42\n2 48\n1 38\n2 47\n1 48\n2 47\n1 41\n2 46\n1 28\n1 49\n1 45\n2 34\n1 43\n2 29\n1 46\n2 45\n2 18",
"output": "630\n13 8 3 18 14 16 10 6 2 5 9 7 1 11"
},
{
"input": "50 27\n2 93\n1 98\n2 62\n1 56\n1 86\n1 42\n2 67\n2 97\n2 59\n1 73\n1 83\n2 96\n1 20\n1 66\n1 84\n1 83\n1 91\n2 97\n1 81\n2 88\n2 63\n1 99\n2 57\n1 39\n1 74\n2 88\n1 30\n2 68\n1 100\n2 57\n1 87\n1 93\n1 83\n1 100\n1 91\n1 14\n1 38\n2 98\n2 85\n2 61\n1 44\n2 93\n2 66\n2 55\n2 74\n1 67\n2 67\n1 85\n2 59\n1 83",
"output": "2055\n34 29 22 2 32 35 17 31 5 48 15 50 33 16 11 19 25 10 46 14 4 38 18 8"
},
{
"input": "1 1\n1 600",
"output": "600\n1"
},
{
"input": "10 14\n2 230\n2 516\n2 527\n2 172\n2 854\n2 61\n1 52\n2 154\n2 832\n2 774",
"output": "3905\n5 9 10 3 2 1 4"
},
{
"input": "8 8\n1 1\n1 1\n1 1\n1 1\n2 100\n2 100\n2 100\n2 100",
"output": "400\n8 7 6 5"
},
{
"input": "8 4\n1 100\n1 100\n1 100\n1 100\n2 1\n2 1\n2 1\n2 1",
"output": "400\n4 3 2 1"
}
] | 1,566,461,067 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 17,100,800 | import functools
def cmp(item1, item2):
if item1.carry / item1.typ > item2.carry / item2.typ:
return -1
else:
return 1
class Vehicles:
def __init__(self, idx, typ, carry):
self.idx = idx
self.typ = typ
self.carry = carry
n, v = map(int, input().split())
a = []
for i in range(n):
typ, carry = map(int, input().split())
a.append(Vehicles(i + 1, typ, carry))
a.sort(key=functools.cmp_to_key(cmp))
res = [0]
path = []
idx = 0
for i in range(n):
if v >= a[i].typ:
v -= a[i].typ
res[0] += a[i].carry
path.append(a[i].idx)
idx = i
if v == 1 and idx < n:
flag = False
tmp = [0, 0]
for i in range(idx + 1, n):
if a[i].typ == 2:
tmp = [a[i].carry, str(a[i].idx)]
flag = True
break
if flag:
path = []
for i in range(idx, -1, -1):
if flag and a[i].typ == 1 and a[i].carry < tmp[0]:
flag = False
res[0] += tmp[0] - a[i].carry
path.append(tmp[1])
else:
path.append(a[i].idx)
print(res[0])
l = len(path)
for i in range(l):
if i == 0:
print(str(path[i]), end='')
else:
print(' ' + str(path[i]), end='') | Title: Lorry
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body.
Input Specification:
The first line contains a pair of integer numbers *n* and *v* (1<=≤<=*n*<=≤<=105; 1<=≤<=*v*<=≤<=109), where *n* is the number of waterborne vehicles in the boat depot, and *v* is the truck body volume of the lorry in cubic metres. The following *n* lines contain the information about the waterborne vehicles, that is a pair of numbers *t**i*,<=*p**i* (1<=≤<=*t**i*<=≤<=2; 1<=≤<=*p**i*<=≤<=104), where *t**i* is the vehicle type (1 – a kayak, 2 – a catamaran), and *p**i* is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file.
Output Specification:
In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them.
Demo Input:
['3 2\n1 2\n2 7\n1 3\n']
Demo Output:
['7\n2\n']
Note:
none | ```python
import functools
def cmp(item1, item2):
if item1.carry / item1.typ > item2.carry / item2.typ:
return -1
else:
return 1
class Vehicles:
def __init__(self, idx, typ, carry):
self.idx = idx
self.typ = typ
self.carry = carry
n, v = map(int, input().split())
a = []
for i in range(n):
typ, carry = map(int, input().split())
a.append(Vehicles(i + 1, typ, carry))
a.sort(key=functools.cmp_to_key(cmp))
res = [0]
path = []
idx = 0
for i in range(n):
if v >= a[i].typ:
v -= a[i].typ
res[0] += a[i].carry
path.append(a[i].idx)
idx = i
if v == 1 and idx < n:
flag = False
tmp = [0, 0]
for i in range(idx + 1, n):
if a[i].typ == 2:
tmp = [a[i].carry, str(a[i].idx)]
flag = True
break
if flag:
path = []
for i in range(idx, -1, -1):
if flag and a[i].typ == 1 and a[i].carry < tmp[0]:
flag = False
res[0] += tmp[0] - a[i].carry
path.append(tmp[1])
else:
path.append(a[i].idx)
print(res[0])
l = len(path)
for i in range(l):
if i == 0:
print(str(path[i]), end='')
else:
print(' ' + str(path[i]), end='')
``` | 0 |
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"
}
] | 1,628,001,929 | 2,147,483,647 | PyPy 3 | OK | TESTS | 62 | 93 | 20,172,800 | n,k = map(int,input().split())
if n < k:
print('YES')
else:
diff = n//k
if diff % 2 == 0:
print('NO')
if diff % 2 != 0:
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:
print('YES')
else:
diff = n//k
if diff % 2 == 0:
print('NO')
if diff % 2 != 0:
print('YES')
``` | 3 |
|
274 | A | k-Multiple Free Set | PROGRAMMING | 1,500 | [
"binary search",
"greedy",
"sortings"
] | null | null | A *k*-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by *k*. That is, there are no two integers *x* and *y* (*x*<=<<=*y*) from the set, such that *y*<==<=*x*·*k*.
You're given a set of *n* distinct positive integers. Your task is to find the size of it's largest *k*-multiple free subset. | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109). The next line contains a list of *n* distinct positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
All the numbers in the lines are separated by single spaces. | On the only line of the output print the size of the largest *k*-multiple free subset of {*a*1,<=*a*2,<=...,<=*a**n*}. | [
"6 2\n2 3 6 5 4 10\n"
] | [
"3\n"
] | In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}. | 500 | [
{
"input": "6 2\n2 3 6 5 4 10",
"output": "3"
},
{
"input": "10 2\n1 2 3 4 5 6 7 8 9 10",
"output": "6"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "100 2\n191 17 61 40 77 95 128 88 26 69 79 10 131 106 142 152 68 39 182 53 83 81 6 89 65 148 33 22 5 47 107 121 52 163 150 158 189 118 75 180 177 176 112 167 140 184 29 166 25 46 169 145 187 123 196 18 115 126 155 100 63 58 159 19 173 113 133 60 130 161 76 157 93 199 50 97 15 67 109 164 99 149 3 137 153 136 56 43 103 170 13 183 194 72 9 181 86 30 91 36",
"output": "79"
},
{
"input": "100 3\n13 38 137 24 46 192 33 8 170 141 118 57 198 133 112 176 40 36 91 130 166 72 123 28 82 180 134 52 64 107 97 79 199 184 158 22 181 163 98 7 88 41 73 87 167 109 15 173 153 70 50 119 139 56 17 152 84 161 11 116 31 187 143 196 27 102 132 126 149 63 146 168 67 48 53 120 20 105 155 10 128 47 23 6 94 3 113 65 44 179 189 99 75 34 111 193 60 145 171 77",
"output": "87"
},
{
"input": "12 400000000\n1 400000000 800000000 2 3 4 5 6 7 8 9 10",
"output": "10"
},
{
"input": "3 1\n1 2 3",
"output": "3"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "10 1\n1 100 300 400 500 500000 1000000 10000000 100000000 1000000000",
"output": "10"
},
{
"input": "2 1\n2 1",
"output": "2"
},
{
"input": "2 1000000000\n1 1000000000",
"output": "1"
},
{
"input": "4 1000\n1 1000 1000000 1000000000",
"output": "2"
},
{
"input": "2 2\n1 3",
"output": "2"
},
{
"input": "2 2\n16 8",
"output": "1"
},
{
"input": "3 2\n8 4 2",
"output": "2"
},
{
"input": "5 1\n1 2 3 4 5",
"output": "5"
},
{
"input": "2 2\n500000000 1000000000",
"output": "1"
},
{
"input": "2 2\n4 2",
"output": "1"
},
{
"input": "10 100000000\n1 2 3 4 5 6 7 8 82000 907431936",
"output": "10"
},
{
"input": "8 65538\n65535 65536 65537 65538 65539 131072 262144 196608",
"output": "8"
},
{
"input": "5 2\n10 8 6 4 2",
"output": "4"
},
{
"input": "2 1000000000\n276447232 100000",
"output": "2"
}
] | 1,618,941,400 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 2,000 | 8,192,000 | n,k=map(int,input().split())
l=list(map(int,input().split()))
u=[]
for i in range(len(l)):
f=0
for g in u:
if 2*g[-1]==l[i]:
g.append(l[i])
f=1
break
if f==0:
u.append([l[i]])
ct=0
for i in u:
if len(i)==1:
ct+=1
else:
if len(i)%2==0:
ct+=len(i)//2
else:
ct+=(len(i)//2)
ct+=1
print(ct)
| Title: k-Multiple Free Set
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A *k*-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by *k*. That is, there are no two integers *x* and *y* (*x*<=<<=*y*) from the set, such that *y*<==<=*x*·*k*.
You're given a set of *n* distinct positive integers. Your task is to find the size of it's largest *k*-multiple free subset.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109). The next line contains a list of *n* distinct positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
All the numbers in the lines are separated by single spaces.
Output Specification:
On the only line of the output print the size of the largest *k*-multiple free subset of {*a*1,<=*a*2,<=...,<=*a**n*}.
Demo Input:
['6 2\n2 3 6 5 4 10\n']
Demo Output:
['3\n']
Note:
In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}. | ```python
n,k=map(int,input().split())
l=list(map(int,input().split()))
u=[]
for i in range(len(l)):
f=0
for g in u:
if 2*g[-1]==l[i]:
g.append(l[i])
f=1
break
if f==0:
u.append([l[i]])
ct=0
for i in u:
if len(i)==1:
ct+=1
else:
if len(i)%2==0:
ct+=len(i)//2
else:
ct+=(len(i)//2)
ct+=1
print(ct)
``` | 0 |
|
922 | A | Cloning Toys | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Imp likes his plush toy a lot.
Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies.
Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly *x* copied toys and *y* original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies. | The only line contains two integers *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=109) — the number of copies and the number of original toys Imp wants to get (including the initial one). | Print "Yes", if the desired configuration is possible, and "No" otherwise.
You can print each letter in arbitrary case (upper or lower). | [
"6 3\n",
"4 2\n",
"1000 1001\n"
] | [
"Yes\n",
"No\n",
"Yes\n"
] | In the first example, Imp has to apply the machine twice to original toys and then twice to copies. | 500 | [
{
"input": "6 3",
"output": "Yes"
},
{
"input": "4 2",
"output": "No"
},
{
"input": "1000 1001",
"output": "Yes"
},
{
"input": "1000000000 999999999",
"output": "Yes"
},
{
"input": "81452244 81452247",
"output": "No"
},
{
"input": "188032448 86524683",
"output": "Yes"
},
{
"input": "365289629 223844571",
"output": "No"
},
{
"input": "247579518 361164458",
"output": "No"
},
{
"input": "424836699 793451637",
"output": "No"
},
{
"input": "602093880 930771525",
"output": "No"
},
{
"input": "779351061 773124120",
"output": "Yes"
},
{
"input": "661640950 836815080",
"output": "No"
},
{
"input": "543930839 974134967",
"output": "No"
},
{
"input": "16155311 406422145",
"output": "No"
},
{
"input": "81601559 445618240",
"output": "No"
},
{
"input": "963891449 582938127",
"output": "No"
},
{
"input": "141148629 351661795",
"output": "No"
},
{
"input": "318405810 783948974",
"output": "No"
},
{
"input": "495662991 921268861",
"output": "No"
},
{
"input": "1 0",
"output": "No"
},
{
"input": "0 1",
"output": "Yes"
},
{
"input": "0 0",
"output": "No"
},
{
"input": "453462237 167520068",
"output": "Yes"
},
{
"input": "630719418 9872663",
"output": "Yes"
},
{
"input": "807976599 442159843",
"output": "No"
},
{
"input": "690266488 579479730",
"output": "No"
},
{
"input": "771581370 589752968",
"output": "No"
},
{
"input": "948838551 727072855",
"output": "No"
},
{
"input": "831128440 790763814",
"output": "No"
},
{
"input": "303352912 928083702",
"output": "No"
},
{
"input": "185642801 65403588",
"output": "Yes"
},
{
"input": "67932690 202723476",
"output": "No"
},
{
"input": "540157163 340043363",
"output": "No"
},
{
"input": "422447052 772330542",
"output": "No"
},
{
"input": "599704233 541054210",
"output": "Yes"
},
{
"input": "481994122 678374097",
"output": "No"
},
{
"input": "48564714 743566477",
"output": "No"
},
{
"input": "225821895 880886365",
"output": "No"
},
{
"input": "403079076 313173543",
"output": "Yes"
},
{
"input": "1000000000 1000000000",
"output": "No"
},
{
"input": "1 1",
"output": "No"
},
{
"input": "1 2",
"output": "Yes"
},
{
"input": "2 1",
"output": "No"
},
{
"input": "2 2",
"output": "No"
},
{
"input": "2 0",
"output": "No"
},
{
"input": "0 2",
"output": "No"
},
{
"input": "1000000000 1",
"output": "No"
},
{
"input": "777777 0",
"output": "No"
},
{
"input": "10 1",
"output": "No"
},
{
"input": "7 0",
"output": "No"
},
{
"input": "3 0",
"output": "No"
},
{
"input": "3 2",
"output": "Yes"
}
] | 1,658,406,807 | 2,147,483,647 | PyPy 3 | OK | TESTS | 52 | 93 | 0 | x, y = map(int, input().split())
print('Yes' if (x - y) % 2 == 1 and y - x <= 1 and y > 0 and (x == 0 or y > 1) else 'No') | Title: Cloning Toys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imp likes his plush toy a lot.
Recently, he found a machine that can clone plush toys. Imp knows that if he applies the machine to an original toy, he additionally gets one more original toy and one copy, and if he applies the machine to a copied toy, he gets two additional copies.
Initially, Imp has only one original toy. He wants to know if it is possible to use machine to get exactly *x* copied toys and *y* original toys? He can't throw toys away, and he can't apply the machine to a copy if he doesn't currently have any copies.
Input Specification:
The only line contains two integers *x* and *y* (0<=≤<=*x*,<=*y*<=≤<=109) — the number of copies and the number of original toys Imp wants to get (including the initial one).
Output Specification:
Print "Yes", if the desired configuration is possible, and "No" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['6 3\n', '4 2\n', '1000 1001\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n']
Note:
In the first example, Imp has to apply the machine twice to original toys and then twice to copies. | ```python
x, y = map(int, input().split())
print('Yes' if (x - y) % 2 == 1 and y - x <= 1 and y > 0 and (x == 0 or y > 1) else 'No')
``` | 3 |
|
660 | C | Hard Process | PROGRAMMING | 1,600 | [
"binary search",
"dp",
"two pointers"
] | null | null | You are given an array *a* with *n* elements. Each element of *a* is either 0 or 1.
Let's denote the length of the longest subsegment of consecutive elements in *a*, consisting of only numbers one, as *f*(*a*). You can change no more than *k* zeroes to ones to maximize *f*(*a*). | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=3·105,<=0<=≤<=*k*<=≤<=*n*) — the number of elements in *a* and the parameter *k*.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1) — the elements of *a*. | On the first line print a non-negative integer *z* — the maximal value of *f*(*a*) after no more than *k* changes of zeroes to ones.
On the second line print *n* integers *a**j* — the elements of the array *a* after the changes.
If there are multiple answers, you can print any one of them. | [
"7 1\n1 0 0 1 1 0 1\n",
"10 2\n1 0 0 1 0 1 0 1 0 1\n"
] | [
"4\n1 0 0 1 1 1 1\n",
"5\n1 0 0 1 1 1 1 1 0 1\n"
] | none | 0 | [
{
"input": "7 1\n1 0 0 1 1 0 1",
"output": "4\n1 0 0 1 1 1 1"
},
{
"input": "10 2\n1 0 0 1 0 1 0 1 0 1",
"output": "5\n1 0 0 1 1 1 1 1 0 1"
},
{
"input": "1 0\n0",
"output": "0\n0"
},
{
"input": "1 0\n0",
"output": "0\n0"
},
{
"input": "7 0\n0 1 0 0 0 1 0",
"output": "1\n0 1 0 0 0 1 0"
},
{
"input": "7 2\n1 0 1 1 0 0 0",
"output": "5\n1 1 1 1 1 0 0"
},
{
"input": "7 3\n1 0 0 0 0 1 0",
"output": "4\n1 1 1 1 0 1 0"
},
{
"input": "3 0\n1 1 1",
"output": "3\n1 1 1"
},
{
"input": "1 0\n1",
"output": "1\n1"
},
{
"input": "2 0\n0 1",
"output": "1\n0 1"
},
{
"input": "1 1\n0",
"output": "1\n1"
},
{
"input": "3 1\n0 0 1",
"output": "2\n0 1 1"
},
{
"input": "2 0\n0 0",
"output": "0\n0 0"
},
{
"input": "3 1\n0 0 0",
"output": "1\n1 0 0"
},
{
"input": "7 7\n1 0 1 0 0 0 0",
"output": "7\n1 1 1 1 1 1 1"
},
{
"input": "1 1\n1",
"output": "1\n1"
},
{
"input": "3 0\n0 0 0",
"output": "0\n0 0 0"
},
{
"input": "7 3\n1 1 1 1 1 1 1",
"output": "7\n1 1 1 1 1 1 1"
},
{
"input": "10 3\n0 0 1 0 0 1 0 0 1 1",
"output": "6\n0 0 1 0 1 1 1 1 1 1"
},
{
"input": "2 1\n0 0",
"output": "1\n1 0"
},
{
"input": "2 2\n1 0",
"output": "2\n1 1"
}
] | 1,623,649,676 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 5 | 92 | 0 | n,k=map(int,input().split())
a=list(map(int,input().split()))
i=0
j=0
countzero=0
ss=0
ee=0
aa=0
while i<n-1:
while countzero<=k:
if a[i]==0:
countzero+=1
i+=1
while countzero>k:
if a[j]==0:
countzero-=1
j+=1
if (i-j+1)>aa:
ss=i
ee=j
aa=(i-j+1)
print(aa)
for k in range(j,i):
a[k]=1
print(*a)
| Title: Hard Process
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a* with *n* elements. Each element of *a* is either 0 or 1.
Let's denote the length of the longest subsegment of consecutive elements in *a*, consisting of only numbers one, as *f*(*a*). You can change no more than *k* zeroes to ones to maximize *f*(*a*).
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=3·105,<=0<=≤<=*k*<=≤<=*n*) — the number of elements in *a* and the parameter *k*.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1) — the elements of *a*.
Output Specification:
On the first line print a non-negative integer *z* — the maximal value of *f*(*a*) after no more than *k* changes of zeroes to ones.
On the second line print *n* integers *a**j* — the elements of the array *a* after the changes.
If there are multiple answers, you can print any one of them.
Demo Input:
['7 1\n1 0 0 1 1 0 1\n', '10 2\n1 0 0 1 0 1 0 1 0 1\n']
Demo Output:
['4\n1 0 0 1 1 1 1\n', '5\n1 0 0 1 1 1 1 1 0 1\n']
Note:
none | ```python
n,k=map(int,input().split())
a=list(map(int,input().split()))
i=0
j=0
countzero=0
ss=0
ee=0
aa=0
while i<n-1:
while countzero<=k:
if a[i]==0:
countzero+=1
i+=1
while countzero>k:
if a[j]==0:
countzero-=1
j+=1
if (i-j+1)>aa:
ss=i
ee=j
aa=(i-j+1)
print(aa)
for k in range(j,i):
a[k]=1
print(*a)
``` | 0 |
|
840 | B | Leha and another game about graph | PROGRAMMING | 2,100 | [
"constructive algorithms",
"data structures",
"dfs and similar",
"dp",
"graphs"
] | null | null | Leha plays a computer game, where is on each level is given a connected graph with *n* vertices and *m* edges. Graph can contain multiple edges, but can not contain self loops. Each vertex has an integer *d**i*, which can be equal to 0, 1 or <=-<=1. To pass the level, he needs to find a «good» subset of edges of the graph or say, that it doesn't exist. Subset is called «good», if by by leaving only edges from this subset in the original graph, we obtain the following: for every vertex i, *d**i*<==<=<=-<=1 or it's degree modulo 2 is equal to *d**i*. Leha wants to pass the game as soon as possible and ask you to help him. In case of multiple correct answers, print any of them. | The first line contains two integers *n*, *m* (1<=≤<=*n*<=≤<=3·105, *n*<=-<=1<=≤<=*m*<=≤<=3·105) — number of vertices and edges.
The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (<=-<=1<=≤<=*d**i*<=≤<=1) — numbers on the vertices.
Each of the next *m* lines contains two integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*) — edges. It's guaranteed, that graph in the input is connected. | Print <=-<=1 in a single line, if solution doesn't exist. Otherwise in the first line *k* — number of edges in a subset. In the next *k* lines indexes of edges. Edges are numerated in order as they are given in the input, starting from 1. | [
"1 0\n1\n",
"4 5\n0 0 0 -1\n1 2\n2 3\n3 4\n1 4\n2 4\n",
"2 1\n1 1\n1 2\n",
"3 3\n0 -1 1\n1 2\n2 3\n1 3\n"
] | [
"-1\n",
"0\n",
"1\n1\n",
"1\n2\n"
] | In the first sample we have single vertex without edges. It's degree is 0 and we can not get 1. | 1,000 | [
{
"input": "1 0\n1",
"output": "-1"
},
{
"input": "4 5\n0 0 0 -1\n1 2\n2 3\n3 4\n1 4\n2 4",
"output": "0"
},
{
"input": "2 1\n1 1\n1 2",
"output": "1\n1"
},
{
"input": "3 3\n0 -1 1\n1 2\n2 3\n1 3",
"output": "1\n2"
},
{
"input": "10 10\n-1 -1 -1 -1 -1 -1 -1 -1 -1 -1\n6 7\n8 3\n6 4\n4 2\n9 2\n5 10\n9 8\n10 7\n5 1\n6 2",
"output": "0"
},
{
"input": "3 2\n1 0 1\n1 2\n2 3",
"output": "2\n2\n1"
}
] | 1,689,249,608 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1689249607.7267241")# 1689249607.726764 | Title: Leha and another game about graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leha plays a computer game, where is on each level is given a connected graph with *n* vertices and *m* edges. Graph can contain multiple edges, but can not contain self loops. Each vertex has an integer *d**i*, which can be equal to 0, 1 or <=-<=1. To pass the level, he needs to find a «good» subset of edges of the graph or say, that it doesn't exist. Subset is called «good», if by by leaving only edges from this subset in the original graph, we obtain the following: for every vertex i, *d**i*<==<=<=-<=1 or it's degree modulo 2 is equal to *d**i*. Leha wants to pass the game as soon as possible and ask you to help him. In case of multiple correct answers, print any of them.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*<=≤<=3·105, *n*<=-<=1<=≤<=*m*<=≤<=3·105) — number of vertices and edges.
The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (<=-<=1<=≤<=*d**i*<=≤<=1) — numbers on the vertices.
Each of the next *m* lines contains two integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*) — edges. It's guaranteed, that graph in the input is connected.
Output Specification:
Print <=-<=1 in a single line, if solution doesn't exist. Otherwise in the first line *k* — number of edges in a subset. In the next *k* lines indexes of edges. Edges are numerated in order as they are given in the input, starting from 1.
Demo Input:
['1 0\n1\n', '4 5\n0 0 0 -1\n1 2\n2 3\n3 4\n1 4\n2 4\n', '2 1\n1 1\n1 2\n', '3 3\n0 -1 1\n1 2\n2 3\n1 3\n']
Demo Output:
['-1\n', '0\n', '1\n1\n', '1\n2\n']
Note:
In the first sample we have single vertex without edges. It's degree is 0 and we can not get 1. | ```python
print("_RANDOM_GUESS_1689249607.7267241")# 1689249607.726764
``` | 0 |
|
667 | B | Coat of Anticubism | PROGRAMMING | 1,100 | [
"constructive algorithms",
"geometry"
] | null | null | As some of you know, cubism is a trend in art, where the problem of constructing volumetrical shape on a plane with a combination of three-dimensional geometric shapes comes to the fore.
A famous sculptor Cicasso, whose self-portrait you can contemplate, hates cubism. He is more impressed by the idea to transmit two-dimensional objects through three-dimensional objects by using his magnificent sculptures. And his new project is connected with this. Cicasso wants to make a coat for the haters of anticubism. To do this, he wants to create a sculpture depicting a well-known geometric primitive — convex polygon.
Cicasso prepared for this a few blanks, which are rods with integer lengths, and now he wants to bring them together. The *i*-th rod is a segment of length *l**i*.
The sculptor plans to make a convex polygon with a nonzero area, using all rods he has as its sides. Each rod should be used as a side to its full length. It is forbidden to cut, break or bend rods. However, two sides may form a straight angle .
Cicasso knows that it is impossible to make a convex polygon with a nonzero area out of the rods with the lengths which he had chosen. Cicasso does not want to leave the unused rods, so the sculptor decides to make another rod-blank with an integer length so that his problem is solvable. Of course, he wants to make it as short as possible, because the materials are expensive, and it is improper deed to spend money for nothing.
Help sculptor! | The first line contains an integer *n* (3<=≤<=*n*<=≤<=105) — a number of rod-blanks.
The second line contains *n* integers *l**i* (1<=≤<=*l**i*<=≤<=109) — lengths of rods, which Cicasso already has. It is guaranteed that it is impossible to make a polygon with *n* vertices and nonzero area using the rods Cicasso already has. | Print the only integer *z* — the minimum length of the rod, so that after adding it it can be possible to construct convex polygon with (*n*<=+<=1) vertices and nonzero area from all of the rods. | [
"3\n1 2 1\n",
"5\n20 4 3 2 1\n"
] | [
"1\n",
"11\n"
] | In the first example triangle with sides {1 + 1 = 2, 2, 1} can be formed from a set of lengths {1, 1, 1, 2}.
In the second example you can make a triangle with lengths {20, 11, 4 + 3 + 2 + 1 = 10}. | 1,000 | [
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "5\n20 4 3 2 1",
"output": "11"
},
{
"input": "7\n77486105 317474713 89523018 332007362 7897847 949616701 54820086",
"output": "70407571"
},
{
"input": "14\n245638694 2941428 4673577 12468 991349408 44735727 14046308 60637707 81525 104620306 88059371 53742651 8489205 3528194",
"output": "360142248"
},
{
"input": "19\n479740 7703374 196076708 180202968 579604 17429 16916 11989886 30832424 6384983 8937497 431 62955 48167457 898566333 29534955 1485775 848444 372839845",
"output": "2404943"
},
{
"input": "35\n306260 278 43508628 54350745 222255 842526 39010821 10627 14916465 3059978 61449 503809 2820 1609513 196062 65695 270869 15079297 2885093 189306 4682268 422616382 1642346 82340 6 2 975464673 1388191 70110665 272855 253160079 1849635 7837751 274070 10394",
"output": "34445194"
},
{
"input": "53\n1014364 40727 75774 243769 314 406417 5272684 14138 10640282 64955 2763 5667043 2121887 204672692 567643 60183 5183 11361359 2792918 199155 174809 16182540 21 392221 19434423 9140891 159733 15438 67903 3816799 616 429181 30392293 413992581 10847741 20771 16366654 1163 414283 156163 55907108 310278 95949614 185865 976650886 197610 87 61264 4586815 107764 26390852 331828 541",
"output": "25390787"
},
{
"input": "3\n1 1 1000000000",
"output": "999999999"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 1000000000",
"output": "999999956"
},
{
"input": "5\n100000000 100000000 100000000 100000000 500000000",
"output": "100000001"
},
{
"input": "3\n300000000 300000000 600000000",
"output": "1"
},
{
"input": "5\n10 4 3 2 1",
"output": "1"
},
{
"input": "3\n800000000 1 1",
"output": "799999999"
},
{
"input": "3\n1000000000 1 1",
"output": "999999999"
}
] | 1,677,322,571 | 2,147,483,647 | Python 3 | OK | TESTS | 51 | 77 | 3,174,400 | def main():
param = 0
sum = 0
geom = []
maximum = 0
index = 0
n = int(input())
geom = list(map(int, input().split()))
for i in range(0, n):
if geom[i] > maximum:
maximum = geom[i]
index = i
geom = geom[:index] + geom[index+1:]
for i in range(0, n - 1):
sum += geom[i]
param = maximum - sum + 1
print(param)
if __name__ == '__main__':
main()
| Title: Coat of Anticubism
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As some of you know, cubism is a trend in art, where the problem of constructing volumetrical shape on a plane with a combination of three-dimensional geometric shapes comes to the fore.
A famous sculptor Cicasso, whose self-portrait you can contemplate, hates cubism. He is more impressed by the idea to transmit two-dimensional objects through three-dimensional objects by using his magnificent sculptures. And his new project is connected with this. Cicasso wants to make a coat for the haters of anticubism. To do this, he wants to create a sculpture depicting a well-known geometric primitive — convex polygon.
Cicasso prepared for this a few blanks, which are rods with integer lengths, and now he wants to bring them together. The *i*-th rod is a segment of length *l**i*.
The sculptor plans to make a convex polygon with a nonzero area, using all rods he has as its sides. Each rod should be used as a side to its full length. It is forbidden to cut, break or bend rods. However, two sides may form a straight angle .
Cicasso knows that it is impossible to make a convex polygon with a nonzero area out of the rods with the lengths which he had chosen. Cicasso does not want to leave the unused rods, so the sculptor decides to make another rod-blank with an integer length so that his problem is solvable. Of course, he wants to make it as short as possible, because the materials are expensive, and it is improper deed to spend money for nothing.
Help sculptor!
Input Specification:
The first line contains an integer *n* (3<=≤<=*n*<=≤<=105) — a number of rod-blanks.
The second line contains *n* integers *l**i* (1<=≤<=*l**i*<=≤<=109) — lengths of rods, which Cicasso already has. It is guaranteed that it is impossible to make a polygon with *n* vertices and nonzero area using the rods Cicasso already has.
Output Specification:
Print the only integer *z* — the minimum length of the rod, so that after adding it it can be possible to construct convex polygon with (*n*<=+<=1) vertices and nonzero area from all of the rods.
Demo Input:
['3\n1 2 1\n', '5\n20 4 3 2 1\n']
Demo Output:
['1\n', '11\n']
Note:
In the first example triangle with sides {1 + 1 = 2, 2, 1} can be formed from a set of lengths {1, 1, 1, 2}.
In the second example you can make a triangle with lengths {20, 11, 4 + 3 + 2 + 1 = 10}. | ```python
def main():
param = 0
sum = 0
geom = []
maximum = 0
index = 0
n = int(input())
geom = list(map(int, input().split()))
for i in range(0, n):
if geom[i] > maximum:
maximum = geom[i]
index = i
geom = geom[:index] + geom[index+1:]
for i in range(0, n - 1):
sum += geom[i]
param = maximum - sum + 1
print(param)
if __name__ == '__main__':
main()
``` | 3 |
|
380 | C | Sereja and Brackets | PROGRAMMING | 2,000 | [
"data structures",
"schedules"
] | null | null | Sereja has a bracket sequence *s*1,<=*s*2,<=...,<=*s**n*, or, in other words, a string *s* of length *n*, consisting of characters "(" and ")".
Sereja needs to answer *m* queries, each of them is described by two integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). The answer to the *i*-th query is the length of the maximum correct bracket subsequence of sequence *s**l**i*,<=*s**l**i*<=+<=1,<=...,<=*s**r**i*. Help Sereja answer all queries.
You can find the definitions for a subsequence and a correct bracket sequence in the notes. | The first line contains a sequence of characters *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*n*<=≤<=106) without any spaces. Each character is either a "(" or a ")". The second line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains a pair of integers. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) — the description of the *i*-th query. | Print the answer to each question on a single line. Print the answers in the order they go in the input. | [
"())(())(())(\n7\n1 1\n2 3\n1 2\n1 12\n8 12\n5 11\n2 10\n"
] | [
"0\n0\n2\n10\n4\n6\n6\n"
] | A subsequence of length |*x*| of string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (where |*s*| is the length of string *s*) is string *x* = *s*<sub class="lower-index">*k*<sub class="lower-index">1</sub></sub>*s*<sub class="lower-index">*k*<sub class="lower-index">2</sub></sub>... *s*<sub class="lower-index">*k*<sub class="lower-index">|*x*|</sub></sub> (1 ≤ *k*<sub class="lower-index">1</sub> < *k*<sub class="lower-index">2</sub> < ... < *k*<sub class="lower-index">|*x*|</sub> ≤ |*s*|).
A correct bracket sequence is a bracket sequence that can be transformed into a correct aryphmetic expression by inserting characters "1" and "+" between the characters of the string. For example, bracket sequences "()()", "(())" are correct (the resulting expressions "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
For the third query required sequence will be «()».
For the fourth query required sequence will be «()(())(())». | 1,500 | [
{
"input": "())(())(())(\n7\n1 1\n2 3\n1 2\n1 12\n8 12\n5 11\n2 10",
"output": "0\n0\n2\n10\n4\n6\n6"
},
{
"input": "(((((()((((((((((()((()(((((\n1\n8 15",
"output": "0"
},
{
"input": "((()((())(((((((((()(()(()(((((((((((((((()(()((((((((((((((()(((((((((((((((((((()(((\n39\n28 56\n39 46\n57 63\n29 48\n51 75\n14 72\n5 70\n51 73\n10 64\n31 56\n50 54\n15 78\n78 82\n1 11\n1 70\n1 19\n10 22\n13 36\n3 10\n34 40\n51 76\n64 71\n36 75\n24 71\n1 63\n5 14\n46 67\n32 56\n39 43\n43 56\n61 82\n2 78\n1 21\n10 72\n49 79\n12 14\n53 79\n15 31\n7 47",
"output": "4\n4\n2\n4\n2\n12\n16\n2\n12\n4\n0\n12\n0\n6\n18\n6\n2\n6\n6\n0\n2\n0\n6\n8\n18\n4\n2\n4\n2\n2\n2\n18\n8\n12\n2\n0\n2\n6\n12"
},
{
"input": "))(()))))())())))))())((()()))))()))))))))))))\n9\n26 42\n21 22\n6 22\n7 26\n43 46\n25 27\n32 39\n22 40\n2 45",
"output": "4\n0\n6\n8\n0\n2\n2\n10\n20"
},
{
"input": "(()((((()(())((((((((()((((((()((((\n71\n15 29\n17 18\n5 26\n7 10\n16 31\n26 35\n2 30\n16 24\n2 24\n7 12\n15 18\n12 13\n25 30\n1 30\n12 13\n16 20\n6 35\n20 28\n18 23\n9 31\n12 35\n14 17\n8 16\n3 10\n12 33\n7 19\n2 33\n7 17\n21 27\n10 30\n29 32\n9 28\n18 32\n28 31\n31 33\n4 26\n15 27\n10 17\n8 14\n11 28\n8 23\n17 33\n4 14\n3 6\n6 34\n19 23\n4 21\n16 27\n14 27\n6 19\n31 32\n29 32\n9 17\n1 21\n2 31\n18 29\n16 26\n15 18\n4 5\n13 20\n9 28\n18 30\n1 32\n2 9\n16 24\n1 20\n4 15\n16 23\n19 34\n5 22\n5 23",
"output": "2\n0\n8\n2\n4\n2\n10\n2\n10\n4\n0\n0\n0\n10\n0\n0\n10\n2\n2\n8\n4\n0\n6\n2\n4\n6\n12\n6\n2\n6\n2\n6\n4\n2\n0\n8\n2\n4\n6\n4\n8\n4\n6\n0\n10\n2\n6\n2\n2\n6\n0\n2\n4\n8\n12\n2\n2\n0\n0\n0\n6\n2\n12\n4\n2\n8\n6\n2\n4\n6\n8"
},
{
"input": "(((())((((()()((((((()((()(((((((((((()((\n6\n20 37\n28 32\n12 18\n7 25\n21 33\n4 5",
"output": "4\n0\n2\n6\n4\n2"
},
{
"input": "(((()((((()()()(()))((((()(((()))()((((()))()((())\n24\n37 41\n13 38\n31 34\n14 16\n29 29\n12 46\n1 26\n15 34\n8 47\n11 23\n6 32\n2 22\n9 27\n17 40\n6 15\n4 49\n12 33\n3 48\n22 47\n19 48\n10 27\n23 25\n4 44\n27 48",
"output": "2\n16\n0\n2\n0\n26\n16\n12\n30\n8\n18\n14\n14\n12\n6\n34\n16\n32\n18\n18\n12\n0\n30\n16"
},
{
"input": ")()((((((((((((((((()(((()()(()((((((()(((((((()()))((((())(((((((((()(((((((((\n51\n29 53\n31 69\n54 59\n3 52\n26 46\n14 62\n6 54\n39 56\n17 27\n46 74\n60 72\n18 26\n38 46\n4 27\n22 52\n44 49\n42 77\n2 20\n39 57\n61 70\n33 54\n10 30\n67 70\n46 66\n17 77\n5 52\n33 77\n26 32\n1 72\n40 78\n38 68\n19 47\n30 53\n19 29\n52 71\n1 11\n22 53\n17 42\n2 51\n4 12\n24 76\n22 34\n21 69\n11 69\n36 52\n17 31\n57 58\n54 62\n23 71\n5 46\n51 53",
"output": "12\n14\n4\n18\n6\n22\n18\n8\n4\n12\n2\n4\n2\n4\n16\n2\n14\n2\n8\n2\n10\n6\n2\n10\n24\n18\n16\n4\n26\n14\n14\n10\n12\n6\n6\n2\n16\n10\n18\n0\n22\n6\n20\n22\n10\n8\n2\n4\n22\n10\n0"
},
{
"input": "(\n1\n1 1",
"output": "0"
},
{
"input": ")\n1\n1 1",
"output": "0"
},
{
"input": "()\n1\n1 2",
"output": "2"
},
{
"input": ")(\n1\n1 2",
"output": "0"
}
] | 1,683,753,323 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 61 | 0 | import sys, math
def merge(left, right):
ans = (
left[0] + right[0] - min(left[0], right[1]),
left[1] + right[1] - min(left[0], right[1]),
left[2] + right[2] + min(left[0], right[1])
)
return ans
def build_iter(tree):
n = len(arr)
# print(n, len(tree))
for i in range(n):
tree[n+i] = (
int(arr[i] == "("),
int(arr[i] == ")"),
0
)
for i in range(n - 1, 0, -1):
tree[i] = merge(tree[2*i], tree[2*i+1])
def query_iter(l, r):
ans = None
n = len(arr)
l += n
r += n
while l <= r:
# print(l, r)
if l % 2 == 1:
# print("\t",'before', l, seg_tree[l], ans)
ans = seg_tree[l] if not ans else merge(ans, seg_tree[l])
# print("\t", "left-aft",ans)
if r % 2 == 0:
# print("\t",'rgt-bef', r, seg_tree[r], ans)
ans = seg_tree[r] if not ans else merge(ans, seg_tree[r])
# print("\t", "rgt-aft",ans)
l = (l+1)//2
r = (r-1)//2
# print(ans)
return ans
def nearestPowerOf2(N):
# Calculate log2 of N
a = int(math.log2(N))
# If 2^a is equal to N, return N
if 2**a == N:
return N
# Return 2^(a + 1)
return 2**(a + 1)
st = sys.stdin.readline()
arr = [*st][:-1]
n = int(sys.stdin.readline())
# print(n,arr)
ln = len(arr)
pow_2 = nearestPowerOf2(ln)
if ln < pow_2:
temp = [None] * (pow_2 - ln)
arr.extend(temp)
seg_tree = [(0, 0, 0)] * (2 * pow_2)
build_iter(seg_tree)
# for i, val in enumerate(seg_tree):
# print(i,val, end="\n")
# print()
for _ in range(n):
l,r = map(int, sys.stdin.readline().split())
# print(l,r)
# print(seg_tree)
ans = query_iter(l - 1, r - 1)
print(ans[2] * 2) | Title: Sereja and Brackets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja has a bracket sequence *s*1,<=*s*2,<=...,<=*s**n*, or, in other words, a string *s* of length *n*, consisting of characters "(" and ")".
Sereja needs to answer *m* queries, each of them is described by two integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). The answer to the *i*-th query is the length of the maximum correct bracket subsequence of sequence *s**l**i*,<=*s**l**i*<=+<=1,<=...,<=*s**r**i*. Help Sereja answer all queries.
You can find the definitions for a subsequence and a correct bracket sequence in the notes.
Input Specification:
The first line contains a sequence of characters *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*n*<=≤<=106) without any spaces. Each character is either a "(" or a ")". The second line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains a pair of integers. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*) — the description of the *i*-th query.
Output Specification:
Print the answer to each question on a single line. Print the answers in the order they go in the input.
Demo Input:
['())(())(())(\n7\n1 1\n2 3\n1 2\n1 12\n8 12\n5 11\n2 10\n']
Demo Output:
['0\n0\n2\n10\n4\n6\n6\n']
Note:
A subsequence of length |*x*| of string *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">|*s*|</sub> (where |*s*| is the length of string *s*) is string *x* = *s*<sub class="lower-index">*k*<sub class="lower-index">1</sub></sub>*s*<sub class="lower-index">*k*<sub class="lower-index">2</sub></sub>... *s*<sub class="lower-index">*k*<sub class="lower-index">|*x*|</sub></sub> (1 ≤ *k*<sub class="lower-index">1</sub> < *k*<sub class="lower-index">2</sub> < ... < *k*<sub class="lower-index">|*x*|</sub> ≤ |*s*|).
A correct bracket sequence is a bracket sequence that can be transformed into a correct aryphmetic expression by inserting characters "1" and "+" between the characters of the string. For example, bracket sequences "()()", "(())" are correct (the resulting expressions "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
For the third query required sequence will be «()».
For the fourth query required sequence will be «()(())(())». | ```python
import sys, math
def merge(left, right):
ans = (
left[0] + right[0] - min(left[0], right[1]),
left[1] + right[1] - min(left[0], right[1]),
left[2] + right[2] + min(left[0], right[1])
)
return ans
def build_iter(tree):
n = len(arr)
# print(n, len(tree))
for i in range(n):
tree[n+i] = (
int(arr[i] == "("),
int(arr[i] == ")"),
0
)
for i in range(n - 1, 0, -1):
tree[i] = merge(tree[2*i], tree[2*i+1])
def query_iter(l, r):
ans = None
n = len(arr)
l += n
r += n
while l <= r:
# print(l, r)
if l % 2 == 1:
# print("\t",'before', l, seg_tree[l], ans)
ans = seg_tree[l] if not ans else merge(ans, seg_tree[l])
# print("\t", "left-aft",ans)
if r % 2 == 0:
# print("\t",'rgt-bef', r, seg_tree[r], ans)
ans = seg_tree[r] if not ans else merge(ans, seg_tree[r])
# print("\t", "rgt-aft",ans)
l = (l+1)//2
r = (r-1)//2
# print(ans)
return ans
def nearestPowerOf2(N):
# Calculate log2 of N
a = int(math.log2(N))
# If 2^a is equal to N, return N
if 2**a == N:
return N
# Return 2^(a + 1)
return 2**(a + 1)
st = sys.stdin.readline()
arr = [*st][:-1]
n = int(sys.stdin.readline())
# print(n,arr)
ln = len(arr)
pow_2 = nearestPowerOf2(ln)
if ln < pow_2:
temp = [None] * (pow_2 - ln)
arr.extend(temp)
seg_tree = [(0, 0, 0)] * (2 * pow_2)
build_iter(seg_tree)
# for i, val in enumerate(seg_tree):
# print(i,val, end="\n")
# print()
for _ in range(n):
l,r = map(int, sys.stdin.readline().split())
# print(l,r)
# print(seg_tree)
ans = query_iter(l - 1, r - 1)
print(ans[2] * 2)
``` | 0 |
|
259 | B | Little Elephant and Magic Square | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Little Elephant loves magic squares very much.
A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.
The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105.
Help the Little Elephant, restore the original magic square, given the Elephant's notes. | The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes.
It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105. | Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.
It is guaranteed that there exists at least one magic square that meets the conditions. | [
"0 1 1\n1 0 1\n1 1 0\n",
"0 3 6\n5 0 5\n4 7 0\n"
] | [
"1 1 1\n1 1 1\n1 1 1\n",
"6 3 6\n5 5 5\n4 7 4\n"
] | none | 1,000 | [
{
"input": "0 1 1\n1 0 1\n1 1 0",
"output": "1 1 1\n1 1 1\n1 1 1"
},
{
"input": "0 3 6\n5 0 5\n4 7 0",
"output": "6 3 6\n5 5 5\n4 7 4"
},
{
"input": "0 4 4\n4 0 4\n4 4 0",
"output": "4 4 4\n4 4 4\n4 4 4"
},
{
"input": "0 54 48\n36 0 78\n66 60 0",
"output": "69 54 48\n36 57 78\n66 60 45"
},
{
"input": "0 17 14\n15 0 15\n16 13 0",
"output": "14 17 14\n15 15 15\n16 13 16"
},
{
"input": "0 97 56\n69 0 71\n84 43 0",
"output": "57 97 56\n69 70 71\n84 43 83"
},
{
"input": "0 1099 1002\n1027 0 1049\n1074 977 0",
"output": "1013 1099 1002\n1027 1038 1049\n1074 977 1063"
},
{
"input": "0 98721 99776\n99575 0 99123\n98922 99977 0",
"output": "99550 98721 99776\n99575 99349 99123\n98922 99977 99148"
},
{
"input": "0 6361 2304\n1433 0 8103\n7232 3175 0",
"output": "5639 6361 2304\n1433 4768 8103\n7232 3175 3897"
},
{
"input": "0 99626 99582\n99766 0 99258\n99442 99398 0",
"output": "99328 99626 99582\n99766 99512 99258\n99442 99398 99696"
},
{
"input": "0 99978 99920\n99950 0 99918\n99948 99890 0",
"output": "99904 99978 99920\n99950 99934 99918\n99948 99890 99964"
},
{
"input": "0 840 666\n612 0 948\n894 720 0",
"output": "834 840 666\n612 780 948\n894 720 726"
},
{
"input": "0 28 10\n12 0 24\n26 8 0",
"output": "16 28 10\n12 18 24\n26 8 20"
},
{
"input": "0 120 83\n98 0 90\n105 68 0",
"output": "79 120 83\n98 94 90\n105 68 109"
},
{
"input": "0 86900 85807\n85836 0 86842\n86871 85778 0",
"output": "86310 86900 85807\n85836 86339 86842\n86871 85778 86368"
},
{
"input": "0 74 78\n78 0 74\n74 78 0",
"output": "76 74 78\n78 76 74\n74 78 76"
},
{
"input": "0 505 681\n605 0 657\n581 757 0",
"output": "707 505 681\n605 631 657\n581 757 555"
},
{
"input": "0 662 918\n822 0 854\n758 1014 0",
"output": "934 662 918\n822 838 854\n758 1014 742"
},
{
"input": "0 93 95\n93 0 97\n95 97 0",
"output": "97 93 95\n93 95 97\n95 97 93"
},
{
"input": "0 709 712\n719 0 695\n702 705 0",
"output": "700 709 712\n719 707 695\n702 705 714"
},
{
"input": "0 7 6\n9 0 1\n4 3 0",
"output": "2 7 6\n9 5 1\n4 3 8"
},
{
"input": "0 9 2\n3 0 7\n8 1 0",
"output": "4 9 2\n3 5 7\n8 1 6"
},
{
"input": "0 1 43\n13 0 61\n31 73 0",
"output": "67 1 43\n13 37 61\n31 73 7"
},
{
"input": "0 100000 100000\n100000 0 100000\n100000 100000 0",
"output": "100000 100000 100000\n100000 100000 100000\n100000 100000 100000"
}
] | 1,686,481,476 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 24 | 124 | 1,843,200 | r1=list(map(int,input().split()))
r2=list(map(int,input().split()))
r3=list(map(int,input().split()))
s1=sum(r1)
s2=sum(r2)
s3=sum(r3)
for i in range(1,10**5+1):
r1[0]=i
s1+=i
r2[1]=s1-s2
s2+=r2[1]
r3[2]=s1-s3
s3+=r3[2]
if r2[1]>0 and r3[2]>0 and s1==s2==s3==(r1[0]+r2[0]+r3[0])==(r1[1]+r2[1]+r3[1])==(r1[2]+r2[2]+r3[2])==(r1[0]+r2[1]+r3[2])==(r1[2]+r2[1]+r3[0]):
print(*r1)
print(*r2)
print(*r3)
break
r1[0]=0
s1-=i
s2-=r2[1]
r2[1]=0
s3-=r3[2]
r3[2]=0 | Title: Little Elephant and Magic Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Elephant loves magic squares very much.
A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.
The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105.
Help the Little Elephant, restore the original magic square, given the Elephant's notes.
Input Specification:
The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes.
It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.
Output Specification:
Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.
It is guaranteed that there exists at least one magic square that meets the conditions.
Demo Input:
['0 1 1\n1 0 1\n1 1 0\n', '0 3 6\n5 0 5\n4 7 0\n']
Demo Output:
['1 1 1\n1 1 1\n1 1 1\n', '6 3 6\n5 5 5\n4 7 4\n']
Note:
none | ```python
r1=list(map(int,input().split()))
r2=list(map(int,input().split()))
r3=list(map(int,input().split()))
s1=sum(r1)
s2=sum(r2)
s3=sum(r3)
for i in range(1,10**5+1):
r1[0]=i
s1+=i
r2[1]=s1-s2
s2+=r2[1]
r3[2]=s1-s3
s3+=r3[2]
if r2[1]>0 and r3[2]>0 and s1==s2==s3==(r1[0]+r2[0]+r3[0])==(r1[1]+r2[1]+r3[1])==(r1[2]+r2[2]+r3[2])==(r1[0]+r2[1]+r3[2])==(r1[2]+r2[1]+r3[0]):
print(*r1)
print(*r2)
print(*r3)
break
r1[0]=0
s1-=i
s2-=r2[1]
r2[1]=0
s3-=r3[2]
r3[2]=0
``` | 3 |
|
888 | D | Almost Identity Permutations | PROGRAMMING | 1,600 | [
"combinatorics",
"dp",
"math"
] | null | null | A permutation *p* of size *n* is an array such that every integer from 1 to *n* occurs exactly once in this array.
Let's call a permutation an almost identity permutation iff there exist at least *n*<=-<=*k* indices *i* (1<=≤<=*i*<=≤<=*n*) such that *p**i*<==<=*i*.
Your task is to count the number of almost identity permutations for given numbers *n* and *k*. | The first line contains two integers *n* and *k* (4<=≤<=*n*<=≤<=1000, 1<=≤<=*k*<=≤<=4). | Print the number of almost identity permutations for given *n* and *k*. | [
"4 1\n",
"4 2\n",
"5 3\n",
"5 4\n"
] | [
"1\n",
"7\n",
"31\n",
"76\n"
] | none | 0 | [
{
"input": "4 1",
"output": "1"
},
{
"input": "4 2",
"output": "7"
},
{
"input": "5 3",
"output": "31"
},
{
"input": "5 4",
"output": "76"
},
{
"input": "200 1",
"output": "1"
},
{
"input": "200 2",
"output": "19901"
},
{
"input": "200 3",
"output": "2646701"
},
{
"input": "200 4",
"output": "584811251"
},
{
"input": "400 1",
"output": "1"
},
{
"input": "400 2",
"output": "79801"
},
{
"input": "400 3",
"output": "21253401"
},
{
"input": "400 4",
"output": "9477912501"
},
{
"input": "600 1",
"output": "1"
},
{
"input": "600 2",
"output": "179701"
},
{
"input": "600 3",
"output": "71820101"
},
{
"input": "600 4",
"output": "48187303751"
},
{
"input": "800 1",
"output": "1"
},
{
"input": "800 2",
"output": "319601"
},
{
"input": "800 3",
"output": "170346801"
},
{
"input": "800 4",
"output": "152620985001"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 2",
"output": "499501"
},
{
"input": "1000 3",
"output": "332833501"
},
{
"input": "1000 4",
"output": "373086956251"
},
{
"input": "4 4",
"output": "24"
}
] | 1,660,810,027 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | from math import factorial as f
n, k = map(int, input().split())
d = {1: 0, 2: 1, 3: 2, 4: 11}
ans = 1
for i in range(2, k + 1):
ans += f(n) // f(n - i) // f(i) * d[i]
print(ans)
| Title: Almost Identity Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation *p* of size *n* is an array such that every integer from 1 to *n* occurs exactly once in this array.
Let's call a permutation an almost identity permutation iff there exist at least *n*<=-<=*k* indices *i* (1<=≤<=*i*<=≤<=*n*) such that *p**i*<==<=*i*.
Your task is to count the number of almost identity permutations for given numbers *n* and *k*.
Input Specification:
The first line contains two integers *n* and *k* (4<=≤<=*n*<=≤<=1000, 1<=≤<=*k*<=≤<=4).
Output Specification:
Print the number of almost identity permutations for given *n* and *k*.
Demo Input:
['4 1\n', '4 2\n', '5 3\n', '5 4\n']
Demo Output:
['1\n', '7\n', '31\n', '76\n']
Note:
none | ```python
from math import factorial as f
n, k = map(int, input().split())
d = {1: 0, 2: 1, 3: 2, 4: 11}
ans = 1
for i in range(2, k + 1):
ans += f(n) // f(n - i) // f(i) * d[i]
print(ans)
``` | 0 |
|
404 | A | Valera and X | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him. | The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper. | Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes. | [
"5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n",
"3\nwsw\nsws\nwsw\n",
"3\nxpx\npxp\nxpe\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nwsw\nsws\nwsw",
"output": "YES"
},
{
"input": "3\nxpx\npxp\nxpe",
"output": "NO"
},
{
"input": "5\nliiil\nilili\niilii\nilili\nliiil",
"output": "YES"
},
{
"input": "7\nbwccccb\nckcccbj\nccbcbcc\ncccbccc\nccbcbcc\ncbcccbc\nbccccdt",
"output": "NO"
},
{
"input": "13\nsooooooooooos\nosoooooooooso\noosooooooosoo\nooosooooosooo\noooosooosoooo\nooooososooooo\noooooosoooooo\nooooososooooo\noooosooosoooo\nooosooooosooo\noosooooooosoo\nosoooooooooso\nsooooooooooos",
"output": "YES"
},
{
"input": "3\naaa\naaa\naaa",
"output": "NO"
},
{
"input": "3\naca\noec\nzba",
"output": "NO"
},
{
"input": "15\nrxeeeeeeeeeeeer\nereeeeeeeeeeere\needeeeeeeeeeoee\neeereeeeeeeewee\neeeereeeeebeeee\nqeeeereeejedyee\neeeeeerereeeeee\neeeeeeereeeeeee\neeeeeerereeeeze\neeeeereeereeeee\neeeereeeeegeeee\neeereeeeeeereee\neereeeeeeqeeved\ncreeeeeeceeeere\nreeerneeeeeeeer",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxxxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxoxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxxxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxoox\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxaxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\noxoxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "3\nxxx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxx\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxxx",
"output": "NO"
},
{
"input": "3\nxax\nxxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naaa\nxax",
"output": "NO"
},
{
"input": "3\naax\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxaa\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\naax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaa",
"output": "NO"
},
{
"input": "3\nxfx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\nafa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaf",
"output": "NO"
},
{
"input": "3\nxox\nxxx\nxxx",
"output": "NO"
},
{
"input": "3\naxa\naax\nxxa",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxox",
"output": "NO"
},
{
"input": "3\nxox\nooo\nxox",
"output": "NO"
},
{
"input": "3\naaa\naab\nbbb",
"output": "NO"
},
{
"input": "3\nxxx\nsxs\nxsx",
"output": "NO"
},
{
"input": "5\nabbba\nbabab\nbbbbb\nbaaab\nabbba",
"output": "NO"
},
{
"input": "5\nabaaa\nbbbbb\nbbabb\nbabab\nabbba",
"output": "NO"
},
{
"input": "5\nxoxox\noxoxo\nooxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxxx",
"output": "NO"
},
{
"input": "5\nxoooo\noxooo\nooxoo\noooxo\noooox",
"output": "NO"
},
{
"input": "5\nxoooo\noxoxx\nooxoo\noxoxo\noxoox",
"output": "NO"
},
{
"input": "3\naaa\nbab\naba",
"output": "NO"
}
] | 1,642,804,611 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | a=int(input())
f=True
for i in range(a):
f=True if (f and ((b:=input())[i]==b[-i-1] and b.count(b[i])==2) or (i==a//2 and b.count(b[i])==1)) else False
print("YES" if f else "NO") | Title: Valera and X
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper.
Output Specification:
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
Demo Input:
['5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n', '3\nwsw\nsws\nwsw\n', '3\nxpx\npxp\nxpe\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none | ```python
a=int(input())
f=True
for i in range(a):
f=True if (f and ((b:=input())[i]==b[-i-1] and b.count(b[i])==2) or (i==a//2 and b.count(b[i])==1)) else False
print("YES" if f else "NO")
``` | 0 |
|
884 | D | Boxes And Balls | PROGRAMMING | 2,300 | [
"data structures",
"greedy"
] | null | null | Ivan has *n* different boxes. The first of them contains some balls of *n* different colors.
Ivan wants to play a strange game. He wants to distribute the balls into boxes in such a way that for every *i* (1<=≤<=*i*<=≤<=*n*) *i*-th box will contain all balls with color *i*.
In order to do this, Ivan will make some turns. Each turn he does the following:
1. Ivan chooses any non-empty box and takes all balls from this box; 1. Then Ivan chooses any *k* empty boxes (the box from the first step becomes empty, and Ivan is allowed to choose it), separates the balls he took on the previous step into *k* non-empty groups and puts each group into one of the boxes. He should put each group into a separate box. He can choose either *k*<==<=2 or *k*<==<=3.
The penalty of the turn is the number of balls Ivan takes from the box during the first step of the turn. And penalty of the game is the total penalty of turns made by Ivan until he distributes all balls to corresponding boxes.
Help Ivan to determine the minimum possible penalty of the game! | The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200000) — the number of boxes and colors.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the number of balls with color *i*. | Print one number — the minimum possible penalty of the game. | [
"3\n1 2 3\n",
"4\n2 3 4 5\n"
] | [
"6\n",
"19\n"
] | In the first example you take all the balls from the first box, choose *k* = 3 and sort all colors to corresponding boxes. Penalty is 6.
In the second example you make two turns:
1. Take all the balls from the first box, choose *k* = 3, put balls of color 3 to the third box, of color 4 — to the fourth box and the rest put back into the first box. Penalty is 14; 1. Take all the balls from the first box, choose *k* = 2, put balls of color 1 to the first box, of color 2 — to the second box. Penalty is 5.
Total penalty is 19. | 0 | [
{
"input": "3\n1 2 3",
"output": "6"
},
{
"input": "4\n2 3 4 5",
"output": "19"
},
{
"input": "6\n1 4 4 4 4 4",
"output": "38"
},
{
"input": "8\n821407370 380061316 428719552 90851747 825473738 704702117 845629927 245820158",
"output": "8176373828"
},
{
"input": "1\n10",
"output": "0"
},
{
"input": "1\n4",
"output": "0"
},
{
"input": "1\n12312",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "2\n3 4",
"output": "7"
}
] | 1,509,644,165 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 15,257,600 | class HeapMin:
def __init__(self):
self.heap = []
self.length = 0
def insert(self,valor):
self.heap.append(valor)
self.up(self.length)
self.length+=1
def remove(self):
if self.length > 0:
ret = self.heap[0]
self.length-=1
self.heap[0] = self.heap[self.length]
self.down(0)
del(self.heap[self.length])
return ret
def up(self,i):
j = (i-1)//2
if(j >= 0):
if(self.heap[i]<self.heap[j]):
aux = self.heap[i]
self.heap[i] = self.heap[j]
self.heap[j] = aux
self.up(j)
def down(self,i):
j = (i+1)*2 - 1
if(j<=self.length):
if(j<self.length):
if(self.heap[j+1] < self.heap[j]):
j+=1
if(self.heap[i] > self.heap[j]):
aux = self.heap[i]
self.heap[i] = self.heap[j]
self.heap[j] = aux
self.down(j)
n = int(input())
boxes = HeapMin()
resp=0
for i in input().split():
boxes.insert(int(i))
if(n%2==0):
boxes.insert(0)
n+=1
while n!=1:
c = boxes.remove() + boxes.remove() + boxes.remove()
resp+=c
boxes.insert(c)
n-=2
print(resp)
| Title: Boxes And Balls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan has *n* different boxes. The first of them contains some balls of *n* different colors.
Ivan wants to play a strange game. He wants to distribute the balls into boxes in such a way that for every *i* (1<=≤<=*i*<=≤<=*n*) *i*-th box will contain all balls with color *i*.
In order to do this, Ivan will make some turns. Each turn he does the following:
1. Ivan chooses any non-empty box and takes all balls from this box; 1. Then Ivan chooses any *k* empty boxes (the box from the first step becomes empty, and Ivan is allowed to choose it), separates the balls he took on the previous step into *k* non-empty groups and puts each group into one of the boxes. He should put each group into a separate box. He can choose either *k*<==<=2 or *k*<==<=3.
The penalty of the turn is the number of balls Ivan takes from the box during the first step of the turn. And penalty of the game is the total penalty of turns made by Ivan until he distributes all balls to corresponding boxes.
Help Ivan to determine the minimum possible penalty of the game!
Input Specification:
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=200000) — the number of boxes and colors.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the number of balls with color *i*.
Output Specification:
Print one number — the minimum possible penalty of the game.
Demo Input:
['3\n1 2 3\n', '4\n2 3 4 5\n']
Demo Output:
['6\n', '19\n']
Note:
In the first example you take all the balls from the first box, choose *k* = 3 and sort all colors to corresponding boxes. Penalty is 6.
In the second example you make two turns:
1. Take all the balls from the first box, choose *k* = 3, put balls of color 3 to the third box, of color 4 — to the fourth box and the rest put back into the first box. Penalty is 14; 1. Take all the balls from the first box, choose *k* = 2, put balls of color 1 to the first box, of color 2 — to the second box. Penalty is 5.
Total penalty is 19. | ```python
class HeapMin:
def __init__(self):
self.heap = []
self.length = 0
def insert(self,valor):
self.heap.append(valor)
self.up(self.length)
self.length+=1
def remove(self):
if self.length > 0:
ret = self.heap[0]
self.length-=1
self.heap[0] = self.heap[self.length]
self.down(0)
del(self.heap[self.length])
return ret
def up(self,i):
j = (i-1)//2
if(j >= 0):
if(self.heap[i]<self.heap[j]):
aux = self.heap[i]
self.heap[i] = self.heap[j]
self.heap[j] = aux
self.up(j)
def down(self,i):
j = (i+1)*2 - 1
if(j<=self.length):
if(j<self.length):
if(self.heap[j+1] < self.heap[j]):
j+=1
if(self.heap[i] > self.heap[j]):
aux = self.heap[i]
self.heap[i] = self.heap[j]
self.heap[j] = aux
self.down(j)
n = int(input())
boxes = HeapMin()
resp=0
for i in input().split():
boxes.insert(int(i))
if(n%2==0):
boxes.insert(0)
n+=1
while n!=1:
c = boxes.remove() + boxes.remove() + boxes.remove()
resp+=c
boxes.insert(c)
n-=2
print(resp)
``` | 0 |
|
393 | A | Nineteen | PROGRAMMING | 0 | [] | null | null | Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string. | The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100. | Print a single integer — the maximum number of "nineteen"s that she can get in her string. | [
"nniinneetteeeenn\n",
"nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n",
"nineteenineteen\n"
] | [
"2",
"2",
"2"
] | none | 500 | [
{
"input": "nniinneetteeeenn",
"output": "2"
},
{
"input": "nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii",
"output": "2"
},
{
"input": "nineteenineteen",
"output": "2"
},
{
"input": "nssemsnnsitjtihtthij",
"output": "0"
},
{
"input": "eehihnttehtherjsihihnrhimihrjinjiehmtjimnrss",
"output": "1"
},
{
"input": "rrrteiehtesisntnjirtitijnjjjthrsmhtneirjimniemmnrhirssjnhetmnmjejjnjjritjttnnrhnjs",
"output": "2"
},
{
"input": "mmrehtretseihsrjmtsenemniehssnisijmsnntesismmtmthnsieijjjnsnhisi",
"output": "2"
},
{
"input": "hshretttnntmmiertrrnjihnrmshnthirnnirrheinnnrjiirshthsrsijtrrtrmnjrrjnresnintnmtrhsnjrinsseimn",
"output": "1"
},
{
"input": "snmmensntritetnmmmerhhrmhnehehtesmhthseemjhmnrti",
"output": "2"
},
{
"input": "rmeetriiitijmrenmeiijt",
"output": "0"
},
{
"input": "ihimeitimrmhriemsjhrtjtijtesmhemnmmrsetmjttthtjhnnmirtimne",
"output": "1"
},
{
"input": "rhtsnmnesieernhstjnmmirthhieejsjttsiierhihhrrijhrrnejsjer",
"output": "2"
},
{
"input": "emmtjsjhretehmiiiestmtmnmissjrstnsnjmhimjmststsitemtttjrnhsrmsenjtjim",
"output": "2"
},
{
"input": "nmehhjrhirniitshjtrrtitsjsntjhrstjehhhrrerhemehjeermhmhjejjesnhsiirheijjrnrjmminneeehtm",
"output": "3"
},
{
"input": "hsntijjetmehejtsitnthietssmeenjrhhetsnjrsethisjrtrhrierjtmimeenjnhnijeesjttrmn",
"output": "3"
},
{
"input": "jnirirhmirmhisemittnnsmsttesjhmjnsjsmntisheneiinsrjsjirnrmnjmjhmistntersimrjni",
"output": "1"
},
{
"input": "neithjhhhtmejjnmieishethmtetthrienrhjmjenrmtejerernmthmsnrthhtrimmtmshm",
"output": "2"
},
{
"input": "sithnrsnemhijsnjitmijjhejjrinejhjinhtisttteermrjjrtsirmessejireihjnnhhemiirmhhjeet",
"output": "3"
},
{
"input": "jrjshtjstteh",
"output": "0"
},
{
"input": "jsihrimrjnnmhttmrtrenetimemjnshnimeiitmnmjishjjneisesrjemeshjsijithtn",
"output": "2"
},
{
"input": "hhtjnnmsemermhhtsstejehsssmnesereehnnsnnremjmmieethmirjjhn",
"output": "2"
},
{
"input": "tmnersmrtsehhntsietttrehrhneiireijnijjejmjhei",
"output": "1"
},
{
"input": "mtstiresrtmesritnjriirehtermtrtseirtjrhsejhhmnsineinsjsin",
"output": "2"
},
{
"input": "ssitrhtmmhtnmtreijteinimjemsiiirhrttinsnneshintjnin",
"output": "1"
},
{
"input": "rnsrsmretjiitrjthhritniijhjmm",
"output": "0"
},
{
"input": "hntrteieimrimteemenserntrejhhmijmtjjhnsrsrmrnsjseihnjmehtthnnithirnhj",
"output": "3"
},
{
"input": "nmmtsmjrntrhhtmimeresnrinstjnhiinjtnjjjnthsintmtrhijnrnmtjihtinmni",
"output": "0"
},
{
"input": "eihstiirnmteejeehimttrijittjsntjejmessstsemmtristjrhenithrrsssihnthheehhrnmimssjmejjreimjiemrmiis",
"output": "2"
},
{
"input": "srthnimimnemtnmhsjmmmjmmrsrisehjseinemienntetmitjtnnneseimhnrmiinsismhinjjnreehseh",
"output": "3"
},
{
"input": "etrsmrjehntjjimjnmsresjnrthjhehhtreiijjminnheeiinseenmmethiemmistsei",
"output": "3"
},
{
"input": "msjeshtthsieshejsjhsnhejsihisijsertenrshhrthjhiirijjneinjrtrmrs",
"output": "1"
},
{
"input": "mehsmstmeejrhhsjihntjmrjrihssmtnensttmirtieehimj",
"output": "1"
},
{
"input": "mmmsermimjmrhrhejhrrejermsneheihhjemnehrhihesnjsehthjsmmjeiejmmnhinsemjrntrhrhsmjtttsrhjjmejj",
"output": "2"
},
{
"input": "rhsmrmesijmmsnsmmhertnrhsetmisshriirhetmjihsmiinimtrnitrseii",
"output": "1"
},
{
"input": "iihienhirmnihh",
"output": "0"
},
{
"input": "ismtthhshjmhisssnmnhe",
"output": "0"
},
{
"input": "rhsmnrmhejshinnjrtmtsssijimimethnm",
"output": "0"
},
{
"input": "eehnshtiriejhiirntminrirnjihmrnittnmmnjejjhjtennremrnssnejtntrtsiejjijisermj",
"output": "3"
},
{
"input": "rnhmeesnhttrjintnhnrhristjrthhrmehrhjmjhjehmstrijemjmmistes",
"output": "2"
},
{
"input": "ssrmjmjeeetrnimemrhimes",
"output": "0"
},
{
"input": "n",
"output": "0"
},
{
"input": "ni",
"output": "0"
},
{
"input": "nine",
"output": "0"
},
{
"input": "nineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteenineteen",
"output": "13"
},
{
"input": "ninetee",
"output": "0"
},
{
"input": "mzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwynd",
"output": "0"
},
{
"input": "zenudggmyopddhszhrbmftgzmjorabhgojdtfnzxjkayjlkgczsyshczutkdch",
"output": "0"
},
{
"input": "rtzxovxqfapkdmelxiyjroohufhbakpmmvaxq",
"output": "0"
},
{
"input": "zninetneeineteeniwnteeennieteenineteenineteenineteenineteenineteenineteenineteenineteeninetzeenz",
"output": "13"
},
{
"input": "nnnnnnniiiiiiiiiiiitttttttttteeeeeeeeeeeeeeeeee",
"output": "3"
},
{
"input": "ttttiiiieeeeeeeeeeeennnnnnnnn",
"output": "4"
},
{
"input": "ttttttttteeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiinnnnnnn",
"output": "3"
},
{
"input": "nnnnnnnnnneeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiiiiiiiiiitttttttttttttttttttt",
"output": "4"
},
{
"input": "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeiiiiiiiiiiiiiiiiittttttttttttttttnnnnnnn",
"output": "3"
},
{
"input": "nineeen",
"output": "0"
},
{
"input": "nineteeeeeeeeeeeeeeeeettttttttttttttttttiiiiiiiiiiiiiiiiii",
"output": "0"
},
{
"input": "nineteenieteenieteenieteenieteenieteenieteen",
"output": "4"
},
{
"input": "nineteenineteenineteenineteenineteen",
"output": "5"
}
] | 1,554,135,137 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 109 | 0 | n=input().split()
x=n[0].count("n")%3
y=n[0].count("n")/3
if(x==0 and n[0].count("i")==y and n[0].count("e")==3*y and n[0].count("t")==y):
print(y)
else:
print("NOt")
| Title: Nineteen
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string.
Input Specification:
The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100.
Output Specification:
Print a single integer — the maximum number of "nineteen"s that she can get in her string.
Demo Input:
['nniinneetteeeenn\n', 'nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n', 'nineteenineteen\n']
Demo Output:
['2', '2', '2']
Note:
none | ```python
n=input().split()
x=n[0].count("n")%3
y=n[0].count("n")/3
if(x==0 and n[0].count("i")==y and n[0].count("e")==3*y and n[0].count("t")==y):
print(y)
else:
print("NOt")
``` | 0 |
|
688 | B | Lovely Palindromes | PROGRAMMING | 1,000 | [
"constructive algorithms",
"math"
] | null | null | Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number? | The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000). | Print the *n*-th even-length palindrome number. | [
"1\n",
"10\n"
] | [
"11\n",
"1001\n"
] | The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001. | 1,000 | [
{
"input": "1",
"output": "11"
},
{
"input": "10",
"output": "1001"
},
{
"input": "11",
"output": "1111"
},
{
"input": "12",
"output": "1221"
},
{
"input": "100",
"output": "100001"
},
{
"input": "1321",
"output": "13211231"
},
{
"input": "2",
"output": "22"
},
{
"input": "3",
"output": "33"
},
{
"input": "4",
"output": "44"
},
{
"input": "5",
"output": "55"
},
{
"input": "6",
"output": "66"
},
{
"input": "7",
"output": "77"
},
{
"input": "8",
"output": "88"
},
{
"input": "9",
"output": "99"
},
{
"input": "13",
"output": "1331"
},
{
"input": "14",
"output": "1441"
},
{
"input": "15",
"output": "1551"
},
{
"input": "16",
"output": "1661"
},
{
"input": "17",
"output": "1771"
},
{
"input": "18",
"output": "1881"
},
{
"input": "19",
"output": "1991"
},
{
"input": "20",
"output": "2002"
},
{
"input": "26550",
"output": "2655005562"
},
{
"input": "16137",
"output": "1613773161"
},
{
"input": "91471",
"output": "9147117419"
},
{
"input": "41242",
"output": "4124224214"
},
{
"input": "30866",
"output": "3086666803"
},
{
"input": "4442231232741003009964183199713851566466079882929007253443848371978048610682180431295671905105980116",
"output": "44422312327410030099641831997138515664660798829290072534438483719780486106821804312956719051059801166110895015091765921340812860168408791738483443527009292889706646651583179913814699003001472321322444"
},
{
"input": "7806526084732010742850470902979267046151204705563201965818390803445852115032119297648697933316807252",
"output": "78065260847320107428504709029792670461512047055632019658183908034458521150321192976486979333168072522527086133397968467929112305112585443080938185691023655074021516407629792090740582470102374806256087"
},
{
"input": "2244399823612183124978103773977739948197709395321563148611877999503840818596183803323258317041875913",
"output": "22443998236121831249781037739777399481977093953215631486118779995038408185961838033232583170418759133195781407138523233083816958180483059997781168413651235939077918499377793773018794213812163289934422"
},
{
"input": "123451",
"output": "123451154321"
}
] | 1,686,667,647 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 1,000 | 102,400 | q=int(input())
a=q
s=''
while a>0:
s=s+str(a%10)
a=a//10
s=str(q)+s
print(s) | Title: Lovely Palindromes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
Output Specification:
Print the *n*-th even-length palindrome number.
Demo Input:
['1\n', '10\n']
Demo Output:
['11\n', '1001\n']
Note:
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001. | ```python
q=int(input())
a=q
s=''
while a>0:
s=s+str(a%10)
a=a//10
s=str(q)+s
print(s)
``` | 0 |
|
755 | E | PolandBall and White-Red graph | PROGRAMMING | 2,400 | [
"constructive algorithms",
"graphs",
"shortest paths"
] | null | null | PolandBall has an undirected simple graph consisting of *n* vertices. Unfortunately, it has no edges. The graph is very sad because of that. PolandBall wanted to make it happier, adding some red edges. Then, he will add white edges in every remaining place. Therefore, the final graph will be a clique in two colors: white and red.
Colorfulness of the graph is a value *min*(*d**r*,<=*d**w*), where *d**r* is the diameter of the red subgraph and *d**w* is the diameter of white subgraph. The diameter of a graph is a largest value *d* such that shortest path between some pair of vertices in it is equal to *d*. If the graph is not connected, we consider its diameter to be -1.
PolandBall wants the final graph to be as neat as possible. He wants the final colorfulness to be equal to *k*. Can you help him and find any graph which satisfies PolandBall's requests? | The only one input line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*k*<=≤<=1000), representing graph's size and sought colorfulness. | If it's impossible to find a suitable graph, print -1.
Otherwise, you can output any graph which fulfills PolandBall's requirements. First, output *m* — the number of red edges in your graph. Then, you should output *m* lines, each containing two integers *a**i* and *b**i*, (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*) which means that there is an undirected red edge between vertices *a**i* and *b**i*. Every red edge should be printed exactly once, you can print the edges and the vertices of every edge in arbitrary order.
Remember that PolandBall's graph should remain simple, so no loops or multiple edges are allowed. | [
"4 1\n",
"5 2\n"
] | [
"-1\n",
"4\n1 2\n2 3\n3 4\n4 5\n"
] | In the first sample case, no graph can fulfill PolandBall's requirements.
In the second sample case, red graph is a path from 1 to 5. Its diameter is 4. However, white graph has diameter 2, because it consists of edges 1-3, 1-4, 1-5, 2-4, 2-5, 3-5. | 2,500 | [
{
"input": "4 1",
"output": "-1"
},
{
"input": "5 2",
"output": "4\n1 2\n2 3\n3 4\n4 5"
},
{
"input": "500 3",
"output": "123755\n1 2\n499 500\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85..."
},
{
"input": "1000 2",
"output": "999\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "10 2",
"output": "9\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10"
},
{
"input": "590 3",
"output": "172580\n1 2\n589 590\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85..."
},
{
"input": "1000 5",
"output": "-1"
},
{
"input": "5 3",
"output": "5\n1 2\n4 5\n2 3\n2 4\n3 4"
},
{
"input": "100 49",
"output": "-1"
},
{
"input": "4 2",
"output": "-1"
},
{
"input": "4 3",
"output": "3\n1 2\n3 4\n2 3"
},
{
"input": "5 4",
"output": "-1"
},
{
"input": "5 1",
"output": "-1"
},
{
"input": "7 2",
"output": "6\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7"
},
{
"input": "1000 1",
"output": "-1"
},
{
"input": "1000 3",
"output": "497505\n1 2\n999 1000\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 8..."
},
{
"input": "1000 4",
"output": "-1"
},
{
"input": "999 1",
"output": "-1"
},
{
"input": "999 2",
"output": "998\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "999 3",
"output": "496508\n1 2\n998 999\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31\n2 32\n2 33\n2 34\n2 35\n2 36\n2 37\n2 38\n2 39\n2 40\n2 41\n2 42\n2 43\n2 44\n2 45\n2 46\n2 47\n2 48\n2 49\n2 50\n2 51\n2 52\n2 53\n2 54\n2 55\n2 56\n2 57\n2 58\n2 59\n2 60\n2 61\n2 62\n2 63\n2 64\n2 65\n2 66\n2 67\n2 68\n2 69\n2 70\n2 71\n2 72\n2 73\n2 74\n2 75\n2 76\n2 77\n2 78\n2 79\n2 80\n2 81\n2 82\n2 83\n2 84\n2 85..."
},
{
"input": "999 4",
"output": "-1"
},
{
"input": "999 5",
"output": "-1"
},
{
"input": "8 2",
"output": "7\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8"
},
{
"input": "9 2",
"output": "8\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9"
},
{
"input": "6 3",
"output": "8\n1 2\n5 6\n2 3\n2 4\n2 5\n3 4\n3 5\n4 5"
},
{
"input": "7 3",
"output": "12\n1 2\n6 7\n2 3\n2 4\n2 5\n2 6\n3 4\n3 5\n3 6\n4 5\n4 6\n5 6"
},
{
"input": "8 3",
"output": "17\n1 2\n7 8\n2 3\n2 4\n2 5\n2 6\n2 7\n3 4\n3 5\n3 6\n3 7\n4 5\n4 6\n4 7\n5 6\n5 7\n6 7"
},
{
"input": "9 3",
"output": "23\n1 2\n8 9\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n3 4\n3 5\n3 6\n3 7\n3 8\n4 5\n4 6\n4 7\n4 8\n5 6\n5 7\n5 8\n6 7\n6 8\n7 8"
},
{
"input": "10 3",
"output": "30\n1 2\n9 10\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n4 5\n4 6\n4 7\n4 8\n4 9\n5 6\n5 7\n5 8\n5 9\n6 7\n6 8\n6 9\n7 8\n7 9\n8 9"
},
{
"input": "527 2",
"output": "526\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
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{
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},
{
"input": "565 2",
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},
{
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"output": "10\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11"
},
{
"input": "237 2",
"output": "236\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "107 2",
"output": "106\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "494 2",
"output": "493\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "40 2",
"output": "39\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40"
},
{
"input": "301 2",
"output": "300\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53\n53 54\n54 55\n55 56\n56 57\n57 58\n58 59\n59 60\n60 61\n61 62\n62 63\n63 64\n64 65\n65 66\n66 67\n67 68\n68 69\n69 70\n70 71\n71 72\n72 73\n73 74\n74 75\n75 76..."
},
{
"input": "101 2",
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},
{
"input": "642 3",
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},
{
"input": "683 3",
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},
{
"input": "750 3",
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},
{
"input": "851 3",
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},
{
"input": "207 3",
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},
{
"input": "196 3",
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},
{
"input": "706 3",
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},
{
"input": "416 3",
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},
{
"input": "2 1",
"output": "-1"
},
{
"input": "2 2",
"output": "-1"
},
{
"input": "2 3",
"output": "-1"
},
{
"input": "2 4",
"output": "-1"
},
{
"input": "3 1",
"output": "-1"
},
{
"input": "3 2",
"output": "-1"
},
{
"input": "3 3",
"output": "-1"
},
{
"input": "3 4",
"output": "-1"
}
] | 1,605,199,234 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 9 | 187 | 3,072,000 | l=input().split()
n=int(l[0])
k=int(l[1])
if(n<=3 or (k!=2 and k!=3)):
print(-1)
quit()
if(k==2):
print(n-1)
for i in range(n-1):
print(i+1,i+2)
else:
print(2*n-5)
print(1,2)
for i in range(n):
if(i==0 or i==2 or i==1):
continue
print(1,i+1)
for i in range(n):
if(i==0 or i==1 or i==3):
continue
print(2,i+1)
| Title: PolandBall and White-Red graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
PolandBall has an undirected simple graph consisting of *n* vertices. Unfortunately, it has no edges. The graph is very sad because of that. PolandBall wanted to make it happier, adding some red edges. Then, he will add white edges in every remaining place. Therefore, the final graph will be a clique in two colors: white and red.
Colorfulness of the graph is a value *min*(*d**r*,<=*d**w*), where *d**r* is the diameter of the red subgraph and *d**w* is the diameter of white subgraph. The diameter of a graph is a largest value *d* such that shortest path between some pair of vertices in it is equal to *d*. If the graph is not connected, we consider its diameter to be -1.
PolandBall wants the final graph to be as neat as possible. He wants the final colorfulness to be equal to *k*. Can you help him and find any graph which satisfies PolandBall's requests?
Input Specification:
The only one input line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*k*<=≤<=1000), representing graph's size and sought colorfulness.
Output Specification:
If it's impossible to find a suitable graph, print -1.
Otherwise, you can output any graph which fulfills PolandBall's requirements. First, output *m* — the number of red edges in your graph. Then, you should output *m* lines, each containing two integers *a**i* and *b**i*, (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*) which means that there is an undirected red edge between vertices *a**i* and *b**i*. Every red edge should be printed exactly once, you can print the edges and the vertices of every edge in arbitrary order.
Remember that PolandBall's graph should remain simple, so no loops or multiple edges are allowed.
Demo Input:
['4 1\n', '5 2\n']
Demo Output:
['-1\n', '4\n1 2\n2 3\n3 4\n4 5\n']
Note:
In the first sample case, no graph can fulfill PolandBall's requirements.
In the second sample case, red graph is a path from 1 to 5. Its diameter is 4. However, white graph has diameter 2, because it consists of edges 1-3, 1-4, 1-5, 2-4, 2-5, 3-5. | ```python
l=input().split()
n=int(l[0])
k=int(l[1])
if(n<=3 or (k!=2 and k!=3)):
print(-1)
quit()
if(k==2):
print(n-1)
for i in range(n-1):
print(i+1,i+2)
else:
print(2*n-5)
print(1,2)
for i in range(n):
if(i==0 or i==2 or i==1):
continue
print(1,i+1)
for i in range(n):
if(i==0 or i==1 or i==3):
continue
print(2,i+1)
``` | 0 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | A flower shop has got *n* bouquets, and the *i*-th bouquet consists of *a**i* flowers. Vasya, the manager of the shop, decided to make large bouquets from these bouquets.
Vasya thinks that a bouquet is large if it is made of two or more initial bouquets, and there is a constraint: the total number of flowers in a large bouquet should be odd. Each of the initial bouquets can be a part of at most one large bouquet. If an initial bouquet becomes a part of a large bouquet, all its flowers are included in the large bouquet.
Determine the maximum possible number of large bouquets Vasya can make. | The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of initial bouquets.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the number of flowers in each of the initial bouquets. | Print the maximum number of large bouquets Vasya can make. | [
"5\n2 3 4 2 7\n",
"6\n2 2 6 8 6 12\n",
"3\n11 4 10\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first example Vasya can make 2 large bouquets. For example, the first bouquet can contain the first and the fifth initial bouquets (the total number of flowers is then equal to 9), and the second bouquet can consist of the second and the third initial bouquets (the total number of flowers is then equal to 7). The fourth initial bouquet is unused in this scheme.
In the second example it is not possible to form a single bouquet with odd number of flowers.
In the third example Vasya can make one large bouquet. For example, he can make it using all three initial bouquets. The size of the large bouquet is then equal to 11 + 4 + 10 = 25. | 0 | [
{
"input": "5\n2 3 4 2 7",
"output": "2"
},
{
"input": "6\n2 2 6 8 6 12",
"output": "0"
},
{
"input": "3\n11 4 10",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n999999",
"output": "0"
},
{
"input": "1\n1000000",
"output": "0"
},
{
"input": "4\n943543 151729 379602 589828",
"output": "2"
},
{
"input": "2\n468463 62253",
"output": "0"
},
{
"input": "3\n352987 849349 967007",
"output": "1"
},
{
"input": "20\n274039 899325 798709 157662 963297 276599 529230 80095 252956 980560 358150 82383 29856 901568 123794 275349 512273 508369 120076 170206",
"output": "10"
},
{
"input": "25\n742168 377547 485672 437223 96307 902863 759104 747933 512899 410317 588598 666688 823202 257684 520631 910066 168864 71499 899972 565350 764848 754913 929040 864132 289976",
"output": "10"
}
] | 1,521,366,706 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 155 | 14,438,400 | n = int(input())
numbers = list(map(int, input().split()))
odd, even = 0, 0
for number in numbers:
if number%2==0:
even += 1
else:
odd += 1
ans = min(odd, even)
odd -= ans
even -= ans
ans += odd // 3
print(ans) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A flower shop has got *n* bouquets, and the *i*-th bouquet consists of *a**i* flowers. Vasya, the manager of the shop, decided to make large bouquets from these bouquets.
Vasya thinks that a bouquet is large if it is made of two or more initial bouquets, and there is a constraint: the total number of flowers in a large bouquet should be odd. Each of the initial bouquets can be a part of at most one large bouquet. If an initial bouquet becomes a part of a large bouquet, all its flowers are included in the large bouquet.
Determine the maximum possible number of large bouquets Vasya can make.
Input Specification:
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of initial bouquets.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the number of flowers in each of the initial bouquets.
Output Specification:
Print the maximum number of large bouquets Vasya can make.
Demo Input:
['5\n2 3 4 2 7\n', '6\n2 2 6 8 6 12\n', '3\n11 4 10\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first example Vasya can make 2 large bouquets. For example, the first bouquet can contain the first and the fifth initial bouquets (the total number of flowers is then equal to 9), and the second bouquet can consist of the second and the third initial bouquets (the total number of flowers is then equal to 7). The fourth initial bouquet is unused in this scheme.
In the second example it is not possible to form a single bouquet with odd number of flowers.
In the third example Vasya can make one large bouquet. For example, he can make it using all three initial bouquets. The size of the large bouquet is then equal to 11 + 4 + 10 = 25. | ```python
n = int(input())
numbers = list(map(int, input().split()))
odd, even = 0, 0
for number in numbers:
if number%2==0:
even += 1
else:
odd += 1
ans = min(odd, even)
odd -= ans
even -= ans
ans += odd // 3
print(ans)
``` | 3 |
|
296 | A | Yaroslav and Permutations | PROGRAMMING | 1,100 | [
"greedy",
"math"
] | null | null | Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements. | In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise. | [
"1\n1\n",
"3\n1 1 2\n",
"4\n7 7 7 7\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs. | 500 | [
{
"input": "1\n1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "YES"
},
{
"input": "4\n7 7 7 7",
"output": "NO"
},
{
"input": "4\n479 170 465 146",
"output": "YES"
},
{
"input": "5\n996 437 605 996 293",
"output": "YES"
},
{
"input": "6\n727 539 896 668 36 896",
"output": "YES"
},
{
"input": "7\n674 712 674 674 674 674 674",
"output": "NO"
},
{
"input": "8\n742 742 742 742 742 289 742 742",
"output": "NO"
},
{
"input": "9\n730 351 806 806 806 630 85 757 967",
"output": "YES"
},
{
"input": "10\n324 539 83 440 834 640 440 440 440 440",
"output": "YES"
},
{
"input": "7\n925 830 925 98 987 162 356",
"output": "YES"
},
{
"input": "68\n575 32 53 351 151 942 725 967 431 108 192 8 338 458 288 754 384 946 910 210 759 222 589 423 947 507 31 414 169 901 592 763 656 411 360 625 538 549 484 596 42 603 351 292 837 375 21 597 22 349 200 669 485 282 735 54 1000 419 939 901 789 128 468 729 894 649 484 808",
"output": "YES"
},
{
"input": "22\n618 814 515 310 617 936 452 601 250 520 557 799 304 225 9 845 610 990 703 196 486 94",
"output": "YES"
},
{
"input": "44\n459 581 449 449 449 449 449 449 449 623 449 449 449 449 449 449 449 449 889 449 203 273 329 449 449 449 449 449 449 845 882 323 22 449 449 893 449 449 449 449 449 870 449 402",
"output": "NO"
},
{
"input": "90\n424 3 586 183 286 89 427 618 758 833 933 170 155 722 190 977 330 369 693 426 556 435 550 442 513 146 61 719 754 140 424 280 997 688 530 550 438 867 950 194 196 298 417 287 106 489 283 456 735 115 702 317 672 787 264 314 356 186 54 913 809 833 946 314 757 322 559 647 983 482 145 197 223 130 162 536 451 174 467 45 660 293 440 254 25 155 511 746 650 187",
"output": "YES"
},
{
"input": "14\n959 203 478 315 788 788 373 834 488 519 774 764 193 103",
"output": "YES"
},
{
"input": "81\n544 528 528 528 528 4 506 528 32 528 528 528 528 528 528 528 528 975 528 528 528 528 528 528 528 528 528 528 528 528 528 20 528 528 528 528 528 528 528 528 852 528 528 120 528 528 61 11 528 528 528 228 528 165 883 528 488 475 628 528 528 528 528 528 528 597 528 528 528 528 528 528 528 528 528 528 528 412 528 521 925",
"output": "NO"
},
{
"input": "89\n354 356 352 355 355 355 352 354 354 352 355 356 355 352 354 356 354 355 355 354 353 352 352 355 355 356 352 352 353 356 352 353 354 352 355 352 353 353 353 354 353 354 354 353 356 353 353 354 354 354 354 353 352 353 355 356 356 352 356 354 353 352 355 354 356 356 356 354 354 356 354 355 354 355 353 352 354 355 352 355 355 354 356 353 353 352 356 352 353",
"output": "YES"
},
{
"input": "71\n284 284 285 285 285 284 285 284 284 285 284 285 284 284 285 284 285 285 285 285 284 284 285 285 284 284 284 285 284 285 284 285 285 284 284 284 285 284 284 285 285 285 284 284 285 284 285 285 284 285 285 284 285 284 284 284 285 285 284 285 284 285 285 285 285 284 284 285 285 284 285",
"output": "NO"
},
{
"input": "28\n602 216 214 825 814 760 814 28 76 814 814 288 814 814 222 707 11 490 814 543 914 705 814 751 976 814 814 99",
"output": "YES"
},
{
"input": "48\n546 547 914 263 986 945 914 914 509 871 324 914 153 571 914 914 914 528 970 566 544 914 914 914 410 914 914 589 609 222 914 889 691 844 621 68 914 36 914 39 630 749 914 258 945 914 727 26",
"output": "YES"
},
{
"input": "56\n516 76 516 197 516 427 174 516 706 813 94 37 516 815 516 516 937 483 16 516 842 516 638 691 516 635 516 516 453 263 516 516 635 257 125 214 29 81 516 51 362 516 677 516 903 516 949 654 221 924 516 879 516 516 972 516",
"output": "YES"
},
{
"input": "46\n314 723 314 314 314 235 314 314 314 314 270 314 59 972 314 216 816 40 314 314 314 314 314 314 314 381 314 314 314 314 314 314 314 789 314 957 114 942 314 314 29 314 314 72 314 314",
"output": "NO"
},
{
"input": "72\n169 169 169 599 694 81 250 529 865 406 817 169 667 169 965 169 169 663 65 169 903 169 942 763 169 807 169 603 169 169 13 169 169 810 169 291 169 169 169 169 169 169 169 713 169 440 169 169 169 169 169 480 169 169 867 169 169 169 169 169 169 169 169 393 169 169 459 169 99 169 601 800",
"output": "NO"
},
{
"input": "100\n317 316 317 316 317 316 317 316 317 316 316 317 317 316 317 316 316 316 317 316 317 317 316 317 316 316 316 316 316 316 317 316 317 317 317 317 317 317 316 316 316 317 316 317 316 317 316 317 317 316 317 316 317 317 316 317 316 317 316 317 316 316 316 317 317 317 317 317 316 317 317 316 316 316 316 317 317 316 317 316 316 316 316 316 316 317 316 316 317 317 317 317 317 317 317 317 317 316 316 317",
"output": "NO"
},
{
"input": "100\n510 510 510 162 969 32 510 511 510 510 911 183 496 875 903 461 510 510 123 578 510 510 510 510 510 755 510 673 510 510 763 510 510 909 510 435 487 959 807 510 368 788 557 448 284 332 510 949 510 510 777 112 857 926 487 510 510 510 678 510 510 197 829 427 698 704 409 509 510 238 314 851 510 651 510 455 682 510 714 635 973 510 443 878 510 510 510 591 510 24 596 510 43 183 510 510 671 652 214 784",
"output": "YES"
},
{
"input": "100\n476 477 474 476 476 475 473 476 474 475 473 477 476 476 474 476 474 475 476 477 473 473 473 474 474 476 473 473 476 476 475 476 473 474 473 473 477 475 475 475 476 475 477 477 477 476 475 475 475 473 476 477 475 476 477 473 474 477 473 475 476 476 474 477 476 474 473 477 473 475 477 473 476 474 477 473 475 477 473 476 476 475 476 475 474 473 477 473 475 473 477 473 473 474 475 473 477 476 477 474",
"output": "YES"
},
{
"input": "100\n498 498 498 498 498 499 498 499 499 499 498 498 498 498 499 498 499 499 498 499 498 498 498 499 499 499 498 498 499 499 498 498 498 499 498 499 498 498 498 499 498 499 498 498 498 498 499 498 498 499 498 498 499 498 499 499 498 499 499 499 498 498 498 498 499 498 499 498 499 499 499 499 498 498 499 499 498 499 499 498 498 499 499 498 498 499 499 499 498 498 499 498 498 498 499 499 499 498 498 499",
"output": "NO"
},
{
"input": "100\n858 53 816 816 816 816 816 816 816 181 816 816 816 816 579 879 816 948 171 816 816 150 866 816 816 816 897 816 816 816 816 816 816 706 816 539 816 816 816 816 816 816 423 487 816 615 254 816 816 816 816 83 816 816 816 816 816 816 816 816 816 816 816 136 775 999 816 816 816 644 816 816 816 816 927 816 802 816 856 816 816 816 816 816 816 816 816 816 816 700 816 816 816 816 982 477 816 891 806 816",
"output": "NO"
},
{
"input": "100\n167 169 169 167 169 169 167 167 167 167 168 166 170 170 169 170 170 170 169 168 166 167 170 169 167 169 168 169 166 170 166 167 170 166 166 167 169 166 166 169 166 167 168 168 170 167 168 166 168 170 167 168 167 169 169 166 168 167 170 168 167 169 168 169 166 168 168 169 169 166 170 168 167 169 170 168 167 169 168 167 168 168 166 169 170 170 166 166 167 170 167 168 167 167 169 169 166 166 169 167",
"output": "YES"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 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": "YES"
},
{
"input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 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": "YES"
},
{
"input": "2\n1 1",
"output": "NO"
},
{
"input": "1\n1000",
"output": "YES"
},
{
"input": "12\n2 2 4 4 4 4 6 6 6 6 6 6",
"output": "YES"
}
] | 1,567,661,453 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 280 | 409,600 | ######################################################################
# Write your code here
import sys
from math import *
input = sys.stdin.readline
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x10000000, resource.RLIM_INFINITY])
#sys.setrecursionlimit(0x100000)
# Write your code here
RI = lambda : [int(x) for x in sys.stdin.readline().strip().split()]
rw = lambda : input().strip().split()
ls = lambda : list(input().strip()) # for strings to list of char
from collections import defaultdict as df
import heapq
#heapq.heapify(li) heappush(li,4) heappop(li)
#import random
#random.shuffle(list)
infinite = float('inf')
#######################################################################
n=int(input())
l=RI()
d=df(int)
for i in l:
d[i]+=1
f=0
m=(n+1)//2
for i in d:
if(d[i]>m):
f=1
break
if(f==0):
print("YES")
else:
print("NO")
| Title: Yaroslav and Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
Output Specification:
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
Demo Input:
['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs. | ```python
######################################################################
# Write your code here
import sys
from math import *
input = sys.stdin.readline
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x10000000, resource.RLIM_INFINITY])
#sys.setrecursionlimit(0x100000)
# Write your code here
RI = lambda : [int(x) for x in sys.stdin.readline().strip().split()]
rw = lambda : input().strip().split()
ls = lambda : list(input().strip()) # for strings to list of char
from collections import defaultdict as df
import heapq
#heapq.heapify(li) heappush(li,4) heappop(li)
#import random
#random.shuffle(list)
infinite = float('inf')
#######################################################################
n=int(input())
l=RI()
d=df(int)
for i in l:
d[i]+=1
f=0
m=(n+1)//2
for i in d:
if(d[i]>m):
f=1
break
if(f==0):
print("YES")
else:
print("NO")
``` | 3 |
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