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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
199 | A | Hexadecimal's theorem | PROGRAMMING | 900 | [
"brute force",
"constructive algorithms",
"implementation",
"number theory"
] | null | null | Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible. | The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number. | Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them. | [
"3\n",
"13\n"
] | [
"1 1 1\n",
"2 3 8\n"
] | none | 500 | [
{
"input": "3",
"output": "1 1 1"
},
{
"input": "13",
"output": "2 3 8"
},
{
"input": "0",
"output": "0 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "2",
"output": "1 1 0"
},
{
"input": "1597",
"output": "233 377 987"
},
{
"input": "0",
"output": "0 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "2",
"output": "1 1 0"
},
{
"input": "3",
"output": "1 1 1"
},
{
"input": "5",
"output": "1 1 3"
},
{
"input": "8",
"output": "1 2 5"
},
{
"input": "13",
"output": "2 3 8"
},
{
"input": "21",
"output": "3 5 13"
},
{
"input": "34",
"output": "5 8 21"
},
{
"input": "55",
"output": "8 13 34"
},
{
"input": "89",
"output": "13 21 55"
},
{
"input": "144",
"output": "21 34 89"
},
{
"input": "233",
"output": "34 55 144"
},
{
"input": "377",
"output": "55 89 233"
},
{
"input": "610",
"output": "89 144 377"
},
{
"input": "987",
"output": "144 233 610"
},
{
"input": "1597",
"output": "233 377 987"
},
{
"input": "2584",
"output": "377 610 1597"
},
{
"input": "4181",
"output": "610 987 2584"
},
{
"input": "6765",
"output": "987 1597 4181"
},
{
"input": "10946",
"output": "1597 2584 6765"
},
{
"input": "17711",
"output": "2584 4181 10946"
},
{
"input": "28657",
"output": "4181 6765 17711"
},
{
"input": "46368",
"output": "6765 10946 28657"
},
{
"input": "75025",
"output": "10946 17711 46368"
},
{
"input": "121393",
"output": "17711 28657 75025"
},
{
"input": "196418",
"output": "28657 46368 121393"
},
{
"input": "317811",
"output": "46368 75025 196418"
},
{
"input": "514229",
"output": "75025 121393 317811"
},
{
"input": "832040",
"output": "121393 196418 514229"
},
{
"input": "1346269",
"output": "196418 317811 832040"
},
{
"input": "2178309",
"output": "317811 514229 1346269"
},
{
"input": "3524578",
"output": "514229 832040 2178309"
},
{
"input": "5702887",
"output": "832040 1346269 3524578"
},
{
"input": "9227465",
"output": "1346269 2178309 5702887"
},
{
"input": "14930352",
"output": "2178309 3524578 9227465"
},
{
"input": "24157817",
"output": "3524578 5702887 14930352"
},
{
"input": "39088169",
"output": "5702887 9227465 24157817"
},
{
"input": "63245986",
"output": "9227465 14930352 39088169"
},
{
"input": "102334155",
"output": "14930352 24157817 63245986"
},
{
"input": "165580141",
"output": "24157817 39088169 102334155"
},
{
"input": "267914296",
"output": "39088169 63245986 165580141"
},
{
"input": "433494437",
"output": "63245986 102334155 267914296"
},
{
"input": "701408733",
"output": "102334155 165580141 433494437"
},
{
"input": "701408733",
"output": "102334155 165580141 433494437"
},
{
"input": "102334155",
"output": "14930352 24157817 63245986"
},
{
"input": "63245986",
"output": "9227465 14930352 39088169"
}
] | 1,679,304,464 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
#include <bits/stdc++.h>
using namespace std;
long long binsearch(vector<long long> vec,long long x){
int l = 0, h = vec.size()-1, m = l+(h-l)/2;
int ans = -1;
while(l<=h){
m = l+(h-l)/2;
if(vec[m] > x){
h = m-1;
}
else if(vec[m] < x){
l = m+1;
}
else{
ans = m;
break;
}
}
return ans;
}
int main(){
vector<long long> vec;
vec.push_back(0);
vec.push_back(1);
long long val = 0;
long long i = 2;
long long c = 1000000000;
while(val < c){
val = vec[i-1]+vec[i-2];
vec.push_back(val);
i++;
//cout<<val<<endl;
}
//cout<<"done"<<endl;
long long n;
cin>>n;
long long id = binsearch(vec,n);
//cout<<id<<endl;
if(id >= 4 ){
cout<<vec[id-1]<<" "<<vec[id-3]<<" "<<vec[id-4]<<endl;
}
if(n == 2){
cout<<1<<" "<<1<<" "<<0<<endl;
}
if(n == 1){
cout<<1<<" "<<0<<" "<<0<<endl;
}
if(n == 0){
cout<<0<<" "<<0<<" "<<0<<endl;
}
}
| Title: Hexadecimal's theorem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.
Input Specification:
The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.
Output Specification:
Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them.
Demo Input:
['3\n', '13\n']
Demo Output:
['1 1 1\n', '2 3 8\n']
Note:
none | ```python
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
long long binsearch(vector<long long> vec,long long x){
int l = 0, h = vec.size()-1, m = l+(h-l)/2;
int ans = -1;
while(l<=h){
m = l+(h-l)/2;
if(vec[m] > x){
h = m-1;
}
else if(vec[m] < x){
l = m+1;
}
else{
ans = m;
break;
}
}
return ans;
}
int main(){
vector<long long> vec;
vec.push_back(0);
vec.push_back(1);
long long val = 0;
long long i = 2;
long long c = 1000000000;
while(val < c){
val = vec[i-1]+vec[i-2];
vec.push_back(val);
i++;
//cout<<val<<endl;
}
//cout<<"done"<<endl;
long long n;
cin>>n;
long long id = binsearch(vec,n);
//cout<<id<<endl;
if(id >= 4 ){
cout<<vec[id-1]<<" "<<vec[id-3]<<" "<<vec[id-4]<<endl;
}
if(n == 2){
cout<<1<<" "<<1<<" "<<0<<endl;
}
if(n == 1){
cout<<1<<" "<<0<<" "<<0<<endl;
}
if(n == 0){
cout<<0<<" "<<0<<" "<<0<<endl;
}
}
``` | -1 |
|
963 | D | Frequency of String | PROGRAMMING | 2,500 | [
"hashing",
"string suffix structures",
"strings"
] | null | null | You are given a string $s$. You should answer $n$ queries. The $i$-th query consists of integer $k_i$ and string $m_i$. The answer for this query is the minimum length of such a string $t$ that $t$ is a substring of $s$ and $m_i$ has at least $k_i$ occurrences as a substring in $t$.
A substring of a string is a continuous segment of characters of the string.
It is guaranteed that for any two queries the strings $m_i$ from these queries are different. | The first line contains string $s$ $(1 \leq \left | s \right | \leq 10^{5})$.
The second line contains an integer $n$ ($1 \leq n \leq 10^5$).
Each of next $n$ lines contains an integer $k_i$ $(1 \leq k_i \leq |s|)$ and a non-empty string $m_i$ — parameters of the query with number $i$, in this order.
All strings in input consists of lowercase English letters. Sum of length of all strings in input doesn't exceed $10^5$. All $m_i$ are distinct. | For each query output the answer for it in a separate line.
If a string $m_{i}$ occurs in $s$ less that $k_{i}$ times, output -1. | [
"aaaaa\n5\n3 a\n3 aa\n2 aaa\n3 aaaa\n1 aaaaa\n",
"abbb\n7\n4 b\n1 ab\n3 bb\n1 abb\n2 bbb\n1 a\n2 abbb\n"
] | [
"3\n4\n4\n-1\n5\n",
"-1\n2\n-1\n3\n-1\n1\n-1\n"
] | none | 2,000 | [] | 1,574,158,405 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | // Frequency of String
#include <iostream>
#include <string>
#include <vector>
#include <map>
using namespace std;
const int N = 1000 * 100 + 5;
const int K = 26;
struct Vertex{
int next[K];
bool leaf = false;
int p = -1;
char pch;
int link = -1;
int _go[K];
int exit_link = -1;
vector<int> acc;
Vertex(int p=-1, char ch='$'): p(p), pch(ch){
for (int i=0;i<K;++i){
next[i] = -1;
_go[i] = -1;
}
}
};
vector<Vertex> t(1);
map<int, pair<int, int> > leafs;
int get_link(int);
int go(int v, char ch){
int c = ch - 'a';
if (t[v]._go[c] == -1){
if (t[v].next[c] != -1)
t[v]._go[c] = t[v].next[c];
else
t[v]._go[c] = v == 0 ? 0 : go(get_link(v), ch);
}
return t[v]._go[c];
}
int get_link(int v){
if (t[v].link == -1){
if (v == 0 || t[v].p == 0)
t[v].link = 0;
else
t[v].link = go(get_link(t[v].p), t[v].pch);
}
return t[v].link;
}
int exit_link(int v){
if (t[v].exit_link == -1){
if (t[v].leaf || v == 0)
t[v].exit_link = v;
else
t[v].exit_link = exit_link(get_link(v));
}
return t[v].exit_link;
}
int add_string(string s){
int v = 0;
for (char ch: s){
int c = ch-'a';
if (t[v].next[c] == -1){
t[v].next[c] = t.size();
t.emplace_back(v, ch);
}
v = t[v].next[c];
}
t[v].leaf = true;
return v;
}
int main(){
string s;
cin >> s;
int n;
cin >> n;
vector<int> inpt(n);
for (int i = 0; i < n; ++i){
int cnt;
cin >> cnt;
string cur;
cin >> cur;
inpt[i] = cur.size();
leafs[add_string(cur)] = make_pair(cnt, i);
}
int v = 0;
for (int ind = 0;ind<s.size(); ++ind){
v = go(v, s[ind]);
int vv = exit_link(v);
while (vv){
t[vv].acc.push_back(ind);
vv = exit_link(get_link(vv));
}
}
vector<int> ans(n);
for (auto leaf: leafs){
int v = leaf.first;
int cnt = leaf.second.first;
int ind = leaf.second.second;
if (t[v].acc.size() < cnt){
ans[ind] = -1;
continue;
}
ans[ind] = s.size();
for(int i=cnt-1;i<t[v].acc.size();i++){
ans[ind] = min(t[v].acc[i] - t[v].acc[i-cnt+1] + inpt[ind], ans[ind]);
}
}
for(int i:ans)
cout << i << endl;
return 0;
}
| Title: Frequency of String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string $s$. You should answer $n$ queries. The $i$-th query consists of integer $k_i$ and string $m_i$. The answer for this query is the minimum length of such a string $t$ that $t$ is a substring of $s$ and $m_i$ has at least $k_i$ occurrences as a substring in $t$.
A substring of a string is a continuous segment of characters of the string.
It is guaranteed that for any two queries the strings $m_i$ from these queries are different.
Input Specification:
The first line contains string $s$ $(1 \leq \left | s \right | \leq 10^{5})$.
The second line contains an integer $n$ ($1 \leq n \leq 10^5$).
Each of next $n$ lines contains an integer $k_i$ $(1 \leq k_i \leq |s|)$ and a non-empty string $m_i$ — parameters of the query with number $i$, in this order.
All strings in input consists of lowercase English letters. Sum of length of all strings in input doesn't exceed $10^5$. All $m_i$ are distinct.
Output Specification:
For each query output the answer for it in a separate line.
If a string $m_{i}$ occurs in $s$ less that $k_{i}$ times, output -1.
Demo Input:
['aaaaa\n5\n3 a\n3 aa\n2 aaa\n3 aaaa\n1 aaaaa\n', 'abbb\n7\n4 b\n1 ab\n3 bb\n1 abb\n2 bbb\n1 a\n2 abbb\n']
Demo Output:
['3\n4\n4\n-1\n5\n', '-1\n2\n-1\n3\n-1\n1\n-1\n']
Note:
none | ```python
// Frequency of String
#include <iostream>
#include <string>
#include <vector>
#include <map>
using namespace std;
const int N = 1000 * 100 + 5;
const int K = 26;
struct Vertex{
int next[K];
bool leaf = false;
int p = -1;
char pch;
int link = -1;
int _go[K];
int exit_link = -1;
vector<int> acc;
Vertex(int p=-1, char ch='$'): p(p), pch(ch){
for (int i=0;i<K;++i){
next[i] = -1;
_go[i] = -1;
}
}
};
vector<Vertex> t(1);
map<int, pair<int, int> > leafs;
int get_link(int);
int go(int v, char ch){
int c = ch - 'a';
if (t[v]._go[c] == -1){
if (t[v].next[c] != -1)
t[v]._go[c] = t[v].next[c];
else
t[v]._go[c] = v == 0 ? 0 : go(get_link(v), ch);
}
return t[v]._go[c];
}
int get_link(int v){
if (t[v].link == -1){
if (v == 0 || t[v].p == 0)
t[v].link = 0;
else
t[v].link = go(get_link(t[v].p), t[v].pch);
}
return t[v].link;
}
int exit_link(int v){
if (t[v].exit_link == -1){
if (t[v].leaf || v == 0)
t[v].exit_link = v;
else
t[v].exit_link = exit_link(get_link(v));
}
return t[v].exit_link;
}
int add_string(string s){
int v = 0;
for (char ch: s){
int c = ch-'a';
if (t[v].next[c] == -1){
t[v].next[c] = t.size();
t.emplace_back(v, ch);
}
v = t[v].next[c];
}
t[v].leaf = true;
return v;
}
int main(){
string s;
cin >> s;
int n;
cin >> n;
vector<int> inpt(n);
for (int i = 0; i < n; ++i){
int cnt;
cin >> cnt;
string cur;
cin >> cur;
inpt[i] = cur.size();
leafs[add_string(cur)] = make_pair(cnt, i);
}
int v = 0;
for (int ind = 0;ind<s.size(); ++ind){
v = go(v, s[ind]);
int vv = exit_link(v);
while (vv){
t[vv].acc.push_back(ind);
vv = exit_link(get_link(vv));
}
}
vector<int> ans(n);
for (auto leaf: leafs){
int v = leaf.first;
int cnt = leaf.second.first;
int ind = leaf.second.second;
if (t[v].acc.size() < cnt){
ans[ind] = -1;
continue;
}
ans[ind] = s.size();
for(int i=cnt-1;i<t[v].acc.size();i++){
ans[ind] = min(t[v].acc[i] - t[v].acc[i-cnt+1] + inpt[ind], ans[ind]);
}
}
for(int i:ans)
cout << i << endl;
return 0;
}
``` | -1 |
|
538 | A | Cutting Banner | PROGRAMMING | 1,400 | [
"brute force",
"implementation"
] | null | null | A large banner with word CODEFORCES was ordered for the 1000-th onsite round of Codeforcesω that takes place on the Miami beach. Unfortunately, the company that made the banner mixed up two orders and delivered somebody else's banner that contains someone else's word. The word on the banner consists only of upper-case English letters.
There is very little time to correct the mistake. All that we can manage to do is to cut out some substring from the banner, i.e. several consecutive letters. After that all the resulting parts of the banner will be glued into a single piece (if the beginning or the end of the original banner was cut out, only one part remains); it is not allowed change the relative order of parts of the banner (i.e. after a substring is cut, several first and last letters are left, it is allowed only to glue the last letters to the right of the first letters). Thus, for example, for example, you can cut a substring out from string 'TEMPLATE' and get string 'TEMPLE' (if you cut out string AT), 'PLATE' (if you cut out TEM), 'T' (if you cut out EMPLATE), etc.
Help the organizers of the round determine whether it is possible to cut out of the banner some substring in such a way that the remaining parts formed word CODEFORCES. | The single line of the input contains the word written on the banner. The word only consists of upper-case English letters. The word is non-empty and its length doesn't exceed 100 characters. It is guaranteed that the word isn't word CODEFORCES. | Print 'YES', if there exists a way to cut out the substring, and 'NO' otherwise (without the quotes). | [
"CODEWAITFORITFORCES\n",
"BOTTOMCODER\n",
"DECODEFORCES\n",
"DOGEFORCES\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "CODEWAITFORITFORCES",
"output": "YES"
},
{
"input": "BOTTOMCODER",
"output": "NO"
},
{
"input": "DECODEFORCES",
"output": "YES"
},
{
"input": "DOGEFORCES",
"output": "NO"
},
{
"input": "ABACABA",
"output": "NO"
},
{
"input": "CODEFORCE",
"output": "NO"
},
{
"input": "C",
"output": "NO"
},
{
"input": "NQTSMZEBLY",
"output": "NO"
},
{
"input": "CODEFZORCES",
"output": "YES"
},
{
"input": "EDYKHVZCNTLJUUOQGHPTIOETQNFLLWEKZOHIUAXELGECABVSBIBGQODQXVYFKBYJWTGBYHVSSNTINKWSINWSMALUSIWNJMTCOOVF",
"output": "NO"
},
{
"input": "OCECFDSRDE",
"output": "NO"
},
{
"input": "MDBUWCZFFZKFMJTTJFXRHTGRPREORKDVUXOEMFYSOMSQGHUKGYCRCVJTNDLFDEWFS",
"output": "NO"
},
{
"input": "CODEFYTORCHES",
"output": "NO"
},
{
"input": "BCODEFORCES",
"output": "YES"
},
{
"input": "CVODEFORCES",
"output": "YES"
},
{
"input": "COAKDEFORCES",
"output": "YES"
},
{
"input": "CODFMWEFORCES",
"output": "YES"
},
{
"input": "CODEVCSYRFORCES",
"output": "YES"
},
{
"input": "CODEFXHHPWCVQORCES",
"output": "YES"
},
{
"input": "CODEFORQWUFJLOFFXTXRCES",
"output": "YES"
},
{
"input": "CODEFORBWFURYIDURNRKRDLHCLXZCES",
"output": "YES"
},
{
"input": "CODEFORCQSYSLYKCDFFUPSAZCJIAENCKZUFJZEINQIES",
"output": "YES"
},
{
"input": "CODEFORCEVENMDBQLSVPQIIBGSHBVOPYZXNWVSTVWDRONUREYJJIJIPMEBPQDCPFS",
"output": "YES"
},
{
"input": "CODEFORCESCFNNPAHNHDIPPBAUSPKJYAQDBVZNLSTSDCREZACVLMRFGVKGVHHZLXOHCTJDBQKIDWBUXDUJARLWGFGFCTTXUCAZB",
"output": "YES"
},
{
"input": "CODJRDPDEFOROES",
"output": "NO"
},
{
"input": "CODEFOGSIUZMZCMWAVQHNYFEKIEZQMAZOVEMDRMOEDBHAXPLBLDYYXCVTOOSJZVSQAKFXTBTZFWAYRZEMDEMVDJTDRXXAQBURCES",
"output": "YES"
},
{
"input": "CODEMKUYHAZSGJBQLXTHUCZZRJJJXUSEBOCNZASOKDZHMSGWZSDFBGHXFLABVPDQBJYXSHHAZAKHSTRGOKJYHRVSSUGDCMFOGCES",
"output": "NO"
},
{
"input": "CODEFORCESCODEFORCESCODEFORCESCODEFORCESCODEFORCESCODEFORCESCODEFORCESCODEFORCESCODEFORCES",
"output": "YES"
},
{
"input": "CCODEFORCESODECODEFORCCODEFORCESODCODEFORCESEFCODEFORCESORCODEFORCESCESCESFORCODEFORCESCES",
"output": "NO"
},
{
"input": "CCODEFORCESC",
"output": "NO"
},
{
"input": "CODEAFORBCES",
"output": "NO"
},
{
"input": "CODERRRRRFORCRRRRES",
"output": "NO"
},
{
"input": "CODELFORCELS",
"output": "NO"
},
{
"input": "CPOPDPEPFPOPRPCPEPS",
"output": "NO"
},
{
"input": "COXDEXFORXCEXS",
"output": "NO"
},
{
"input": "CODAAAAAFORCES",
"output": "NO"
},
{
"input": "CAOADEFORCES",
"output": "NO"
},
{
"input": "FORCESXCODE",
"output": "NO"
},
{
"input": "FORCESACODE",
"output": "NO"
},
{
"input": "ACAOADAEFORCES",
"output": "NO"
},
{
"input": "CCODEFORCESS",
"output": "NO"
},
{
"input": "ZCODEFORCEZ",
"output": "NO"
},
{
"input": "CODXEFORXCES",
"output": "NO"
},
{
"input": "CODEFORCEZ",
"output": "NO"
},
{
"input": "CODEFORCEST",
"output": "YES"
},
{
"input": "AXODEFORCES",
"output": "NO"
},
{
"input": "RCODEFORCESR",
"output": "NO"
},
{
"input": "CODECODEFORCESFORCES",
"output": "YES"
},
{
"input": "TTTWWWCODEFORCES",
"output": "YES"
},
{
"input": "CODEFORRCEST",
"output": "NO"
},
{
"input": "UJYTYUCODEFORCES",
"output": "YES"
},
{
"input": "CODEXXXXXXXXXXXXXXXXXXCODEFORCESXXXXXXXXXXXXXXXXXXXXXFORCES",
"output": "YES"
},
{
"input": "COXEDYFORCES",
"output": "NO"
},
{
"input": "UJYTYCODEFORCES",
"output": "YES"
},
{
"input": "UJYTCODEFORCES",
"output": "YES"
}
] | 1,479,203,115 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 11 | 93 | 0 | masukkan = input()
checker = "CODEFORCES"
string = ""
for y in checker:
if y in masukkan:
index = masukkan.index(y)
masukkan = masukkan[index:]
if y == "S":
print("YES")
else:
print("NO")
break
| Title: Cutting Banner
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A large banner with word CODEFORCES was ordered for the 1000-th onsite round of Codeforcesω that takes place on the Miami beach. Unfortunately, the company that made the banner mixed up two orders and delivered somebody else's banner that contains someone else's word. The word on the banner consists only of upper-case English letters.
There is very little time to correct the mistake. All that we can manage to do is to cut out some substring from the banner, i.e. several consecutive letters. After that all the resulting parts of the banner will be glued into a single piece (if the beginning or the end of the original banner was cut out, only one part remains); it is not allowed change the relative order of parts of the banner (i.e. after a substring is cut, several first and last letters are left, it is allowed only to glue the last letters to the right of the first letters). Thus, for example, for example, you can cut a substring out from string 'TEMPLATE' and get string 'TEMPLE' (if you cut out string AT), 'PLATE' (if you cut out TEM), 'T' (if you cut out EMPLATE), etc.
Help the organizers of the round determine whether it is possible to cut out of the banner some substring in such a way that the remaining parts formed word CODEFORCES.
Input Specification:
The single line of the input contains the word written on the banner. The word only consists of upper-case English letters. The word is non-empty and its length doesn't exceed 100 characters. It is guaranteed that the word isn't word CODEFORCES.
Output Specification:
Print 'YES', if there exists a way to cut out the substring, and 'NO' otherwise (without the quotes).
Demo Input:
['CODEWAITFORITFORCES\n', 'BOTTOMCODER\n', 'DECODEFORCES\n', 'DOGEFORCES\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n']
Note:
none | ```python
masukkan = input()
checker = "CODEFORCES"
string = ""
for y in checker:
if y in masukkan:
index = masukkan.index(y)
masukkan = masukkan[index:]
if y == "S":
print("YES")
else:
print("NO")
break
``` | 0 |
|
900 | B | Position in Fraction | PROGRAMMING | 1,300 | [
"math",
"number theory"
] | null | null | You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point. | The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9). | Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1. | [
"1 2 0\n",
"2 3 7\n"
] | [
"2",
"-1"
] | The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction. | 1,000 | [
{
"input": "1 2 0",
"output": "2"
},
{
"input": "2 3 7",
"output": "-1"
},
{
"input": "1 100000 1",
"output": "5"
},
{
"input": "1 7 7",
"output": "6"
},
{
"input": "99999 100000 8",
"output": "-1"
},
{
"input": "44102 73848 2",
"output": "132"
},
{
"input": "7 31 3",
"output": "15"
},
{
"input": "8880 81608 9",
"output": "161"
},
{
"input": "4942 62768 5",
"output": "122"
},
{
"input": "69168 84860 4",
"output": "107"
},
{
"input": "971 1883 3",
"output": "130"
},
{
"input": "1636 3269 6",
"output": "150"
},
{
"input": "6873 7769 3",
"output": "163"
},
{
"input": "13805 15538 3",
"output": "164"
},
{
"input": "10958 21926 3",
"output": "117"
},
{
"input": "8 51 0",
"output": "10"
},
{
"input": "1 10 1",
"output": "1"
},
{
"input": "1 9 0",
"output": "-1"
},
{
"input": "53 101 6",
"output": "-1"
},
{
"input": "1 10001 9",
"output": "5"
},
{
"input": "25102 31579 2",
"output": "174"
},
{
"input": "38790 39359 0",
"output": "212"
},
{
"input": "47117 78718 0",
"output": "213"
},
{
"input": "1 57 0",
"output": "1"
},
{
"input": "1 3 0",
"output": "-1"
},
{
"input": "1 100 0",
"output": "1"
},
{
"input": "2 3 0",
"output": "-1"
},
{
"input": "99971 99989 0",
"output": "9"
},
{
"input": "567 1580 0",
"output": "6"
},
{
"input": "45 97 0",
"output": "39"
},
{
"input": "35 111 4",
"output": "-1"
},
{
"input": "1 2 5",
"output": "1"
},
{
"input": "1 7 0",
"output": "-1"
}
] | 1,652,365,318 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | # Python3 program to calculate fraction
# of two numbers
# Function to return the required
# fraction in string format
def calculateFraction(num, den) :
# If the numerator is zero, answer is 0
if (num == 0):
return "0"
# If any one (out of numerator and denominator)
# is -ve, sign of resultant answer -ve.
sign = -1 if (num < 0) ^ (den < 0) else 1
num = abs(num)
den = abs(den)
# Calculate the absolute part
# (before decimal point).
initial = num // den
# Output string to store the answer
res = ""
# Append sign
if (sign == -1):
res += "-"
# Append the initial part
res += str(initial)
# If completely divisible, return answer.
if (num % den == 0):
return res
res += "."
# Initialize Remainder
rem = num % den
mp = {}
# Position at which fraction starts
# repeating if it exists
index = 0
repeating = False
while (rem > 0 and not repeating) :
# If this remainder is already seen,
# then there exists a repeating fraction.
if ( rem in mp):
# Index to insert parentheses
index = mp[rem]
repeating = True
break
else:
mp[rem] = len(res)
rem = rem * 10
# Calculate quotient, append it to result
# and calculate next remainder
temp = rem // den
res += str(temp )
rem = rem % den
# If repeating fraction exists,
# insert parentheses.
if (repeating) :
res += ")"
x = res[:index]
x += "("
x += res[index:]
res = x
# Return result.
return res
def searchDigit(a,b,c):
decimal = calculateFraction(a,b)
try:
# Si el numero se repite no ponemos eros de más al final
decimal.index(')')
except:
# SI el numero no se repite, ponemos ceros al final
decimal += "000"
decimal = decimal[2:]
try:
print(decimal.index(str(c))+1)
except:
print(-1)
| Title: Position in Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a fraction . You need to find the first occurrence of digit *c* into decimal notation of the fraction after decimal point.
Input Specification:
The first contains three single positive integers *a*, *b*, *c* (1<=≤<=*a*<=<<=*b*<=≤<=105, 0<=≤<=*c*<=≤<=9).
Output Specification:
Print position of the first occurrence of digit *c* into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Demo Input:
['1 2 0\n', '2 3 7\n']
Demo Output:
['2', '-1']
Note:
The fraction in the first example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/896357459a466614a0542f34c9cfb0cef1afc9ed.png" style="max-width: 100.0%;max-height: 100.0%;"/>. The first zero stands on second position.
The fraction in the second example has the following decimal notation: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/130ba579a8276fc53a1917606eee9db58817f28d.png" style="max-width: 100.0%;max-height: 100.0%;"/>. There is no digit 7 in decimal notation of the fraction. | ```python
# Python3 program to calculate fraction
# of two numbers
# Function to return the required
# fraction in string format
def calculateFraction(num, den) :
# If the numerator is zero, answer is 0
if (num == 0):
return "0"
# If any one (out of numerator and denominator)
# is -ve, sign of resultant answer -ve.
sign = -1 if (num < 0) ^ (den < 0) else 1
num = abs(num)
den = abs(den)
# Calculate the absolute part
# (before decimal point).
initial = num // den
# Output string to store the answer
res = ""
# Append sign
if (sign == -1):
res += "-"
# Append the initial part
res += str(initial)
# If completely divisible, return answer.
if (num % den == 0):
return res
res += "."
# Initialize Remainder
rem = num % den
mp = {}
# Position at which fraction starts
# repeating if it exists
index = 0
repeating = False
while (rem > 0 and not repeating) :
# If this remainder is already seen,
# then there exists a repeating fraction.
if ( rem in mp):
# Index to insert parentheses
index = mp[rem]
repeating = True
break
else:
mp[rem] = len(res)
rem = rem * 10
# Calculate quotient, append it to result
# and calculate next remainder
temp = rem // den
res += str(temp )
rem = rem % den
# If repeating fraction exists,
# insert parentheses.
if (repeating) :
res += ")"
x = res[:index]
x += "("
x += res[index:]
res = x
# Return result.
return res
def searchDigit(a,b,c):
decimal = calculateFraction(a,b)
try:
# Si el numero se repite no ponemos eros de más al final
decimal.index(')')
except:
# SI el numero no se repite, ponemos ceros al final
decimal += "000"
decimal = decimal[2:]
try:
print(decimal.index(str(c))+1)
except:
print(-1)
``` | 0 |
|
139 | A | Petr and Book | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book. | The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero. | Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. | [
"100\n15 20 20 15 10 30 45\n",
"2\n1 0 0 0 0 0 0\n"
] | [
"6\n",
"1\n"
] | Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book. | 500 | [
{
"input": "100\n15 20 20 15 10 30 45",
"output": "6"
},
{
"input": "2\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100\n100 200 100 200 300 400 500",
"output": "1"
},
{
"input": "3\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "1\n1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "20\n5 3 7 2 1 6 4",
"output": "6"
},
{
"input": "10\n5 1 1 1 1 1 5",
"output": "6"
},
{
"input": "50\n10 1 10 1 10 1 10",
"output": "1"
},
{
"input": "77\n11 11 11 11 11 11 10",
"output": "1"
},
{
"input": "1\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n100 100 100 100 100 100 100",
"output": "3"
},
{
"input": "999\n10 20 10 20 30 20 10",
"output": "3"
},
{
"input": "433\n109 58 77 10 39 125 15",
"output": "7"
},
{
"input": "1\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n1 0 1 0 1 0 1",
"output": "1"
},
{
"input": "997\n1 1 0 0 1 0 1",
"output": "1"
},
{
"input": "1000\n1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1000\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "1000\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "1000\n1 0 0 1 0 0 1",
"output": "1"
},
{
"input": "509\n105 23 98 0 7 0 155",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2\n1 1 0 0 0 0 0",
"output": "2"
},
{
"input": "1\n0 0 0 0 0 1 0",
"output": "6"
},
{
"input": "10\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 6 0",
"output": "6"
},
{
"input": "3\n0 1 0 0 0 0 0",
"output": "2"
},
{
"input": "10\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "28\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "100\n5 5 5 5 5 5 5",
"output": "6"
},
{
"input": "4\n1 0 0 0 0 0 1",
"output": "7"
},
{
"input": "2\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "7\n0 0 0 0 0 0 7",
"output": "7"
},
{
"input": "7\n2 1 1 1 1 1 0",
"output": "6"
},
{
"input": "2\n0 0 1 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1 0",
"output": "6"
},
{
"input": "5\n1 1 1 0 0 1 1",
"output": "7"
},
{
"input": "100\n10 20 30 10 10 10 10",
"output": "7"
},
{
"input": "1\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "70\n10 10 10 10 10 10 10",
"output": "7"
},
{
"input": "22\n1 2 3 4 5 6 10",
"output": "7"
},
{
"input": "5\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "2\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 0 0 0 0 0 2",
"output": "7"
},
{
"input": "10\n1 2 2 1 2 1 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "4\n0 1 1 0 0 0 0",
"output": "3"
},
{
"input": "100\n0 0 0 0 0 1 0",
"output": "6"
}
] | 1,699,018,986 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 49 | 124 | 1,536,000 | n=int(input())
a=list(map(int,input().split()))
k=0
t=0
i=0
while t<n:
if i==7:
i=0
t+=a[i]
i+=1
print(i) | Title: Petr and Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
Input Specification:
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Output Specification:
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Demo Input:
['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n']
Demo Output:
['6\n', '1\n']
Note:
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book. | ```python
n=int(input())
a=list(map(int,input().split()))
k=0
t=0
i=0
while t<n:
if i==7:
i=0
t+=a[i]
i+=1
print(i)
``` | 3 |
|
895 | B | XK Segments | PROGRAMMING | 1,700 | [
"binary search",
"math",
"sortings",
"two pointers"
] | null | null | While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array *a* and integer *x*. He should find the number of different ordered pairs of indexes (*i*,<=*j*) such that *a**i*<=≤<=*a**j* and there are exactly *k* integers *y* such that *a**i*<=≤<=*y*<=≤<=*a**j* and *y* is divisible by *x*.
In this problem it is meant that pair (*i*,<=*j*) is equal to (*j*,<=*i*) only if *i* is equal to *j*. For example pair (1,<=2) is not the same as (2,<=1). | The first line contains 3 integers *n*,<=*x*,<=*k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*x*<=≤<=109,<=0<=≤<=*k*<=≤<=109), where *n* is the size of the array *a* and *x* and *k* are numbers from the statement.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*. | Print one integer — the answer to the problem. | [
"4 2 1\n1 3 5 7\n",
"4 2 0\n5 3 1 7\n",
"5 3 1\n3 3 3 3 3\n"
] | [
"3\n",
"4\n",
"25\n"
] | In first sample there are only three suitable pairs of indexes — (1, 2), (2, 3), (3, 4).
In second sample there are four suitable pairs of indexes(1, 1), (2, 2), (3, 3), (4, 4).
In third sample every pair (*i*, *j*) is suitable, so the answer is 5 * 5 = 25. | 1,000 | [
{
"input": "4 2 1\n1 3 5 7",
"output": "3"
},
{
"input": "4 2 0\n5 3 1 7",
"output": "4"
},
{
"input": "5 3 1\n3 3 3 3 3",
"output": "25"
},
{
"input": "5 3 4\n24 13 1 24 24",
"output": "4"
},
{
"input": "4 2 2\n1 3 5 7",
"output": "2"
},
{
"input": "5 1 0\n10 12 14 20 30",
"output": "0"
},
{
"input": "7 3 1\n3 3 6 9 12 15 18",
"output": "9"
},
{
"input": "2 5 0\n3 4",
"output": "3"
},
{
"input": "3 5 0\n4 4 4",
"output": "9"
},
{
"input": "3 6 0\n3 4 5",
"output": "6"
},
{
"input": "10 2 1\n2 2 2 2 2 2 2 1 2 1",
"output": "80"
},
{
"input": "100 2 1\n1 2 2 2 1 2 1 2 2 2 2 1 2 1 1 2 1 2 2 1 1 2 1 1 2 2 1 1 2 1 2 2 2 2 2 1 1 1 2 1 2 1 2 1 2 1 2 2 1 2 2 2 1 1 1 2 2 1 1 2 2 1 1 1 1 1 1 2 2 1 1 2 2 1 2 2 1 2 1 2 2 1 2 1 1 1 2 2 1 1 2 1 2 2 1 1 2 2 2 1",
"output": "5200"
},
{
"input": "1 13 1\n13",
"output": "1"
},
{
"input": "2 1 1000000000\n1 1000000000",
"output": "1"
},
{
"input": "2 1000000000 1\n1 1000000000",
"output": "2"
},
{
"input": "1 1 1\n1",
"output": "1"
},
{
"input": "5 1000000000 1\n1000000000 1 1 1 1",
"output": "5"
},
{
"input": "2 1000 1000000\n1 1000000000",
"output": "1"
},
{
"input": "6 2 10\n1 1 1 20 20 20",
"output": "9"
},
{
"input": "1 1000000000 1\n1000000000",
"output": "1"
},
{
"input": "4 65536 65536\n1 2 3 4",
"output": "0"
},
{
"input": "1 600000000 1\n500000000",
"output": "0"
},
{
"input": "5 3 2\n2 3 4 5 8",
"output": "2"
}
] | 1,512,313,479 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 62 | 5,529,600 | from bisect import bisect_left
from bisect import bisect_right
from math import ceil
def next_line(a, x):
return x * ceil(a / x)
def num_of(arr, a, x, k):
return bisect_left(arr, next_line(a, x) + k * x) - bisect_left(arr, a)
def find(arr, a, x, k):
return num_of(arr, a, x, k) - num_of(arr, a, x, k-1)
n, x, k = input().split(' ')
n, x, k = int(n), int(x), int(k)
t = input().split(' ')
t = [int(a) for a in t]
t.sort()
#n, x, k = 9, 3, 2
#t = [3, 3, 4, 5, 6, 6, 6, 8, 8]
sum = 0
for a in t:
sum += find(t, a, x, k)
print(sum) | Title: XK Segments
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array *a* and integer *x*. He should find the number of different ordered pairs of indexes (*i*,<=*j*) such that *a**i*<=≤<=*a**j* and there are exactly *k* integers *y* such that *a**i*<=≤<=*y*<=≤<=*a**j* and *y* is divisible by *x*.
In this problem it is meant that pair (*i*,<=*j*) is equal to (*j*,<=*i*) only if *i* is equal to *j*. For example pair (1,<=2) is not the same as (2,<=1).
Input Specification:
The first line contains 3 integers *n*,<=*x*,<=*k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*x*<=≤<=109,<=0<=≤<=*k*<=≤<=109), where *n* is the size of the array *a* and *x* and *k* are numbers from the statement.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
Output Specification:
Print one integer — the answer to the problem.
Demo Input:
['4 2 1\n1 3 5 7\n', '4 2 0\n5 3 1 7\n', '5 3 1\n3 3 3 3 3\n']
Demo Output:
['3\n', '4\n', '25\n']
Note:
In first sample there are only three suitable pairs of indexes — (1, 2), (2, 3), (3, 4).
In second sample there are four suitable pairs of indexes(1, 1), (2, 2), (3, 3), (4, 4).
In third sample every pair (*i*, *j*) is suitable, so the answer is 5 * 5 = 25. | ```python
from bisect import bisect_left
from bisect import bisect_right
from math import ceil
def next_line(a, x):
return x * ceil(a / x)
def num_of(arr, a, x, k):
return bisect_left(arr, next_line(a, x) + k * x) - bisect_left(arr, a)
def find(arr, a, x, k):
return num_of(arr, a, x, k) - num_of(arr, a, x, k-1)
n, x, k = input().split(' ')
n, x, k = int(n), int(x), int(k)
t = input().split(' ')
t = [int(a) for a in t]
t.sort()
#n, x, k = 9, 3, 2
#t = [3, 3, 4, 5, 6, 6, 6, 8, 8]
sum = 0
for a in t:
sum += find(t, a, x, k)
print(sum)
``` | 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"
}
] | 1,656,426,326 | 2,147,483,647 | PyPy 3 | OK | TESTS | 35 | 154 | 0 | import math
ip = input()
num = ip.split(" ")
l = int(num[0])
b = int(num[1])
op = math.floor((l * b) / 2)
print(op)
| 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
import math
ip = input()
num = ip.split(" ")
l = int(num[0])
b = int(num[1])
op = math.floor((l * b) / 2)
print(op)
``` | 3.9615 |
165 | B | Burning Midnight Oil | PROGRAMMING | 1,500 | [
"binary search",
"implementation"
] | null | null | One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of *n* lines of code. Vasya is already exhausted, so he works like that: first he writes *v* lines of code, drinks a cup of tea, then he writes as much as lines, drinks another cup of tea, then he writes lines and so on: , , , ...
The expression is regarded as the integral part from dividing number *a* by number *b*.
The moment the current value equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished.
Vasya is wondering, what minimum allowable value *v* can take to let him write not less than *n* lines of code before he falls asleep. | The input consists of two integers *n* and *k*, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1<=≤<=*n*<=≤<=109, 2<=≤<=*k*<=≤<=10. | Print the only integer — the minimum value of *v* that lets Vasya write the program in one night. | [
"7 2\n",
"59 9\n"
] | [
"4\n",
"54\n"
] | In the first sample the answer is *v* = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task.
In the second sample the answer is *v* = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that's even more than *n* = 59. | 1,000 | [
{
"input": "7 2",
"output": "4"
},
{
"input": "59 9",
"output": "54"
},
{
"input": "1 9",
"output": "1"
},
{
"input": "11 2",
"output": "7"
},
{
"input": "747 2",
"output": "376"
},
{
"input": "6578 2",
"output": "3293"
},
{
"input": "37212 2",
"output": "18609"
},
{
"input": "12357 2",
"output": "6181"
},
{
"input": "7998332 2",
"output": "3999172"
},
{
"input": "86275251 2",
"output": "43137632"
},
{
"input": "75584551 2",
"output": "37792280"
},
{
"input": "6 3",
"output": "5"
},
{
"input": "43 4",
"output": "33"
},
{
"input": "811 3",
"output": "543"
},
{
"input": "3410 4",
"output": "2560"
},
{
"input": "21341 4",
"output": "16009"
},
{
"input": "696485 4",
"output": "522368"
},
{
"input": "8856748 3",
"output": "5904504"
},
{
"input": "2959379 4",
"output": "2219538"
},
{
"input": "831410263 3",
"output": "554273516"
},
{
"input": "2 5",
"output": "2"
},
{
"input": "19 6",
"output": "17"
},
{
"input": "715 7",
"output": "615"
},
{
"input": "9122 5",
"output": "7300"
},
{
"input": "89117 6",
"output": "74268"
},
{
"input": "689973 7",
"output": "591408"
},
{
"input": "3024524 5",
"output": "2419624"
},
{
"input": "67127156 6",
"output": "55939302"
},
{
"input": "412262167 7",
"output": "353367574"
},
{
"input": "6 8",
"output": "6"
},
{
"input": "59 9",
"output": "54"
},
{
"input": "246 10",
"output": "222"
},
{
"input": "5314 8",
"output": "4651"
},
{
"input": "15309 9",
"output": "13609"
},
{
"input": "35648 10",
"output": "32085"
},
{
"input": "3018012 8",
"output": "2640764"
},
{
"input": "92153348 9",
"output": "81914089"
},
{
"input": "177583558 10",
"output": "159825206"
},
{
"input": "1000000000 2",
"output": "500000008"
},
{
"input": "1000000000 3",
"output": "666666672"
},
{
"input": "1000000000 4",
"output": "750000005"
},
{
"input": "1000000000 5",
"output": "800000003"
},
{
"input": "1000000000 6",
"output": "833333338"
},
{
"input": "1000000000 7",
"output": "857142861"
},
{
"input": "1000000000 8",
"output": "875000004"
},
{
"input": "1000000000 9",
"output": "888888894"
},
{
"input": "1000000000 10",
"output": "900000001"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "2 10",
"output": "2"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "6 8",
"output": "6"
},
{
"input": "987862820 9",
"output": "878100288"
},
{
"input": "979591791 9",
"output": "870748262"
},
{
"input": "948889213 9",
"output": "843457081"
},
{
"input": "8 9",
"output": "8"
},
{
"input": "999999999 10",
"output": "900000000"
}
] | 1,676,581,370 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 154 | 2,867,200 | import math;
def process(guess, n, k):
denom = k
totalLines = guess
while (math.floor(guess / denom) > 0):
# print("-> ", math.floor(guess / denom))
totalLines += math.floor(guess / denom)
denom *= k
# print("totalLines: ", totalLines)
return totalLines
def main():
n = int(input())
k = int(input())
res = -1
low = 0
high = math.pow(10, 3)
while low <= high:
guess = math.floor((low + high) / 2)
temp = process(guess, n, k)
# print("guess: ", guess, "temp: ", temp)
if temp < n:
# too low
# print("low")
low = guess + 1
elif temp >= n:
high = guess - 1
res = guess
print(res)
main()
| Title: Burning Midnight Oil
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day a highly important task was commissioned to Vasya — writing a program in a night. The program consists of *n* lines of code. Vasya is already exhausted, so he works like that: first he writes *v* lines of code, drinks a cup of tea, then he writes as much as lines, drinks another cup of tea, then he writes lines and so on: , , , ...
The expression is regarded as the integral part from dividing number *a* by number *b*.
The moment the current value equals 0, Vasya immediately falls asleep and he wakes up only in the morning, when the program should already be finished.
Vasya is wondering, what minimum allowable value *v* can take to let him write not less than *n* lines of code before he falls asleep.
Input Specification:
The input consists of two integers *n* and *k*, separated by spaces — the size of the program in lines and the productivity reduction coefficient, 1<=≤<=*n*<=≤<=109, 2<=≤<=*k*<=≤<=10.
Output Specification:
Print the only integer — the minimum value of *v* that lets Vasya write the program in one night.
Demo Input:
['7 2\n', '59 9\n']
Demo Output:
['4\n', '54\n']
Note:
In the first sample the answer is *v* = 4. Vasya writes the code in the following portions: first 4 lines, then 2, then 1, and then Vasya falls asleep. Thus, he manages to write 4 + 2 + 1 = 7 lines in a night and complete the task.
In the second sample the answer is *v* = 54. Vasya writes the code in the following portions: 54, 6. The total sum is 54 + 6 = 60, that's even more than *n* = 59. | ```python
import math;
def process(guess, n, k):
denom = k
totalLines = guess
while (math.floor(guess / denom) > 0):
# print("-> ", math.floor(guess / denom))
totalLines += math.floor(guess / denom)
denom *= k
# print("totalLines: ", totalLines)
return totalLines
def main():
n = int(input())
k = int(input())
res = -1
low = 0
high = math.pow(10, 3)
while low <= high:
guess = math.floor((low + high) / 2)
temp = process(guess, n, k)
# print("guess: ", guess, "temp: ", temp)
if temp < n:
# too low
# print("low")
low = guess + 1
elif temp >= n:
high = guess - 1
res = guess
print(res)
main()
``` | -1 |
|
553 | A | Kyoya and Colored Balls | PROGRAMMING | 1,500 | [
"combinatorics",
"dp",
"math"
] | null | null | Kyoya Ootori has a bag with *n* colored balls that are colored with *k* different colors. The colors are labeled from 1 to *k*. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color *i* before drawing the last ball of color *i*<=+<=1 for all *i* from 1 to *k*<=-<=1. Now he wonders how many different ways this can happen. | The first line of input will have one integer *k* (1<=≤<=*k*<=≤<=1000) the number of colors.
Then, *k* lines will follow. The *i*-th line will contain *c**i*, the number of balls of the *i*-th color (1<=≤<=*c**i*<=≤<=1000).
The total number of balls doesn't exceed 1000. | A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1<=000<=000<=007. | [
"3\n2\n2\n1\n",
"4\n1\n2\n3\n4\n"
] | [
"3\n",
"1680\n"
] | In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are: | 250 | [
{
"input": "3\n2\n2\n1",
"output": "3"
},
{
"input": "4\n1\n2\n3\n4",
"output": "1680"
},
{
"input": "10\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "12520708"
},
{
"input": "5\n10\n10\n10\n10\n10",
"output": "425711769"
},
{
"input": "11\n291\n381\n126\n39\n19\n20\n3\n1\n20\n45\n2",
"output": "902382672"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "13\n67\n75\n76\n80\n69\n86\n75\n86\n81\n84\n73\n72\n76",
"output": "232242896"
},
{
"input": "25\n35\n43\n38\n33\n47\n44\n40\n36\n41\n42\n33\n30\n49\n42\n62\n39\n40\n35\n43\n31\n42\n46\n42\n34\n33",
"output": "362689152"
},
{
"input": "47\n20\n21\n16\n18\n24\n20\n25\n13\n20\n22\n26\n24\n17\n18\n21\n22\n21\n23\n17\n15\n24\n19\n18\n21\n20\n19\n26\n25\n20\n17\n17\n17\n26\n32\n20\n21\n25\n28\n24\n21\n21\n17\n28\n20\n20\n31\n19",
"output": "295545118"
},
{
"input": "3\n343\n317\n337",
"output": "691446102"
},
{
"input": "1\n5",
"output": "1"
}
] | 1,537,899,234 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 124 | 307,200 | def factorial(n):
if n == 0:
return 1
else:
return n * factorial(n-1)
def nCr(n, r):
return int(factorial(n)/(factorial(r)*factorial(n-r)))
def orden(b, pos):
if pos == 0:
return nCr(b[pos+1]+b[pos]-1, b[pos])
else:
sum = 0
for i in range(0, pos+2):
sum += b[i]
return nCr(sum-1-b[pos-1], b[pos])
colors = int(input())
b = [0]*colors
for i in range(0, colors):
b[i] = int(input())
mult = 1
for i in range(0, colors):
if i != colors-1:
mult *= orden(b, i)
else:
mult *= b[colors-1]
print(mult)
| Title: Kyoya and Colored Balls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori has a bag with *n* colored balls that are colored with *k* different colors. The colors are labeled from 1 to *k*. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color *i* before drawing the last ball of color *i*<=+<=1 for all *i* from 1 to *k*<=-<=1. Now he wonders how many different ways this can happen.
Input Specification:
The first line of input will have one integer *k* (1<=≤<=*k*<=≤<=1000) the number of colors.
Then, *k* lines will follow. The *i*-th line will contain *c**i*, the number of balls of the *i*-th color (1<=≤<=*c**i*<=≤<=1000).
The total number of balls doesn't exceed 1000.
Output Specification:
A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1<=000<=000<=007.
Demo Input:
['3\n2\n2\n1\n', '4\n1\n2\n3\n4\n']
Demo Output:
['3\n', '1680\n']
Note:
In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are: | ```python
def factorial(n):
if n == 0:
return 1
else:
return n * factorial(n-1)
def nCr(n, r):
return int(factorial(n)/(factorial(r)*factorial(n-r)))
def orden(b, pos):
if pos == 0:
return nCr(b[pos+1]+b[pos]-1, b[pos])
else:
sum = 0
for i in range(0, pos+2):
sum += b[i]
return nCr(sum-1-b[pos-1], b[pos])
colors = int(input())
b = [0]*colors
for i in range(0, colors):
b[i] = int(input())
mult = 1
for i in range(0, colors):
if i != colors-1:
mult *= orden(b, i)
else:
mult *= b[colors-1]
print(mult)
``` | 0 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn! | The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length. | Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise. | [
"aaba\nabaa\n",
"aabb\nabab\n"
] | [
"YES\n",
"NO\n"
] | In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | 0 | [
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "ab\nbb",
"output": "NO"
},
{
"input": "zzaa\naazz",
"output": "YES"
},
{
"input": "azza\nzaaz",
"output": "YES"
},
{
"input": "abc\nabc",
"output": "YES"
},
{
"input": "abc\nacb",
"output": "NO"
},
{
"input": "azzz\nzzaz",
"output": "YES"
},
{
"input": "abcd\ndcab",
"output": "YES"
},
{
"input": "abcd\ncdab",
"output": "YES"
},
{
"input": "abcd\ndcba",
"output": "YES"
},
{
"input": "abcd\nacbd",
"output": "NO"
},
{
"input": "oloaxgddgujq\noloaxgujqddg",
"output": "YES"
},
{
"input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta",
"output": "YES"
},
{
"input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz",
"output": "YES"
},
{
"input": "snyaydaeobufdg\nsnyaydaeobufdg",
"output": "YES"
},
{
"input": "baaaaa\nabaaaa",
"output": "NO"
},
{
"input": "hhiisug\nmzdjwju",
"output": "NO"
},
{
"input": "bbbabbabaaab\naaaabbabbbbb",
"output": "NO"
},
{
"input": "bbaaab\naababb",
"output": "NO"
},
{
"input": "aabbaaaa\naaaaabab",
"output": "NO"
},
{
"input": "aab\naba",
"output": "NO"
},
{
"input": "abcddd\nbacddd",
"output": "NO"
},
{
"input": "qgiufelsfhanx\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "aabaababaaba\naababaaababa",
"output": "NO"
},
{
"input": "nocdqzdriyyil\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "zdmctxl\nkojqhgw",
"output": "NO"
},
{
"input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw",
"output": "NO"
},
{
"input": "abc\nbac",
"output": "NO"
},
{
"input": "ottceez\npcstdvz",
"output": "NO"
}
] | 1,630,465,827 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 92 | 22,118,400 | def streq(a, ai, af, b, bi, bf, l):
eq = True
for i in range(l):
if a[ai+i] != b[bi+i]:
eq = False
if eq:
return True
if l % 2 != 0:
return False
half = l//2
return (streq(a, ai, half, b, bi, half, half) and streq(a, half, af, b, half, bf, half)) or (streq(a, ai, half, b, half, bf, half) and streq(a, half, af, b, bi, half, half))
a = input()
b = input()
l = len(a)
if streq(a, 0, l, b, 0, l, l):
print("YES")
else:
print("NO") | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | ```python
def streq(a, ai, af, b, bi, bf, l):
eq = True
for i in range(l):
if a[ai+i] != b[bi+i]:
eq = False
if eq:
return True
if l % 2 != 0:
return False
half = l//2
return (streq(a, ai, half, b, bi, half, half) and streq(a, half, af, b, half, bf, half)) or (streq(a, ai, half, b, half, bf, half) and streq(a, half, af, b, bi, half, half))
a = input()
b = input()
l = len(a)
if streq(a, 0, l, b, 0, l, l):
print("YES")
else:
print("NO")
``` | 0 |
|
722 | B | Verse Pattern | PROGRAMMING | 1,200 | [
"implementation",
"strings"
] | null | null | You are given a text consisting of *n* lines. Each line contains some space-separated words, consisting of lowercase English letters.
We define a syllable as a string that contains exactly one vowel and any arbitrary number (possibly none) of consonants. In English alphabet following letters are considered to be vowels: 'a', 'e', 'i', 'o', 'u' and 'y'.
Each word of the text that contains at least one vowel can be divided into syllables. Each character should be a part of exactly one syllable. For example, the word "mamma" can be divided into syllables as "ma" and "mma", "mam" and "ma", and "mamm" and "a". Words that consist of only consonants should be ignored.
The verse patterns for the given text is a sequence of *n* integers *p*1,<=*p*2,<=...,<=*p**n*. Text matches the given verse pattern if for each *i* from 1 to *n* one can divide words of the *i*-th line in syllables in such a way that the total number of syllables is equal to *p**i*.
You are given the text and the verse pattern. Check, if the given text matches the given verse pattern. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the text.
The second line contains integers *p*1,<=...,<=*p**n* (0<=≤<=*p**i*<=≤<=100) — the verse pattern.
Next *n* lines contain the text itself. Text consists of lowercase English letters and spaces. It's guaranteed that all lines are non-empty, each line starts and ends with a letter and words are separated by exactly one space. The length of each line doesn't exceed 100 characters. | If the given text matches the given verse pattern, then print "YES" (without quotes) in the only line of the output. Otherwise, print "NO" (without quotes). | [
"3\n2 2 3\nintel\ncode\nch allenge\n",
"4\n1 2 3 1\na\nbcdefghi\njklmnopqrstu\nvwxyz\n",
"4\n13 11 15 15\nto be or not to be that is the question\nwhether tis nobler in the mind to suffer\nthe slings and arrows of outrageous fortune\nor to take arms against a sea of troubles\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first sample, one can split words into syllables in the following way:
Since the word "ch" in the third line doesn't contain vowels, we can ignore it. As the result we get 2 syllabels in first two lines and 3 syllables in the third one. | 500 | [
{
"input": "3\n2 2 3\nintel\ncode\nch allenge",
"output": "YES"
},
{
"input": "4\n1 2 3 1\na\nbcdefghi\njklmnopqrstu\nvwxyz",
"output": "NO"
},
{
"input": "4\n13 11 15 15\nto be or not to be that is the question\nwhether tis nobler in the mind to suffer\nthe slings and arrows of outrageous fortune\nor to take arms against a sea of troubles",
"output": "YES"
},
{
"input": "5\n2 2 1 1 1\nfdbie\naaj\ni\ni n\nshi",
"output": "YES"
},
{
"input": "5\n2 11 10 7 9\nhy of\nyur pjyacbatdoylojayu\nemd ibweioiimyxya\nyocpyivudobua\nuiraueect impxqhzpty e",
"output": "NO"
},
{
"input": "5\n6 9 7 3 10\nabtbdaa\nom auhz ub iaravozegs\ncieulibsdhj ufki\nadu pnpurt\nh naony i jaysjsjxpwuuc",
"output": "NO"
},
{
"input": "2\n26 35\ngouojxaoobw iu bkaadyo degnjkubeabt kbap thwki dyebailrhnoh ooa\npiaeaebaocptyswuc wezesazipu osebhaonouygasjrciyiqaejtqsioubiuakg umynbsvw xpfqdwxo",
"output": "NO"
},
{
"input": "5\n1 0 0 1 1\ngqex\nw\nh\nzsvu\nqcqd",
"output": "NO"
},
{
"input": "5\n0 0 0 0 0\njtv\nl\nqg\ntp\nfgd",
"output": "YES"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0\nj t fr\nn\nnhcgx\np\nmb hmhtz\ndbjc\ncwdxj\nn j whkbt\nzk m cwh\nqr n",
"output": "YES"
},
{
"input": "5\n4 5 1 0 0\noa\nqfohq\ni l\naik\nx",
"output": "NO"
},
{
"input": "10\n2 9 0 3 2 4 1 2 4 2\nxtwl oy\nafgeju fi\nr hy\nddsowagw\nxoredo f\nwufnxy k uh\nod\nlejrinw\nsueecohfjl\nedufg",
"output": "NO"
},
{
"input": "10\n1 1 0 0 0 4 0 4 0 0\na bn\nhnwss f\nd s bn\nbdzxzgsxq\nghh v\neimblv i er\nca kn k\nzm ffc zcb\nn\nz hkhvfkwhg",
"output": "NO"
},
{
"input": "5\n0 10 6 6 0\nfgthrxst\nsohnweymewnnmbobj\nj\nfwwt acdtfvkpv khbxokn\nhndovkkgfhnhqod",
"output": "NO"
},
{
"input": "5\n3 2 2 4 2\ni yu\niu\noa\naiio\nuo",
"output": "YES"
},
{
"input": "5\n11 12 11 4 6\nuuuayoiaoiy\nuaiee iai eieu\nooayaayeuee\noii o\noea uuo",
"output": "YES"
},
{
"input": "3\n2 3 2\nintel\ncode\nch allenge",
"output": "NO"
},
{
"input": "2\n1 2\ncode\na",
"output": "NO"
},
{
"input": "2\n1 1\nbababa\nbababa",
"output": "NO"
},
{
"input": "1\n1\naa",
"output": "NO"
},
{
"input": "1\n1\naaa",
"output": "NO"
},
{
"input": "2\n2 3\naee\nae",
"output": "NO"
}
] | 1,503,040,363 | 1,182 | Python 3 | OK | TESTS | 54 | 62 | 0 | n = int(input())
no = list(map(int,input().split()))
num = [0 for i in range(n)]
for i in range(n):
x = input()
for a in x:
if a == 'a' or a == 'e' or a == 'i' or a == 'o' or a == 'u' or a == 'y':
num[i]+=1
if no == num:
print("YES")
else :
print("NO") | Title: Verse Pattern
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a text consisting of *n* lines. Each line contains some space-separated words, consisting of lowercase English letters.
We define a syllable as a string that contains exactly one vowel and any arbitrary number (possibly none) of consonants. In English alphabet following letters are considered to be vowels: 'a', 'e', 'i', 'o', 'u' and 'y'.
Each word of the text that contains at least one vowel can be divided into syllables. Each character should be a part of exactly one syllable. For example, the word "mamma" can be divided into syllables as "ma" and "mma", "mam" and "ma", and "mamm" and "a". Words that consist of only consonants should be ignored.
The verse patterns for the given text is a sequence of *n* integers *p*1,<=*p*2,<=...,<=*p**n*. Text matches the given verse pattern if for each *i* from 1 to *n* one can divide words of the *i*-th line in syllables in such a way that the total number of syllables is equal to *p**i*.
You are given the text and the verse pattern. Check, if the given text matches the given verse pattern.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the text.
The second line contains integers *p*1,<=...,<=*p**n* (0<=≤<=*p**i*<=≤<=100) — the verse pattern.
Next *n* lines contain the text itself. Text consists of lowercase English letters and spaces. It's guaranteed that all lines are non-empty, each line starts and ends with a letter and words are separated by exactly one space. The length of each line doesn't exceed 100 characters.
Output Specification:
If the given text matches the given verse pattern, then print "YES" (without quotes) in the only line of the output. Otherwise, print "NO" (without quotes).
Demo Input:
['3\n2 2 3\nintel\ncode\nch allenge\n', '4\n1 2 3 1\na\nbcdefghi\njklmnopqrstu\nvwxyz\n', '4\n13 11 15 15\nto be or not to be that is the question\nwhether tis nobler in the mind to suffer\nthe slings and arrows of outrageous fortune\nor to take arms against a sea of troubles\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample, one can split words into syllables in the following way:
Since the word "ch" in the third line doesn't contain vowels, we can ignore it. As the result we get 2 syllabels in first two lines and 3 syllables in the third one. | ```python
n = int(input())
no = list(map(int,input().split()))
num = [0 for i in range(n)]
for i in range(n):
x = input()
for a in x:
if a == 'a' or a == 'e' or a == 'i' or a == 'o' or a == 'u' or a == 'y':
num[i]+=1
if no == num:
print("YES")
else :
print("NO")
``` | 3 |
|
898 | A | Rounding | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it. | The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has. | Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer. | [
"5\n",
"113\n",
"1000000000\n",
"5432359\n"
] | [
"0\n",
"110\n",
"1000000000\n",
"5432360\n"
] | In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10. | 500 | [
{
"input": "5",
"output": "0"
},
{
"input": "113",
"output": "110"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "5432359",
"output": "5432360"
},
{
"input": "999999994",
"output": "999999990"
},
{
"input": "10",
"output": "10"
},
{
"input": "9",
"output": "10"
},
{
"input": "1",
"output": "0"
},
{
"input": "0",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "6",
"output": "10"
},
{
"input": "7",
"output": "10"
},
{
"input": "8",
"output": "10"
},
{
"input": "19",
"output": "20"
},
{
"input": "100",
"output": "100"
},
{
"input": "997",
"output": "1000"
},
{
"input": "9994",
"output": "9990"
},
{
"input": "10002",
"output": "10000"
},
{
"input": "100000",
"output": "100000"
},
{
"input": "99999",
"output": "100000"
},
{
"input": "999999999",
"output": "1000000000"
},
{
"input": "999999998",
"output": "1000000000"
},
{
"input": "999999995",
"output": "999999990"
},
{
"input": "999999990",
"output": "999999990"
},
{
"input": "1000000",
"output": "1000000"
},
{
"input": "1000010",
"output": "1000010"
},
{
"input": "10000010",
"output": "10000010"
},
{
"input": "100000011",
"output": "100000010"
},
{
"input": "400000003",
"output": "400000000"
},
{
"input": "234234",
"output": "234230"
},
{
"input": "675621",
"output": "675620"
},
{
"input": "43532",
"output": "43530"
},
{
"input": "4576453",
"output": "4576450"
},
{
"input": "65754674",
"output": "65754670"
},
{
"input": "3245526",
"output": "3245530"
},
{
"input": "123445",
"output": "123440"
},
{
"input": "234217",
"output": "234220"
},
{
"input": "23451218",
"output": "23451220"
},
{
"input": "1231239",
"output": "1231240"
},
{
"input": "1923140",
"output": "1923140"
},
{
"input": "307910310",
"output": "307910310"
},
{
"input": "780961030",
"output": "780961030"
},
{
"input": "103509421",
"output": "103509420"
},
{
"input": "576560141",
"output": "576560140"
},
{
"input": "48851642",
"output": "48851640"
},
{
"input": "226935072",
"output": "226935070"
},
{
"input": "844450763",
"output": "844450760"
},
{
"input": "22534183",
"output": "22534180"
},
{
"input": "640049874",
"output": "640049870"
},
{
"input": "818133304",
"output": "818133300"
},
{
"input": "730616285",
"output": "730616280"
},
{
"input": "613732415",
"output": "613732410"
},
{
"input": "380991216",
"output": "380991220"
},
{
"input": "559074636",
"output": "559074640"
},
{
"input": "176590327",
"output": "176590330"
},
{
"input": "354673757",
"output": "354673760"
},
{
"input": "267156738",
"output": "267156740"
},
{
"input": "150272868",
"output": "150272870"
},
{
"input": "62755859",
"output": "62755860"
},
{
"input": "945871979",
"output": "945871980"
},
{
"input": "46",
"output": "50"
},
{
"input": "999",
"output": "1000"
},
{
"input": "1397",
"output": "1400"
}
] | 1,676,053,202 | 2,147,483,647 | PyPy 3-64 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<iostream>
using namespace std;
int solve(){
int n;
cin >> n;
if(n % 10 == 0){
return n;
}
else if(n % 10 == 5){
return + 5;
}
else{
return n + 10 - n % 10;
}
}
int main(){
cout << solve();
} | Title: Rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has a non-negative integer *n*. He wants to round it to nearest integer, which ends up with 0. If *n* already ends up with 0, Vasya considers it already rounded.
For example, if *n*<==<=4722 answer is 4720. If *n*<==<=5 Vasya can round it to 0 or to 10. Both ways are correct.
For given *n* find out to which integer will Vasya round it.
Input Specification:
The first line contains single integer *n* (0<=≤<=*n*<=≤<=109) — number that Vasya has.
Output Specification:
Print result of rounding *n*. Pay attention that in some cases answer isn't unique. In that case print any correct answer.
Demo Input:
['5\n', '113\n', '1000000000\n', '5432359\n']
Demo Output:
['0\n', '110\n', '1000000000\n', '5432360\n']
Note:
In the first example *n* = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10. | ```python
#include<iostream>
using namespace std;
int solve(){
int n;
cin >> n;
if(n % 10 == 0){
return n;
}
else if(n % 10 == 5){
return + 5;
}
else{
return n + 10 - n % 10;
}
}
int main(){
cout << solve();
}
``` | -1 |
|
746 | B | Decoding | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Polycarp is mad about coding, that is why he writes Sveta encoded messages. He calls the median letter in a word the letter which is in the middle of the word. If the word's length is even, the median letter is the left of the two middle letters. In the following examples, the median letter is highlighted: contest, info. If the word consists of single letter, then according to above definition this letter is the median letter.
Polycarp encodes each word in the following way: he writes down the median letter of the word, then deletes it and repeats the process until there are no letters left. For example, he encodes the word volga as logva.
You are given an encoding *s* of some word, your task is to decode it. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=2000) — the length of the encoded word.
The second line contains the string *s* of length *n* consisting of lowercase English letters — the encoding. | Print the word that Polycarp encoded. | [
"5\nlogva\n",
"2\nno\n",
"4\nabba\n"
] | [
"volga\n",
"no\n",
"baba\n"
] | In the first example Polycarp encoded the word volga. At first, he wrote down the letter l from the position 3, after that his word looked like voga. After that Polycarp wrote down the letter o from the position 2, his word became vga. Then Polycarp wrote down the letter g which was at the second position, the word became va. Then he wrote down the letter v, then the letter a. Thus, the encoding looked like logva.
In the second example Polycarp encoded the word no. He wrote down the letter n, the word became o, and he wrote down the letter o. Thus, in this example, the word and its encoding are the same.
In the third example Polycarp encoded the word baba. At first, he wrote down the letter a, which was at the position 2, after that the word looked like bba. Then he wrote down the letter b, which was at the position 2, his word looked like ba. After that he wrote down the letter b, which was at the position 1, the word looked like a, and he wrote down that letter a. Thus, the encoding is abba. | 1,000 | [
{
"input": "5\nlogva",
"output": "volga"
},
{
"input": "2\nno",
"output": "no"
},
{
"input": "4\nabba",
"output": "baba"
},
{
"input": "51\nkfsmpaeviowvkdbuhdagquxxqniselafnfbrgbhmsugcbbnlrvv",
"output": "vlbcumbrfflsnxugdudvovamfkspeiwkbhaqxqieanbghsgbnrv"
},
{
"input": "1\nw",
"output": "w"
},
{
"input": "2\ncb",
"output": "cb"
},
{
"input": "3\nqok",
"output": "oqk"
},
{
"input": "4\naegi",
"output": "gaei"
},
{
"input": "5\noqquy",
"output": "uqoqy"
},
{
"input": "6\nulhpnm",
"output": "nhulpm"
},
{
"input": "7\nijvxljt",
"output": "jxjivlt"
},
{
"input": "8\nwwmiwkeo",
"output": "ewmwwiko"
},
{
"input": "9\ngmwqmpfow",
"output": "opqmgwmfw"
},
{
"input": "10\nhncmexsslh",
"output": "lsechnmxsh"
},
{
"input": "20\nrtcjbjlbtjfmvzdqutuw",
"output": "uudvftlbcrtjjbjmzqtw"
},
{
"input": "21\ngjyiqoebcnpsdegxnsauh",
"output": "usxesnboijgyqecpdgnah"
},
{
"input": "30\nudotcwvcwxajkadxqvxvwgmwmnqrby",
"output": "bqmmwxqdkawvcoudtwcxjaxvvgwnry"
},
{
"input": "31\nipgfrxxcgckksfgexlicjvtnhvrfbmb",
"output": "mfvnvclefkccxfpigrxgksgxijthrbb"
},
{
"input": "50\nwobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy",
"output": "vsolrruoeqehviaqtycivhrbwoevvkhujmhagaholrmsatdjjy"
},
{
"input": "200\nhvayscqiwpcfykibwyudkzuzdkgqqvbnrfeupjefevlvojngmlcjwzijrkzbsaovabkvvwmjgoonyhuiphwmqdoiuueuyqtychbsklflnvghipdgaxhuhiiqlqocpvhldgvnsrtcwxpidrjffwvwcirluyyxzxrglheczeuouklzkvnyubsvgvmdbrylimztotdbmjph",
"output": "pmdoziybmgsunkluuzelrzyurcvfjdpwtsvdhpolihhadignfkbctyeuoqwpuyogmvkaoszriwcmnoleeperbqgdukuwiycwqsahvycipfkbydzzkqvnfujfvvjgljzjkbavbvwjonhihmdiuuqyhsllvhpgxuiqqcvlgnrcxirfwwilyxxghceokzvybvvdrlmttbjh"
},
{
"input": "201\nrpkghhfibtmlkpdiklegblbuyshfirheatjkfoqkfayfbxeeqijwqdwkkrkbdxlhzkhyiifemsghwovorlqedngldskfbhmwrnzmtjuckxoqdszmsdnbuqnlqzswdfhagasmfswanifrjjcuwdsplytvmnfarchgqteedgfpumkssindxndliozojzlpznwedodzwrrus",
"output": "urzoenpzoolndismpgetgcanvypdujriasmaafwzlqbdmsqxcjmnwhfslneloohseiykhxbrkdwiexfakokterfsulglipltihgprkhfbmkdkebbyhihajfqfybeqjqwkkdlzhifmgwvrqdgdkbmrztukodzsnunqsdhgsfwnfjcwsltmfrhqedfuksnxdizjlzwddwrs"
},
{
"input": "500\naopxumqciwxewxvlxzebsztskjvjzwyewjztqrsuvamtvklhqrbodtncqdchjrlpywvmtgnkkwtvpggktewdgvnhydkexwoxkgltaesrtifbwpciqsvrgjtqrdnyqkgqwrryacluaqmgdwxinqieiblolyekcbzahlhxdwqcgieyfgmicvgbbitbzhejkshjunzjteyyfngigjwyqqndtjrdykzrnrpinkwtrlchhxvycrhstpecadszilicrqdeyyidohqvzfnsqfyuemigacysxvtrgxyjcvejkjstsnatfqlkeytxgsksgpcooypsmqgcluzwofaupegxppbupvtumjerohdteuenwcmqaoazohkilgpkjavcrjcslhzkyjcgfzxxzjfufichxcodcawonkxhbqgfimmlycswdzwbnmjwhbwihfoftpcqplncavmbxuwnsabiyvpcrhfgtqyaguoaigknushbqjwqmmyvsxwabrub",
"output": "ubwsymwqhukiogytfrpybswxmanpctohwhjnwdsymigbxnwcoxcffzxfcyzlcrvjplkoaamweedoemtpbpgpaozlgmpocgkgtelfasskecygtxyaieyqnzqoiydriisaethcvhcrwnpnzyrtnqwggfytzuhkeztbgcmfegqdhhzcelliinxdmalarwgqnrtgvqcwftsalkoxkyngwtgptkntvyljcqndbqlvmvsqzwyzvktsexvwxiqupaoxmcwexlzbzsjjwejtruatkhrotcdhrpwmgkwvgkedvhdewxgteribpisrjqdykqrycuqgwiqeboykbalxwciygivbibhjsjnjeynijyqdjdkrriktlhxyrspcdzlcqeydhvfsfumgcsvrxjvjjtntqkyxsspoysqcuwfuexpuvujrhtuncqozhigkacjshkjgzxjuihcdaokhqfmlcwzbmwbiffpqlcvbunaivchgqauagnsbjqmvxarb"
},
{
"input": "501\noilesjbgowlnayckhpoaitijewsyhgavnthycaecwnvzpxgjqfjyxnjcjknvvsmjbjwtcoyfbegmnnheeamvtfjkigqoanhvgdfrjchdqgowrstlmrjmcsuuwvvoeucfyhnxivosrxblfoqwikfxjnnyejdiihpenfcahtjwcnzwvxxseicvdfgqhtvefswznuyohmmljlnxubhevywpmnitnkhecsgccpstxkmdzabsnwxkokdfsogzbpnfvgudvqlstxojzfzugxbfrozveaiofrzksegdelxsdhcjlqwqlgjcqiujptoenxozhkqhcpkarretqzfkwuvbmfdcdoqliyzmlfrfhzrnkbhofuctkpeacqejwvdrlyvepudrlzncbhwrgmxrczphsoymvtuzqjscvbtqpymogupgzctepccravjcrfsadueyraqvwasravkubebojqspdntewnjohvccamvoxdauyakvehjhabpdyzyme",
"output": "mzdbhhvaudomcvonenpqoeuvravayuafcvrcecguoyqbcjztmohzrmrhczrueyrvjqaptuobnzffmylocfbukztraphkzxetjicgqqjhsldekroavofxufjxsqdgfpzofkkwsadktpcsektnpyebxllmouzsetqfvisxwnwtafehijynxkwolxsvxhfuovusmrlswgdcrdvnogkfvaenmefotjjsvkcnyfjxzncaytvgywjtapkynwgjeiolsbolachoiieshanhcewvpgqjxjjnvmbwcybgnhemtjiqahgfjhqortmjcuwvecyniorbfqifjnedipnchjczvxecdghvfwnyhmjnuhvwminhcgcsxmzbnxodsgbnvuvltozzgbrzeifzsgexdclwljquponohqckreqfwvmddqizlrhrkhfckecewdlvpdlnbwgxcpsyvuqsvtpmgpztpcajrsderqwsakbbjsdtwjhcavxaykejapyye"
}
] | 1,628,410,442 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 62 | 20,172,800 | encrypted = input()
original = ''
shift = 1 ^ (len(encrypted) % 2)
for i in range(len(encrypted)):
if (i + shift) % 2 == 0:
original += encrypted[i]
else:
original = encrypted[i] + original
print(original) | Title: Decoding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is mad about coding, that is why he writes Sveta encoded messages. He calls the median letter in a word the letter which is in the middle of the word. If the word's length is even, the median letter is the left of the two middle letters. In the following examples, the median letter is highlighted: contest, info. If the word consists of single letter, then according to above definition this letter is the median letter.
Polycarp encodes each word in the following way: he writes down the median letter of the word, then deletes it and repeats the process until there are no letters left. For example, he encodes the word volga as logva.
You are given an encoding *s* of some word, your task is to decode it.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=2000) — the length of the encoded word.
The second line contains the string *s* of length *n* consisting of lowercase English letters — the encoding.
Output Specification:
Print the word that Polycarp encoded.
Demo Input:
['5\nlogva\n', '2\nno\n', '4\nabba\n']
Demo Output:
['volga\n', 'no\n', 'baba\n']
Note:
In the first example Polycarp encoded the word volga. At first, he wrote down the letter l from the position 3, after that his word looked like voga. After that Polycarp wrote down the letter o from the position 2, his word became vga. Then Polycarp wrote down the letter g which was at the second position, the word became va. Then he wrote down the letter v, then the letter a. Thus, the encoding looked like logva.
In the second example Polycarp encoded the word no. He wrote down the letter n, the word became o, and he wrote down the letter o. Thus, in this example, the word and its encoding are the same.
In the third example Polycarp encoded the word baba. At first, he wrote down the letter a, which was at the position 2, after that the word looked like bba. Then he wrote down the letter b, which was at the position 2, his word looked like ba. After that he wrote down the letter b, which was at the position 1, the word looked like a, and he wrote down that letter a. Thus, the encoding is abba. | ```python
encrypted = input()
original = ''
shift = 1 ^ (len(encrypted) % 2)
for i in range(len(encrypted)):
if (i + shift) % 2 == 0:
original += encrypted[i]
else:
original = encrypted[i] + original
print(original)
``` | 0 |
|
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"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,647,697,302 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 46 | 0 | n = list(map(int, input()))
m = list(map(int, input()))
a = []
for i in range(len(m)):
b = n[i] + m[i]
if b == 2:
a.append(str(0))
else:
a.append(str(b))
print("".join(a))
| 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
n = list(map(int, input()))
m = list(map(int, input()))
a = []
for i in range(len(m)):
b = n[i] + m[i]
if b == 2:
a.append(str(0))
else:
a.append(str(b))
print("".join(a))
``` | 3.9885 |
743 | A | Vladik and flights | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy",
"implementation"
] | null | null | Vladik is a competitive programmer. This year he is going to win the International Olympiad in Informatics. But it is not as easy as it sounds: the question Vladik face now is to find the cheapest way to get to the olympiad.
Vladik knows *n* airports. All the airports are located on a straight line. Each airport has unique id from 1 to *n*, Vladik's house is situated next to the airport with id *a*, and the place of the olympiad is situated next to the airport with id *b*. It is possible that Vladik's house and the place of the olympiad are located near the same airport.
To get to the olympiad, Vladik can fly between any pair of airports any number of times, but he has to start his route at the airport *a* and finish it at the airport *b*.
Each airport belongs to one of two companies. The cost of flight from the airport *i* to the airport *j* is zero if both airports belong to the same company, and |*i*<=-<=*j*| if they belong to different companies.
Print the minimum cost Vladik has to pay to get to the olympiad. | The first line contains three integers *n*, *a*, and *b* (1<=≤<=*n*<=≤<=105, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the number of airports, the id of the airport from which Vladik starts his route and the id of the airport which he has to reach.
The second line contains a string with length *n*, which consists only of characters 0 and 1. If the *i*-th character in this string is 0, then *i*-th airport belongs to first company, otherwise it belongs to the second. | Print single integer — the minimum cost Vladik has to pay to get to the olympiad. | [
"4 1 4\n1010\n",
"5 5 2\n10110\n"
] | [
"1",
"0"
] | In the first example Vladik can fly to the airport 2 at first and pay |1 - 2| = 1 (because the airports belong to different companies), and then fly from the airport 2 to the airport 4 for free (because the airports belong to the same company). So the cost of the whole flight is equal to 1. It's impossible to get to the olympiad for free, so the answer is equal to 1.
In the second example Vladik can fly directly from the airport 5 to the airport 2, because they belong to the same company. | 500 | [
{
"input": "4 1 4\n1010",
"output": "1"
},
{
"input": "5 5 2\n10110",
"output": "0"
},
{
"input": "10 9 5\n1011111001",
"output": "1"
},
{
"input": "7 3 7\n1110111",
"output": "0"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "10 3 3\n1001011011",
"output": "0"
},
{
"input": "1 1 1\n0",
"output": "0"
},
{
"input": "10 5 8\n1000001110",
"output": "1"
},
{
"input": "10 1 10\n0000011111",
"output": "1"
},
{
"input": "4 1 4\n0011",
"output": "1"
},
{
"input": "10 3 7\n0000011111",
"output": "1"
},
{
"input": "5 1 5\n11010",
"output": "1"
},
{
"input": "6 1 6\n111000",
"output": "1"
},
{
"input": "2 1 2\n01",
"output": "1"
},
{
"input": "10 10 1\n0000011111",
"output": "1"
},
{
"input": "6 1 6\n000111",
"output": "1"
},
{
"input": "10 2 10\n0000011111",
"output": "1"
},
{
"input": "8 1 8\n11110000",
"output": "1"
},
{
"input": "6 1 5\n100000",
"output": "1"
},
{
"input": "16 4 12\n0000000011111111",
"output": "1"
},
{
"input": "6 1 5\n111000",
"output": "1"
},
{
"input": "8 2 7\n11110000",
"output": "1"
},
{
"input": "6 2 5\n111000",
"output": "1"
},
{
"input": "9 9 1\n111000000",
"output": "1"
},
{
"input": "2 2 1\n01",
"output": "1"
},
{
"input": "5 2 5\n00001",
"output": "1"
},
{
"input": "5 1 5\n10000",
"output": "1"
},
{
"input": "6 1 6\n011111",
"output": "1"
},
{
"input": "5 5 1\n11110",
"output": "1"
},
{
"input": "2 1 2\n10",
"output": "1"
},
{
"input": "4 2 4\n0001",
"output": "1"
},
{
"input": "10 1 10\n1111100000",
"output": "1"
},
{
"input": "8 4 5\n00001111",
"output": "1"
},
{
"input": "4 4 1\n0111",
"output": "1"
},
{
"input": "8 1 8\n00101001",
"output": "1"
}
] | 1,483,280,432 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 26 | 124 | 4,812,800 | n, a, b = map(int, input().split())
a, b = a-1, b-1
id = input()
if id[a] == id[b]:
print(0)
else:
k = n
aid = id[a]
bid = id[b]
for i in range(n):
if id[i] == aid:
k = min(k, abs(b-i))
elif id[i] == bid:
k = min(k, abs(a-i))
print(k)
| Title: Vladik and flights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vladik is a competitive programmer. This year he is going to win the International Olympiad in Informatics. But it is not as easy as it sounds: the question Vladik face now is to find the cheapest way to get to the olympiad.
Vladik knows *n* airports. All the airports are located on a straight line. Each airport has unique id from 1 to *n*, Vladik's house is situated next to the airport with id *a*, and the place of the olympiad is situated next to the airport with id *b*. It is possible that Vladik's house and the place of the olympiad are located near the same airport.
To get to the olympiad, Vladik can fly between any pair of airports any number of times, but he has to start his route at the airport *a* and finish it at the airport *b*.
Each airport belongs to one of two companies. The cost of flight from the airport *i* to the airport *j* is zero if both airports belong to the same company, and |*i*<=-<=*j*| if they belong to different companies.
Print the minimum cost Vladik has to pay to get to the olympiad.
Input Specification:
The first line contains three integers *n*, *a*, and *b* (1<=≤<=*n*<=≤<=105, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the number of airports, the id of the airport from which Vladik starts his route and the id of the airport which he has to reach.
The second line contains a string with length *n*, which consists only of characters 0 and 1. If the *i*-th character in this string is 0, then *i*-th airport belongs to first company, otherwise it belongs to the second.
Output Specification:
Print single integer — the minimum cost Vladik has to pay to get to the olympiad.
Demo Input:
['4 1 4\n1010\n', '5 5 2\n10110\n']
Demo Output:
['1', '0']
Note:
In the first example Vladik can fly to the airport 2 at first and pay |1 - 2| = 1 (because the airports belong to different companies), and then fly from the airport 2 to the airport 4 for free (because the airports belong to the same company). So the cost of the whole flight is equal to 1. It's impossible to get to the olympiad for free, so the answer is equal to 1.
In the second example Vladik can fly directly from the airport 5 to the airport 2, because they belong to the same company. | ```python
n, a, b = map(int, input().split())
a, b = a-1, b-1
id = input()
if id[a] == id[b]:
print(0)
else:
k = n
aid = id[a]
bid = id[b]
for i in range(n):
if id[i] == aid:
k = min(k, abs(b-i))
elif id[i] == bid:
k = min(k, abs(a-i))
print(k)
``` | 0 |
|
78 | B | Easter Eggs | PROGRAMMING | 1,200 | [
"constructive algorithms",
"implementation"
] | B. Easter Eggs | 2 | 256 | The Easter Rabbit laid *n* eggs in a circle and is about to paint them.
Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied:
- Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors.
Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible. | The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100). | Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet.
If there are several answers, print any of them. | [
"8\n",
"13\n"
] | [
"ROYGRBIV\n",
"ROYGBIVGBIVYG\n"
] | The way the eggs will be painted in the first sample is shown on the picture: | 1,000 | [
{
"input": "8",
"output": "ROYGBIVG"
},
{
"input": "13",
"output": "ROYGBIVOYGBIV"
},
{
"input": "7",
"output": "ROYGBIV"
},
{
"input": "10",
"output": "ROYGBIVYGB"
},
{
"input": "14",
"output": "ROYGBIVROYGBIV"
},
{
"input": "50",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "9",
"output": "ROYGBIVGB"
},
{
"input": "11",
"output": "ROYGBIVYGBI"
},
{
"input": "12",
"output": "ROYGBIVOYGBI"
},
{
"input": "15",
"output": "ROYGBIVROYGBIVG"
},
{
"input": "16",
"output": "ROYGBIVROYGBIVGB"
},
{
"input": "17",
"output": "ROYGBIVROYGBIVYGB"
},
{
"input": "18",
"output": "ROYGBIVROYGBIVYGBI"
},
{
"input": "19",
"output": "ROYGBIVROYGBIVOYGBI"
},
{
"input": "20",
"output": "ROYGBIVROYGBIVOYGBIV"
},
{
"input": "21",
"output": "ROYGBIVROYGBIVROYGBIV"
},
{
"input": "22",
"output": "ROYGBIVROYGBIVROYGBIVG"
},
{
"input": "23",
"output": "ROYGBIVROYGBIVROYGBIVGB"
},
{
"input": "24",
"output": "ROYGBIVROYGBIVROYGBIVYGB"
},
{
"input": "25",
"output": "ROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "28",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIV"
},
{
"input": "29",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "34",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVOYGBIV"
},
{
"input": "43",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "61",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBI"
},
{
"input": "79",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVGB"
},
{
"input": "81",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "92",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "95",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "96",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBI"
},
{
"input": "97",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBIV"
},
{
"input": "98",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIV"
},
{
"input": "99",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "100",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVGB"
}
] | 1,574,146,166 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 186 | 0 | a = int(input())
s = "ROYGBIV"
s4 = s[:4]
s3 = s[4:]
if a > 7:
ls = a // 4
ds = a - ls
all_aggs = ls * s4 + s3[:ds]
print(all_aggs)
else:
print(s[:a])
| Title: Easter Eggs
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Easter Rabbit laid *n* eggs in a circle and is about to paint them.
Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied:
- Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors.
Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible.
Input Specification:
The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100).
Output Specification:
Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet.
If there are several answers, print any of them.
Demo Input:
['8\n', '13\n']
Demo Output:
['ROYGRBIV\n', 'ROYGBIVGBIVYG\n']
Note:
The way the eggs will be painted in the first sample is shown on the picture: | ```python
a = int(input())
s = "ROYGBIV"
s4 = s[:4]
s3 = s[4:]
if a > 7:
ls = a // 4
ds = a - ls
all_aggs = ls * s4 + s3[:ds]
print(all_aggs)
else:
print(s[:a])
``` | 0 |
560 | A | Currency System in Geraldion | PROGRAMMING | 1,000 | [
"implementation",
"sortings"
] | null | null | A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum? | The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes. | Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1. | [
"5\n1 2 3 4 5\n"
] | [
"-1\n"
] | none | 500 | [
{
"input": "5\n1 2 3 4 5",
"output": "-1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "10\n371054 506438 397130 1 766759 208409 769264 549213 641270 771837",
"output": "-1"
},
{
"input": "10\n635370 154890 909382 220996 276501 716105 538714 140162 171960 271264",
"output": "1"
},
{
"input": "50\n110876 835020 859879 999908 712969 788264 287153 921820 330355 499311 209594 484829 296329 940051 174081 931503 1 780512 390075 97866 124255 950067 697612 244256 782385 789882 37608 82153 399889 598867 416717 377988 535636 511221 792568 683271 131077 290194 496712 330720 587436 563481 645817 942562 654093 980561 382937 48293 582608 116156",
"output": "-1"
},
{
"input": "50\n474421 421097 217233 156339 27075 733996 281778 863492 184707 956857 288561 70997 393786 337382 663642 131184 637 273801 799870 295017 392338 842567 161819 297705 102013 930684 375703 838048 154915 138503 629056 256591 893619 19263 787927 684541 320265 841090 421423 490879 394582 493952 619247 633202 612928 50907 276653 407819 489945 153173",
"output": "1"
},
{
"input": "1\n1",
"output": "-1"
},
{
"input": "1\n1000000",
"output": "1"
},
{
"input": "2\n3 2",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
}
] | 1,488,768,365 | 2,147,483,647 | Python 3 | OK | TESTS | 16 | 62 | 4,915,200 | '''input
5
1 2 3 4 5
'''
n = int(input())
l = list(map(int, input().split()))
print(-1 if 1 in l else 1) | Title: Currency System in Geraldion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
Output Specification:
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
Demo Input:
['5\n1 2 3 4 5\n']
Demo Output:
['-1\n']
Note:
none | ```python
'''input
5
1 2 3 4 5
'''
n = int(input())
l = list(map(int, input().split()))
print(-1 if 1 in l else 1)
``` | 3 |
|
883 | F | Lost in Transliteration | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | There are some ambiguities when one writes Berland names with the letters of the Latin alphabet.
For example, the Berland sound u can be written in the Latin alphabet as "u", and can be written as "oo". For this reason, two words "ulyana" and "oolyana" denote the same name.
The second ambiguity is about the Berland sound h: one can use both "h" and "kh" to write it. For example, the words "mihail" and "mikhail" denote the same name.
There are *n* users registered on the Polycarp's website. Each of them indicated a name represented by the Latin letters. How many distinct names are there among them, if two ambiguities described above are taken into account?
Formally, we assume that two words denote the same name, if using the replacements "u" "oo" and "h" "kh", you can make the words equal. One can make replacements in both directions, in any of the two words an arbitrary number of times. A letter that resulted from the previous replacement can participate in the next replacements.
For example, the following pairs of words denote the same name:
- "koouper" and "kuooper". Making the replacements described above, you can make both words to be equal: "koouper" "kuuper" and "kuooper" "kuuper". - "khun" and "kkkhoon". With the replacements described above you can make both words to be equal: "khun" "khoon" and "kkkhoon" "kkhoon" "khoon".
For a given list of words, find the minimal number of groups where the words in each group denote the same name. | The first line contains integer number *n* (2<=≤<=*n*<=≤<=400) — number of the words in the list.
The following *n* lines contain words, one word per line. Each word consists of only lowercase Latin letters. The length of each word is between 1 and 20 letters inclusive. | Print the minimal number of groups where the words in each group denote the same name. | [
"10\nmihail\noolyana\nkooooper\nhoon\nulyana\nkoouper\nmikhail\nkhun\nkuooper\nkkkhoon\n",
"9\nhariton\nhkariton\nbuoi\nkkkhariton\nboooi\nbui\nkhariton\nboui\nboi\n",
"2\nalex\nalex\n"
] | [
"4\n",
"5\n",
"1\n"
] | There are four groups of words in the first example. Words in each group denote same name:
1. "mihail", "mikhail" 1. "oolyana", "ulyana" 1. "kooooper", "koouper" 1. "hoon", "khun", "kkkhoon"
There are five groups of words in the second example. Words in each group denote same name:
1. "hariton", "kkkhariton", "khariton" 1. "hkariton" 1. "buoi", "boooi", "boui" 1. "bui" 1. "boi"
In the third example the words are equal, so they denote the same name. | 0 | [
{
"input": "10\nmihail\noolyana\nkooooper\nhoon\nulyana\nkoouper\nmikhail\nkhun\nkuooper\nkkkhoon",
"output": "4"
},
{
"input": "9\nhariton\nhkariton\nbuoi\nkkkhariton\nboooi\nbui\nkhariton\nboui\nboi",
"output": "5"
},
{
"input": "2\nalex\nalex",
"output": "1"
},
{
"input": "40\nuok\nkuu\nku\no\nkku\nuh\nu\nu\nhh\nk\nkh\nh\nh\nou\nokh\nukk\nou\nuhk\nuo\nuko\nu\nuu\nh\nh\nhk\nuhu\nuoh\nooo\nk\nh\nuk\nk\nkku\nh\nku\nok\nk\nkuu\nou\nhh",
"output": "21"
},
{
"input": "40\noooo\nhu\no\nhoh\nkhk\nuuh\nhu\nou\nuuoh\no\nkouk\nuouo\nu\nok\nuu\nuuuo\nhoh\nuu\nkuu\nh\nu\nkkoh\nkhh\nuoh\nouuk\nkuo\nk\nu\nuku\nh\nu\nk\nhuho\nku\nh\noo\nuh\nk\nuo\nou",
"output": "25"
},
{
"input": "100\nuh\nu\nou\nhk\nokh\nuou\nk\no\nuhh\nk\noku\nk\nou\nhuh\nkoo\nuo\nkk\nkok\nhhu\nuu\noou\nk\nk\noh\nhk\nk\nu\no\nuo\no\no\no\nhoh\nkuo\nhuh\nkhu\nuu\nk\noku\nk\nh\nuu\nuo\nhuo\noo\nhu\nukk\nok\no\noh\nuo\nkko\nok\nouh\nkoh\nhhu\nku\nko\nhho\nkho\nkho\nkhk\nho\nhk\nuko\nukh\nh\nkh\nkk\nuku\nkkk\no\nuo\no\nouh\nou\nuhk\nou\nk\nh\nkko\nuko\no\nu\nho\nu\nooo\nuo\no\nko\noh\nkh\nuk\nohk\noko\nuko\nh\nh\noo\no",
"output": "36"
},
{
"input": "101\nukuu\nh\nouuo\no\nkkuo\nko\nu\nh\nhku\nh\nh\nhuo\nuhoh\nkuu\nhu\nhkko\nuhuk\nkoho\nh\nhukk\noohu\nkk\nkko\nou\noou\nh\nuuu\nuh\nkhuk\nokoo\nouou\nuo\nkk\noo\nhuok\no\nu\nhok\nhu\nhhuu\nkuu\nooho\noku\nhuoh\nhhkh\nuuuh\nouo\nhou\nhhu\nh\no\nokou\nuo\nh\nukk\nu\nhook\nh\noouk\nokuo\nkuuu\nk\nuuk\nu\nukk\nkk\nu\nuhk\nh\nk\nokuu\nuoho\nkhuk\nhukk\nhoo\nouko\nu\nuu\nu\nh\nhuo\nh\nukk\nhk\nk\nuoh\nhk\nko\nou\nho\nu\nhhhk\nkuo\nhuo\nhkh\nku\nhok\nho\nkok\nhk\nouuh",
"output": "50"
},
{
"input": "2\nkkkhkkh\nhh",
"output": "1"
},
{
"input": "2\nkkhookkhoo\nhuhu",
"output": "1"
}
] | 1,508,759,941 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 62 | 5,529,600 | n = int(input())
i = 1
words = []
words2 = []
res = set()
while i <= n:
words.append(input())
i += 1
for i in range(n):
words2 = words[i]
while 'u' in words2:
words2 = words2.replace('u', 'oo')
while 'kh' in words2:
words2 = words2.replace('kh', 'h')
res.add(words2)
print(len(res)) | Title: Lost in Transliteration
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are some ambiguities when one writes Berland names with the letters of the Latin alphabet.
For example, the Berland sound u can be written in the Latin alphabet as "u", and can be written as "oo". For this reason, two words "ulyana" and "oolyana" denote the same name.
The second ambiguity is about the Berland sound h: one can use both "h" and "kh" to write it. For example, the words "mihail" and "mikhail" denote the same name.
There are *n* users registered on the Polycarp's website. Each of them indicated a name represented by the Latin letters. How many distinct names are there among them, if two ambiguities described above are taken into account?
Formally, we assume that two words denote the same name, if using the replacements "u" "oo" and "h" "kh", you can make the words equal. One can make replacements in both directions, in any of the two words an arbitrary number of times. A letter that resulted from the previous replacement can participate in the next replacements.
For example, the following pairs of words denote the same name:
- "koouper" and "kuooper". Making the replacements described above, you can make both words to be equal: "koouper" "kuuper" and "kuooper" "kuuper". - "khun" and "kkkhoon". With the replacements described above you can make both words to be equal: "khun" "khoon" and "kkkhoon" "kkhoon" "khoon".
For a given list of words, find the minimal number of groups where the words in each group denote the same name.
Input Specification:
The first line contains integer number *n* (2<=≤<=*n*<=≤<=400) — number of the words in the list.
The following *n* lines contain words, one word per line. Each word consists of only lowercase Latin letters. The length of each word is between 1 and 20 letters inclusive.
Output Specification:
Print the minimal number of groups where the words in each group denote the same name.
Demo Input:
['10\nmihail\noolyana\nkooooper\nhoon\nulyana\nkoouper\nmikhail\nkhun\nkuooper\nkkkhoon\n', '9\nhariton\nhkariton\nbuoi\nkkkhariton\nboooi\nbui\nkhariton\nboui\nboi\n', '2\nalex\nalex\n']
Demo Output:
['4\n', '5\n', '1\n']
Note:
There are four groups of words in the first example. Words in each group denote same name:
1. "mihail", "mikhail" 1. "oolyana", "ulyana" 1. "kooooper", "koouper" 1. "hoon", "khun", "kkkhoon"
There are five groups of words in the second example. Words in each group denote same name:
1. "hariton", "kkkhariton", "khariton" 1. "hkariton" 1. "buoi", "boooi", "boui" 1. "bui" 1. "boi"
In the third example the words are equal, so they denote the same name. | ```python
n = int(input())
i = 1
words = []
words2 = []
res = set()
while i <= n:
words.append(input())
i += 1
for i in range(n):
words2 = words[i]
while 'u' in words2:
words2 = words2.replace('u', 'oo')
while 'kh' in words2:
words2 = words2.replace('kh', 'h')
res.add(words2)
print(len(res))
``` | 3 |
|
281 | A | Word Capitalization | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged. | A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103. | Output the given word after capitalization. | [
"ApPLe\n",
"konjac\n"
] | [
"ApPLe\n",
"Konjac\n"
] | none | 500 | [
{
"input": "ApPLe",
"output": "ApPLe"
},
{
"input": "konjac",
"output": "Konjac"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "A"
},
{
"input": "z",
"output": "Z"
},
{
"input": "ABACABA",
"output": "ABACABA"
},
{
"input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX",
"output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX"
},
{
"input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO",
"output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO"
},
{
"input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD",
"output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD"
},
{
"input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW",
"output": "QUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW"
},
{
"input": "kHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL",
"output": "KHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL"
},
{
"input": "aHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC",
"output": "AHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC"
},
{
"input": "sLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN",
"output": "SLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN"
},
{
"input": "iRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE",
"output": "IRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE"
},
{
"input": "cRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP",
"output": "CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP"
},
{
"input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG",
"output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG"
},
{
"input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV",
"output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV"
},
{
"input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI",
"output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI"
},
{
"input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX",
"output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX"
},
{
"input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ",
"output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ"
},
{
"input": "P",
"output": "P"
},
{
"input": "Xyzzy",
"output": "Xyzzy"
},
{
"input": "Zzz",
"output": "Zzz"
},
{
"input": "Zp",
"output": "Zp"
}
] | 1,696,631,532 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 92 | 0 | string=input()
print(string[0].capitalize()+string[1:]) | Title: Word Capitalization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
Input Specification:
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output Specification:
Output the given word after capitalization.
Demo Input:
['ApPLe\n', 'konjac\n']
Demo Output:
['ApPLe\n', 'Konjac\n']
Note:
none | ```python
string=input()
print(string[0].capitalize()+string[1:])
``` | 3 |
|
157 | A | Game Outcome | PROGRAMMING | 800 | [
"brute force"
] | null | null | Sherlock Holmes and Dr. Watson played some game on a checkered board *n*<=×<=*n* in size. During the game they put numbers on the board's squares by some tricky rules we don't know. However, the game is now over and each square of the board contains exactly one number. To understand who has won, they need to count the number of winning squares. To determine if the particular square is winning you should do the following. Calculate the sum of all numbers on the squares that share this column (including the given square) and separately calculate the sum of all numbers on the squares that share this row (including the given square). A square is considered winning if the sum of the column numbers is strictly greater than the sum of the row numbers.
For instance, lets game was ended like is shown in the picture. Then the purple cell is winning, because the sum of its column numbers equals 8<=+<=3<=+<=6<=+<=7<==<=24, sum of its row numbers equals 9<=+<=5<=+<=3<=+<=2<==<=19, and 24<=><=19. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=30). Each of the following *n* lines contain *n* space-separated integers. The *j*-th number on the *i*-th line represents the number on the square that belongs to the *j*-th column and the *i*-th row on the board. All number on the board are integers from 1 to 100. | Print the single number — the number of the winning squares. | [
"1\n1\n",
"2\n1 2\n3 4\n",
"4\n5 7 8 4\n9 5 3 2\n1 6 6 4\n9 5 7 3\n"
] | [
"0\n",
"2\n",
"6\n"
] | In the first example two upper squares are winning.
In the third example three left squares in the both middle rows are winning: | 500 | [
{
"input": "1\n1",
"output": "0"
},
{
"input": "2\n1 2\n3 4",
"output": "2"
},
{
"input": "4\n5 7 8 4\n9 5 3 2\n1 6 6 4\n9 5 7 3",
"output": "6"
},
{
"input": "2\n1 1\n1 1",
"output": "0"
},
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "4"
},
{
"input": "3\n1 2 3\n3 1 2\n2 3 1",
"output": "0"
},
{
"input": "4\n1 2 3 4\n8 7 6 5\n9 10 11 12\n16 15 14 13",
"output": "8"
},
{
"input": "1\n53",
"output": "0"
},
{
"input": "5\n1 98 22 9 39\n10 9 44 49 66\n79 17 23 8 47\n59 69 72 47 14\n94 91 98 19 54",
"output": "13"
},
{
"input": "1\n31",
"output": "0"
},
{
"input": "1\n92",
"output": "0"
},
{
"input": "5\n61 45 70 19 48\n52 29 98 21 74\n21 66 12 6 55\n62 75 66 62 57\n94 74 9 86 24",
"output": "13"
},
{
"input": "2\n73 99\n13 100",
"output": "2"
},
{
"input": "4\n89 79 14 89\n73 24 58 89\n62 88 69 65\n58 92 18 83",
"output": "10"
},
{
"input": "5\n99 77 32 20 49\n93 81 63 7 58\n37 1 17 35 53\n18 94 38 80 23\n91 50 42 61 63",
"output": "12"
},
{
"input": "4\n81 100 38 54\n8 64 39 59\n6 12 53 65\n79 50 99 71",
"output": "8"
},
{
"input": "5\n42 74 45 85 14\n68 94 11 3 89\n68 67 97 62 66\n65 76 96 18 84\n61 98 28 94 74",
"output": "12"
},
{
"input": "9\n53 80 94 41 58 49 88 24 42\n85 11 32 64 40 56 63 95 73\n17 85 60 41 13 71 54 67 87\n38 14 21 81 66 59 52 33 86\n29 34 46 18 19 80 10 44 51\n4 27 65 75 77 21 15 49 50\n35 68 86 98 98 62 69 52 71\n43 28 56 91 89 21 14 57 79\n27 27 29 26 15 76 21 70 78",
"output": "40"
},
{
"input": "7\n80 81 45 81 72 19 65\n31 24 15 52 47 1 14\n81 35 42 24 96 59 46\n16 2 59 56 60 98 76\n20 95 10 68 68 56 93\n60 16 68 77 89 52 43\n11 22 43 36 99 2 11",
"output": "21"
},
{
"input": "9\n33 80 34 56 56 33 27 74 57\n14 69 78 44 56 70 26 73 47\n13 42 17 33 78 83 94 70 37\n96 78 92 6 16 68 8 31 46\n67 97 21 10 44 64 15 77 28\n34 44 83 96 63 52 29 27 79\n23 23 57 54 35 16 5 64 36\n29 71 36 78 47 81 72 97 36\n24 83 70 58 36 82 42 44 26",
"output": "41"
},
{
"input": "9\n57 70 94 69 77 59 88 63 83\n6 79 46 5 9 43 20 39 48\n46 35 58 22 17 3 81 82 34\n77 10 40 53 71 84 14 58 56\n6 92 77 81 13 20 77 29 40\n59 53 3 97 21 97 22 11 64\n52 91 82 20 6 3 99 17 44\n79 25 43 69 85 55 95 61 31\n89 24 50 84 54 93 54 60 87",
"output": "46"
},
{
"input": "5\n77 44 22 21 20\n84 3 35 86 35\n97 50 1 44 92\n4 88 56 20 3\n32 56 26 17 80",
"output": "13"
},
{
"input": "7\n62 73 50 63 66 92 2\n27 13 83 84 88 81 47\n60 41 25 2 68 32 60\n7 94 18 98 41 25 72\n69 37 4 10 82 49 91\n76 26 67 27 30 49 18\n44 78 6 1 41 94 80",
"output": "26"
},
{
"input": "9\n40 70 98 28 44 78 15 73 20\n25 74 46 3 27 59 33 96 19\n100 47 99 68 68 67 66 87 31\n26 39 8 91 58 20 91 69 81\n77 43 90 60 17 91 78 85 68\n41 46 47 50 96 18 69 81 26\n10 58 2 36 54 64 69 10 65\n6 86 26 7 88 20 43 92 59\n61 76 13 23 49 28 22 79 8",
"output": "44"
},
{
"input": "8\n44 74 25 81 32 33 55 58\n36 13 28 28 20 65 87 58\n8 35 52 59 34 15 33 16\n2 22 42 29 11 66 30 72\n33 47 8 61 31 64 59 63\n79 36 38 42 12 21 92 36\n56 47 44 6 6 1 37 2\n79 88 79 53 50 69 94 39",
"output": "31"
},
{
"input": "5\n4 91 100 8 48\n78 56 61 49 83\n12 21 95 77 78\n40 20 91 79 25\n32 88 94 28 55",
"output": "10"
},
{
"input": "5\n23 70 5 36 69\n83 18 19 98 40\n84 91 18 51 35\n17 18 35 47 59\n29 72 35 87 27",
"output": "13"
},
{
"input": "12\n8 42 23 20 39 5 23 86 26 65 93 82\n48 35 12 4 59 19 19 28 38 81 97 99\n93 24 31 44 97 50 44 99 50 7 10 64\n79 43 65 29 84 43 46 41 89 16 6 1\n34 90 33 1 7 12 46 84 67 30 1 58\n58 21 100 66 56 22 7 24 72 73 86 37\n2 17 85 6 2 73 85 44 43 79 34 65\n3 53 29 76 87 2 27 19 11 42 71 38\n69 82 73 52 44 23 92 10 13 72 59 16\n73 32 37 93 21 94 43 39 27 53 14 15\n86 16 90 91 14 50 73 61 77 36 93 90\n22 56 30 52 81 70 12 92 75 27 38 12",
"output": "77"
},
{
"input": "3\n41 94 58\n73 61 8\n34 88 89",
"output": "5"
},
{
"input": "3\n1 2 3\n1 1 1\n1 1 1",
"output": "4"
},
{
"input": "2\n7 3\n9 5",
"output": "2"
},
{
"input": "3\n4 3 2\n2 2 2\n2 2 2",
"output": "4"
}
] | 1,563,874,543 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 8 | 218 | 0 | a=[]
x=[]
y=[]
count=0
b=int(input())
if b==1:
asdf=input()
print("0")
else:
for i in range (b):
temp=list(map(int,input().split()))
a.append(temp)
y.append(sum(temp))
for i in range (b):
temp=0
for j in range(b):
temp+=a[j][i]
x.append(temp)
for i in x:
for j in y:
if i<j:
count+=1
print(count)
| Title: Game Outcome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sherlock Holmes and Dr. Watson played some game on a checkered board *n*<=×<=*n* in size. During the game they put numbers on the board's squares by some tricky rules we don't know. However, the game is now over and each square of the board contains exactly one number. To understand who has won, they need to count the number of winning squares. To determine if the particular square is winning you should do the following. Calculate the sum of all numbers on the squares that share this column (including the given square) and separately calculate the sum of all numbers on the squares that share this row (including the given square). A square is considered winning if the sum of the column numbers is strictly greater than the sum of the row numbers.
For instance, lets game was ended like is shown in the picture. Then the purple cell is winning, because the sum of its column numbers equals 8<=+<=3<=+<=6<=+<=7<==<=24, sum of its row numbers equals 9<=+<=5<=+<=3<=+<=2<==<=19, and 24<=><=19.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=30). Each of the following *n* lines contain *n* space-separated integers. The *j*-th number on the *i*-th line represents the number on the square that belongs to the *j*-th column and the *i*-th row on the board. All number on the board are integers from 1 to 100.
Output Specification:
Print the single number — the number of the winning squares.
Demo Input:
['1\n1\n', '2\n1 2\n3 4\n', '4\n5 7 8 4\n9 5 3 2\n1 6 6 4\n9 5 7 3\n']
Demo Output:
['0\n', '2\n', '6\n']
Note:
In the first example two upper squares are winning.
In the third example three left squares in the both middle rows are winning: | ```python
a=[]
x=[]
y=[]
count=0
b=int(input())
if b==1:
asdf=input()
print("0")
else:
for i in range (b):
temp=list(map(int,input().split()))
a.append(temp)
y.append(sum(temp))
for i in range (b):
temp=0
for j in range(b):
temp+=a[j][i]
x.append(temp)
for i in x:
for j in y:
if i<j:
count+=1
print(count)
``` | 0 |
|
331 | A2 | Oh Sweet Beaverette | PROGRAMMING | 1,500 | [
"data structures",
"sortings"
] | null | null | — Oh my sweet Beaverette, would you fancy a walk along a wonderful woodland belt with me?
— Of course, my Smart Beaver! Let us enjoy the splendid view together. How about Friday night?
At this point the Smart Beaver got rushing. Everything should be perfect by Friday, so he needed to prepare the belt to the upcoming walk. He needed to cut down several trees.
Let's consider the woodland belt as a sequence of trees. Each tree *i* is described by the esthetic appeal *a**i* — some trees are very esthetically pleasing, others are 'so-so', and some trees are positively ugly!
The Smart Beaver calculated that he needed the following effects to win the Beaverette's heart:
- The first objective is to please the Beaverette: the sum of esthetic appeal of the remaining trees must be maximum possible; - the second objective is to surprise the Beaverette: the esthetic appeal of the first and the last trees in the resulting belt must be the same; - and of course, the walk should be successful: there must be at least two trees in the woodland belt left.
Now help the Smart Beaver! Which trees does he need to cut down to win the Beaverette's heart? | The first line contains a single integer *n* — the initial number of trees in the woodland belt, 2<=≤<=*n*. The second line contains space-separated integers *a**i* — the esthetic appeals of each tree. All esthetic appeals do not exceed 109 in their absolute value.
- to get 30 points, you need to solve the problem with constraints: *n*<=≤<=100 (subproblem A1); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=3·105 (subproblems A1+A2). | In the first line print two integers — the total esthetic appeal of the woodland belt after the Smart Beaver's intervention and the number of the cut down trees *k*.
In the next line print *k* integers — the numbers of the trees the Beaver needs to cut down. Assume that the trees are numbered from 1 to *n* from left to right.
If there are multiple solutions, print any of them. It is guaranteed that at least two trees have equal esthetic appeal. | [
"5\n1 2 3 1 2\n",
"5\n1 -2 3 1 -2\n"
] | [
"8 1\n1 ",
"5 2\n2 5 "
] | none | 70 | [
{
"input": "5\n1 2 3 1 2",
"output": "8 1\n1 "
},
{
"input": "5\n1 -2 3 1 -2",
"output": "5 2\n2 5 "
},
{
"input": "2\n0 0",
"output": "0 0"
},
{
"input": "3\n0 -1 0",
"output": "0 1\n2 "
},
{
"input": "3\n1 1 1",
"output": "3 0"
},
{
"input": "4\n-1 1 1 -1",
"output": "2 2\n1 4 "
},
{
"input": "4\n-1 1 -1 1",
"output": "2 2\n1 3 "
},
{
"input": "2\n-1 -1",
"output": "-2 0"
},
{
"input": "3\n-1 0 -1",
"output": "-2 0"
},
{
"input": "6\n-1 3 3 5 5 -1",
"output": "14 0"
},
{
"input": "2\n-1000000000 -1000000000",
"output": "-2000000000 0"
},
{
"input": "3\n-1000000000 -1000000000 -1000000000",
"output": "-2000000000 1\n3 "
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "3000000000 0"
},
{
"input": "10\n-589330597 -126288833 -126288833 -834860352 -834860352 -834860352 -834860352 -21170405 -834860352 -834860352",
"output": "-252577666 8\n1 4 5 6 7 8 9 10 "
},
{
"input": "20\n-808998072 733614990 579897311 -337992089 579897311 120800519 -337992089 -803027570 733614990 -686536765 733614990 -803027570 -803027570 733614990 120800519 -803027570 -686536765 579897311 -808998072 -686536765",
"output": "4215055101 13\n1 4 7 8 10 12 13 15 16 17 18 19 20 "
}
] | 1,651,504,755 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS2 | 4 | 92 | 0 | from math import inf
def solve_task5():
n = int(input())
trees = tuple(map(int, input().split(" ")))
d = {}
max_rating = -1 * inf
begin_tree = 0
end_tree = 0
current_sum = 0
for i in range(n):
if trees[i] in d:
d[trees[i]][1], d[trees[i]][2] = i, current_sum - d[trees[i]][2]
else:
d[trees[i]] = [i, -1, current_sum]
if trees[i] > 0:
current_sum += trees[i]
for key in d.keys():
if d[key][1] == -1:
continue
if key > 0:
temp_sum = d[key][2] + key
else:
temp_sum = d[key][2] + (key * 2)
if temp_sum > max_rating:
max_rating, begin_tree, end_tree = temp_sum, d[key][0], d[key][1]
deleted_trees = list(range(begin_tree)) + list(range(end_tree + 1, n))
for i in range(begin_tree + 1, end_tree):
if trees[i] < 0:
deleted_trees.append(i)
deleted_trees.sort()
deleted_trees = [el + 1 for el in deleted_trees]
k = len(deleted_trees)
return " ".join([str(max_rating), str(k)]) + "\n" + " ".join([str(el) for el in deleted_trees])
result = solve_task5()
print(result)
| Title: Oh Sweet Beaverette
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
— Oh my sweet Beaverette, would you fancy a walk along a wonderful woodland belt with me?
— Of course, my Smart Beaver! Let us enjoy the splendid view together. How about Friday night?
At this point the Smart Beaver got rushing. Everything should be perfect by Friday, so he needed to prepare the belt to the upcoming walk. He needed to cut down several trees.
Let's consider the woodland belt as a sequence of trees. Each tree *i* is described by the esthetic appeal *a**i* — some trees are very esthetically pleasing, others are 'so-so', and some trees are positively ugly!
The Smart Beaver calculated that he needed the following effects to win the Beaverette's heart:
- The first objective is to please the Beaverette: the sum of esthetic appeal of the remaining trees must be maximum possible; - the second objective is to surprise the Beaverette: the esthetic appeal of the first and the last trees in the resulting belt must be the same; - and of course, the walk should be successful: there must be at least two trees in the woodland belt left.
Now help the Smart Beaver! Which trees does he need to cut down to win the Beaverette's heart?
Input Specification:
The first line contains a single integer *n* — the initial number of trees in the woodland belt, 2<=≤<=*n*. The second line contains space-separated integers *a**i* — the esthetic appeals of each tree. All esthetic appeals do not exceed 109 in their absolute value.
- to get 30 points, you need to solve the problem with constraints: *n*<=≤<=100 (subproblem A1); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=3·105 (subproblems A1+A2).
Output Specification:
In the first line print two integers — the total esthetic appeal of the woodland belt after the Smart Beaver's intervention and the number of the cut down trees *k*.
In the next line print *k* integers — the numbers of the trees the Beaver needs to cut down. Assume that the trees are numbered from 1 to *n* from left to right.
If there are multiple solutions, print any of them. It is guaranteed that at least two trees have equal esthetic appeal.
Demo Input:
['5\n1 2 3 1 2\n', '5\n1 -2 3 1 -2\n']
Demo Output:
['8 1\n1 ', '5 2\n2 5 ']
Note:
none | ```python
from math import inf
def solve_task5():
n = int(input())
trees = tuple(map(int, input().split(" ")))
d = {}
max_rating = -1 * inf
begin_tree = 0
end_tree = 0
current_sum = 0
for i in range(n):
if trees[i] in d:
d[trees[i]][1], d[trees[i]][2] = i, current_sum - d[trees[i]][2]
else:
d[trees[i]] = [i, -1, current_sum]
if trees[i] > 0:
current_sum += trees[i]
for key in d.keys():
if d[key][1] == -1:
continue
if key > 0:
temp_sum = d[key][2] + key
else:
temp_sum = d[key][2] + (key * 2)
if temp_sum > max_rating:
max_rating, begin_tree, end_tree = temp_sum, d[key][0], d[key][1]
deleted_trees = list(range(begin_tree)) + list(range(end_tree + 1, n))
for i in range(begin_tree + 1, end_tree):
if trees[i] < 0:
deleted_trees.append(i)
deleted_trees.sort()
deleted_trees = [el + 1 for el in deleted_trees]
k = len(deleted_trees)
return " ".join([str(max_rating), str(k)]) + "\n" + " ".join([str(el) for el in deleted_trees])
result = solve_task5()
print(result)
``` | 0 |
|
919 | B | Perfect Number | PROGRAMMING | 1,100 | [
"binary search",
"brute force",
"dp",
"implementation",
"number theory"
] | null | null | We consider a positive integer perfect, if and only if the sum of its digits is exactly $10$. Given a positive integer $k$, your task is to find the $k$-th smallest perfect positive integer. | A single line with a positive integer $k$ ($1 \leq k \leq 10\,000$). | A single number, denoting the $k$-th smallest perfect integer. | [
"1\n",
"2\n"
] | [
"19\n",
"28\n"
] | The first perfect integer is $19$ and the second one is $28$. | 750 | [
{
"input": "1",
"output": "19"
},
{
"input": "2",
"output": "28"
},
{
"input": "13",
"output": "136"
},
{
"input": "101",
"output": "1432"
},
{
"input": "1023",
"output": "100270"
},
{
"input": "9999",
"output": "10800010"
},
{
"input": "10000",
"output": "10800100"
},
{
"input": "2333",
"output": "310060"
},
{
"input": "9139",
"output": "10134010"
},
{
"input": "9859",
"output": "10422001"
},
{
"input": "5987",
"output": "2221201"
},
{
"input": "3",
"output": "37"
},
{
"input": "4",
"output": "46"
},
{
"input": "5",
"output": "55"
},
{
"input": "7",
"output": "73"
},
{
"input": "9",
"output": "91"
},
{
"input": "10",
"output": "109"
},
{
"input": "15",
"output": "154"
},
{
"input": "51",
"output": "613"
},
{
"input": "99",
"output": "1414"
},
{
"input": "138",
"output": "2224"
},
{
"input": "233",
"output": "5050"
},
{
"input": "666",
"output": "27100"
},
{
"input": "1234",
"output": "110206"
},
{
"input": "3567",
"output": "1033003"
},
{
"input": "4445",
"output": "1221301"
},
{
"input": "5689",
"output": "2114002"
},
{
"input": "6666",
"output": "3102004"
},
{
"input": "7777",
"output": "5300200"
},
{
"input": "8888",
"output": "10110061"
},
{
"input": "9411",
"output": "10214200"
},
{
"input": "5539",
"output": "2101114"
},
{
"input": "6259",
"output": "2511100"
},
{
"input": "2387",
"output": "312220"
},
{
"input": "8515",
"output": "10030114"
}
] | 1,632,026,208 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 0 | 2,000 | 6,963,200 | def sum_digits(n):
if n == 0:
return 0
else:
return n % 10 + sum_digits(n // 10)
perfect_numbers = []
MAX = 15000000
for i in range(MAX):
if sum_digits(i) == 10:
perfect_numbers.append(i)
print(perfect_numbers[int(input()) - 1])
# print(len(perfect_numbers))
| Title: Perfect Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We consider a positive integer perfect, if and only if the sum of its digits is exactly $10$. Given a positive integer $k$, your task is to find the $k$-th smallest perfect positive integer.
Input Specification:
A single line with a positive integer $k$ ($1 \leq k \leq 10\,000$).
Output Specification:
A single number, denoting the $k$-th smallest perfect integer.
Demo Input:
['1\n', '2\n']
Demo Output:
['19\n', '28\n']
Note:
The first perfect integer is $19$ and the second one is $28$. | ```python
def sum_digits(n):
if n == 0:
return 0
else:
return n % 10 + sum_digits(n // 10)
perfect_numbers = []
MAX = 15000000
for i in range(MAX):
if sum_digits(i) == 10:
perfect_numbers.append(i)
print(perfect_numbers[int(input()) - 1])
# print(len(perfect_numbers))
``` | 0 |
|
344 | A | Magnets | PROGRAMMING | 800 | [
"implementation"
] | null | null | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | On the single line of the output print the number of groups of magnets. | [
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] | [
"3\n",
"2\n"
] | The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets. | 500 | [
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",
"output": "3"
},
{
"input": "1\n01",
"output": "1"
},
{
"input": "2\n01\n01",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "2"
},
{
"input": "3\n01\n01\n01",
"output": "1"
},
{
"input": "3\n10\n10\n01",
"output": "2"
},
{
"input": "3\n01\n10\n10",
"output": "2"
},
{
"input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10",
"output": "55"
}
] | 1,693,205,098 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 2 | 436 | 17,612,800 | t= int(input())
count=1
import pdb
#pdb.set_trace()
i=0
c=str(input())
while(i<=t):
if(t-i==1):
count+=1
i+=1
d=str(input())
if (d==c):
c=d
i+=2
continue
else:
c=d
count+=1
i+=2
print(count)
| Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets. | ```python
t= int(input())
count=1
import pdb
#pdb.set_trace()
i=0
c=str(input())
while(i<=t):
if(t-i==1):
count+=1
i+=1
d=str(input())
if (d==c):
c=d
i+=2
continue
else:
c=d
count+=1
i+=2
print(count)
``` | -1 |
|
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,697,113,600 | 2,147,483,647 | PyPy 3 | OK | TESTS | 30 | 184 | 0 | st = input()
l = 0
u = 0
for el in st:
if el.isupper():
u += 1
else:
l += 1
if l >= u:
print(st.lower())
else:
print(st.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
st = input()
l = 0
u = 0
for el in st:
if el.isupper():
u += 1
else:
l += 1
if l >= u:
print(st.lower())
else:
print(st.upper())
``` | 3.954 |
126 | B | Password | PROGRAMMING | 1,700 | [
"binary search",
"dp",
"hashing",
"string suffix structures",
"strings"
] | null | null | Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend. | You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters. | Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes. | [
"fixprefixsuffix\n",
"abcdabc\n"
] | [
"fix",
"Just a legend"
] | none | 1,000 | [
{
"input": "fixprefixsuffix",
"output": "fix"
},
{
"input": "abcdabc",
"output": "Just a legend"
},
{
"input": "qwertyqwertyqwerty",
"output": "qwerty"
},
{
"input": "papapapap",
"output": "papap"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaa"
},
{
"input": "ghbdtn",
"output": "Just a legend"
},
{
"input": "a",
"output": "Just a legend"
},
{
"input": "aa",
"output": "Just a legend"
},
{
"input": "ab",
"output": "Just a legend"
},
{
"input": "aaa",
"output": "a"
},
{
"input": "aba",
"output": "Just a legend"
},
{
"input": "aab",
"output": "Just a legend"
},
{
"input": "abb",
"output": "Just a legend"
},
{
"input": "abc",
"output": "Just a legend"
},
{
"input": "aaabaabaaaaab",
"output": "Just a legend"
},
{
"input": "aabaaabaaaaab",
"output": "aab"
},
{
"input": "aaabaaaabab",
"output": "Just a legend"
},
{
"input": "abcabcabcabcabc",
"output": "abcabcabc"
},
{
"input": "aaaaabaaaa",
"output": "aaaa"
},
{
"input": "aaaabaaaaaaa",
"output": "aaaa"
},
{
"input": "ghghghgxghghghg",
"output": "ghghg"
},
{
"input": "kincenvizh",
"output": "Just a legend"
},
{
"input": "amcksgurlgqzqizdauqminfzshiweejkevbazyzylrrghumnvqeqqdedyopgtvxakqwpvxntxgrkrcxabhrgoxngrwrxrvcguuyw",
"output": "Just a legend"
},
{
"input": "kwuaizneqxfflhmyruotjlkqksinoanvkyvqptkkntnpjdyzicceelgooajdgpkneuhyvhdtmasiglplajxolxovlhkwuaizneqx",
"output": "Just a legend"
},
{
"input": "nfbdzgdlbjhrlvfryyjbvtsmzacxglcvukmyexdgpuiwvqbnfbdzgdlbjhrlvfryyjbtuomcwbwvlhefnfbdzgdlbjhrlvfryyjb",
"output": "nfbdzgdlbjhrlvfryyjb"
},
{
"input": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpmvbxbyszdugvmoyjeeqsywcvuqskcjotyhvdijyslgfvyujxthngnonasgnbzzygyikheelirdfsvuicbiazfdeqecbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgpccdtdteohtjjuicfjfcbzpynlqhecyqqrjpmexgfqywnfmusldgzvgnzdmuuujtkdfnlytzgstquwqpuccxbtbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp",
"output": "btbdpnzdenxueteteytvkwnegodyhmdwhmrmbftrifytzudumzlacwytsuzefqfpjesdblxhpcsidgksisjdmhyxmmugrjmtfcgxmcgp"
},
{
"input": "abcazabca",
"output": "a"
}
] | 1,699,816,182 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 60 | 7,884,800 | word = input()
# print(word)
pi = [0]*1000000
pi[0] = -1
pos = -1
lenght = len(word)
for i in range(1, lenght):
while (word[i] != word[pos+1]) and (pos != -1):
now = pi[pos]
if word[i] == word[pos+1]:
pi[i] = pos
else:
pi[i] = -1
max = pi[lenght - 1]
m = pi[max]
if max == -1:
m = -1
for i in range(1, lenght-1):
if pi[i] == max:
m = max
# Saida
if m == -1: print("Just a legend")
else: print(word[:m+1])
| Title: Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.
Input Specification:
You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
Output Specification:
Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.
Demo Input:
['fixprefixsuffix\n', 'abcdabc\n']
Demo Output:
['fix', 'Just a legend']
Note:
none | ```python
word = input()
# print(word)
pi = [0]*1000000
pi[0] = -1
pos = -1
lenght = len(word)
for i in range(1, lenght):
while (word[i] != word[pos+1]) and (pos != -1):
now = pi[pos]
if word[i] == word[pos+1]:
pi[i] = pos
else:
pi[i] = -1
max = pi[lenght - 1]
m = pi[max]
if max == -1:
m = -1
for i in range(1, lenght-1):
if pi[i] == max:
m = max
# Saida
if m == -1: print("Just a legend")
else: print(word[:m+1])
``` | 0 |
|
133 | B | Unary | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Unary is a minimalistic Brainfuck dialect in which programs are written using only one token.
Brainfuck programs use 8 commands: "+", "-", "[", "]", "<", ">", "." and "," (their meaning is not important for the purposes of this problem). Unary programs are created from Brainfuck programs using the following algorithm. First, replace each command with a corresponding binary code, using the following conversion table:
- ">" <=→<= 1000, - "<" <=→<= 1001, - "+" <=→<= 1010, - "-" <=→<= 1011, - "." <=→<= 1100, - "," <=→<= 1101, - "[" <=→<= 1110, - "]" <=→<= 1111.
Next, concatenate the resulting binary codes into one binary number in the same order as in the program. Finally, write this number using unary numeral system — this is the Unary program equivalent to the original Brainfuck one.
You are given a Brainfuck program. Your task is to calculate the size of the equivalent Unary program, and print it modulo 1000003 (106<=+<=3). | The input will consist of a single line *p* which gives a Brainfuck program. String *p* will contain between 1 and 100 characters, inclusive. Each character of *p* will be "+", "-", "[", "]", "<", ">", "." or ",". | Output the size of the equivalent Unary program modulo 1000003 (106<=+<=3). | [
",.\n",
"++++[>,.<-]\n"
] | [
"220\n",
"61425\n"
] | To write a number *n* in unary numeral system, one simply has to write 1 *n* times. For example, 5 written in unary system will be 11111.
In the first example replacing Brainfuck commands with binary code will give us 1101 1100. After we concatenate the codes, we'll get 11011100 in binary system, or 220 in decimal. That's exactly the number of tokens in the equivalent Unary program. | 1,000 | [
{
"input": ",.",
"output": "220"
},
{
"input": "++++[>,.<-]",
"output": "61425"
},
{
"input": "[-],<],<<,<[,>,+>[[<>.,[>-[-[<><>><<<<]>,.-].>-[[>+,>,[,-,.-,-[[]>..<>,<[+,-<]-++.<+.]<,[[.<<-><<<],",
"output": "43789"
},
{
"input": "+",
"output": "10"
},
{
"input": "-",
"output": "11"
},
{
"input": "<",
"output": "9"
},
{
"input": ">",
"output": "8"
},
{
"input": ".",
"output": "12"
},
{
"input": ",",
"output": "13"
},
{
"input": "[",
"output": "14"
},
{
"input": "]",
"output": "15"
},
{
"input": ",]+>.],,+->+>-[]][><,-]><]++<.,-[.>.<+.[.<,[-,,[<]+>]->>]>]-+-+<][].,.]+][[<,-.+][+<<-+.],,,<,.]-].-",
"output": "859903"
},
{
"input": "][-+>,>[,<[<+-,[+[-.<+,<[.,<+<,>+],.]><+<,+<..[[[>,[<>+-<<[>,[>-->[>+[<+<[-<]]]<>.+-,.+++-+++-+>-.]+",
"output": "235230"
},
{
"input": "+]+<-]-<,>[,]<[][+<[+]>[[,",
"output": "221907"
},
{
"input": ".>]+,>->,.>[+>+<-.-+<<>-,..+-<.,>]>.<<,+-[].,],<,..-<[-",
"output": "223676"
},
{
"input": ">.><]..>,,<<-[.,]]+,+,>[<>>+]+++--,>.[+,,+,+[><+,+[<,-]<-,..[,,.[[><]]<[<.-++][.[]][<",
"output": "916864"
},
{
"input": "]+<+[,.[,]-,.][]..[.<[<-]]]+.<[]]>>]-+]-+-.>-.].,[+[]><-.[[]++<",
"output": "86015"
},
{
"input": "-[.<>].[,>,]>++<+].>,<<],,,]++<[<+,,,,[.]<[-[,,]-..+<++].----]++><,+.,>+,+[,-[<.]-+++][-]<+.<",
"output": "170107"
},
{
"input": "<.,+.><[,.+<[,.,<-,[>,",
"output": "982288"
},
{
"input": "[,+.-.<],,]-]-[[,[]+,[.]][>],,]<[>,<+<<>>].>]>][->+>",
"output": "411338"
},
{
"input": "+]]],,>],][],<+.[->,>..<-+]][>><.+>[][.]<,>-..-,..-]>-]+>,><+<<.+>.,++]<]],],<+-<.",
"output": "113966"
},
{
"input": ".<>.+]>],>><",
"output": "228058"
},
{
"input": "-[.<++]-,-]-,[<<+[,-+]+[[...,[-...,<>+[]>][+.],[-[>>-->---+-+]>>><-++]]-++>][,],<[[,+],++<---<[",
"output": "709697"
},
{
"input": "]<><]>,>]-]],[,>+[->,,[<-+,.][[++[,+.<[,[-][[>.]<.].+-,,]]+[->]]-][>[].,>.,],,>,]-]]<+[,>>-]+]",
"output": "283602"
},
{
"input": "<-[>[,.+>-]<-[[]+[.]--<-[[]->[.<<,,.,+[.][].,<<]],,+[.>+.>+-<.-+[-,-<][+,[>[.,.+,[+[]+<-.>-<>",
"output": "204479"
},
{
"input": "+,+><[>..,]-.[.++[>-<<-,-.>,.>]+,<,]]<[>,-+,.-[+,[.[<.-[].+>[,>-.>>]-[-][+,>>-,+<-,<.+-.+[.,",
"output": "537427"
},
{
"input": ">]-[.-+[,,]].]+,][[>>[+][,<+,>.<[],.>+[]-[,[[+],..>..<[>.,,,+]]<+++<][[>..>>+-]+][--],]<[]",
"output": "952413"
},
{
"input": ",><[-]-,],+<<]>.][]][+]>.[-]]>++-.+[.<[,.-,<,[,,>,],,>-<+],>->-[<<.,>>,<][,<-->+-..+.,>>.",
"output": "11994"
},
{
"input": "[.[[+.<<>,+,>],<][+-],>.]<+]>><<][+-,][.>[-,.>--][-[]>]-<>,+<<>+,]][.>>.<,>.<..]>]<][-.[",
"output": "386152"
},
{
"input": "-,]]]+[]-,+]>][>[[->,..-.,[[.<,,.,+[].[[[-.][.<.,.<.>[.,+.,<[-]-[--<,>+-,.,.[.,]+.>>--,",
"output": "533116"
},
{
"input": "]+,]>>+-+++<[].][[.]->,+]]>>,<>>+<+,>]",
"output": "694915"
},
{
"input": ".[.+<,->[++,]]++[[<-.]][.<.<]<-,>]]>.",
"output": "626679"
},
{
"input": "+<.[[<,]<-<[<[-]<<.>]]]<--.<,-++<<<[,<.>+<+[>-,.->,<[>-><<>-<[.,+<][+],>,],],<[[,+.],<,.-,-",
"output": "7032"
},
{
"input": ".,,>-,<-+,-<[,<>",
"output": "900168"
},
{
"input": ">[[<][[><]+.+.[..],.<,<[],]<[>]-.-+<+->]],",
"output": "419600"
},
{
"input": "].<.<.,++[>--[++[><",
"output": "983198"
},
{
"input": ",]--++..<>.+.,-[-.],,<++.+<<-+[<,,.,++],>[+>",
"output": "647820"
},
{
"input": ".<],>>[[+.+]><<<>,,+][.,-+-+<>-[,+><].+-+<[],+-+]<].>]<+-.][,,+>],[,[+",
"output": "898085"
},
{
"input": ">,>+,-,+<-[[]][-,[<][]>.+]+<].>]+][]][,...<,-,]",
"output": "586457"
},
{
"input": "+[-][]..+,<<+,++<<][<,]<[][+,+,++[+-],->],-.--<-[.]+]-+]<][,.>.+[<+]<+<>-",
"output": "240679"
},
{
"input": "-.+[.<[[<],.-<-[+-->.-->>[<<.[>,]>->[<.[-++>..,.[.",
"output": "185396"
},
{
"input": "<+[[],+,+[]-<]<<.+><,.<[.[-+>.+-]><+[]<]>[>]<<[<>.+[-><>]->>>,>.[[.>-+>]+],",
"output": "915891"
},
{
"input": "[-.].+<<]---+[+-+-[,[[[,]-<[-[[><>]",
"output": "765140"
},
{
"input": "[[>>[>[],+>-..]<]>-<-]<>].-[,,,.[+.-].-",
"output": "416600"
},
{
"input": "[,[.+-,,.>+-[+[][,[][,.-+>+]]<.,,.]<+><.[<,",
"output": "96775"
},
{
"input": "[>+,.+<<>..-+[>,><.-,--[+[>+>+[].[-[,][..<<[<,-<+-,<][][,>]++]+-<,,]++>.].[-[-[[,<[>><->]->+[+-",
"output": "89776"
},
{
"input": ">+,][>]]]+[-..<<<+]>>.+-++.+<.,>>-[+-,+.+>]<.>-[<>]<<+[>].[++[].[++++,<[+-<<[+<[]-+][>[-+.,,],<<,>+",
"output": "701493"
},
{
"input": "><",
"output": "137"
},
{
"input": ">]<-.+>>..<-,[-+.]+<<>[-,.],,,[,-+>>>>>.-]>,+<.+[,<>><",
"output": "481849"
},
{
"input": ">-[+>[++[,]-<<,.-->]+[<[-<>-]<,]<.+][]].]++]]+<,...>-[><,-",
"output": "739433"
},
{
"input": ">[][+...+[.-[,,>,[,-.].--[..>+<>[]<,],,<<,<>[<<.+>-[]+><]+,[+[",
"output": "356953"
},
{
"input": "<,+<-+[[-<[-,]",
"output": "570514"
},
{
"input": "<+.,,<[+-.+[<>[>.]+<[[<]<,<].-<-",
"output": "975622"
},
{
"input": ",-,[,,,.-]+]]>-<[+[.]]][[>-<[.[<->+.>[++[.><[+<].],]>,.,<+.--[",
"output": "243567"
},
{
"input": "]<[--]>[-.,,.[[.,-<.]+->+++,+<+]]<>>+].]>[>].[+,><->>++,[,><,,.<-]-+-+<].][-[<].<-->-+>]<]>,",
"output": "589486"
},
{
"input": "<].]->+,.],-][--<.--++",
"output": "425408"
},
{
"input": "-,,....<.<-<.[[,]<+-,,+.>].<>[<>,.+]>,,>.>[<>.+<,><,>",
"output": "474184"
},
{
"input": "].[+.,]].+->][.+-[[,--.-+.,,+]>+[-[,+.><-+[[--<[.->]<>+-]<.,]]>.<>-+,<.+[]>>[<+.][[",
"output": "9085"
},
{
"input": ">>.<<[-,.<.].",
"output": "518117"
},
{
"input": "-],,<<><.,.+][.[++-+..,,>-><>+,[,,,.[>.-+]]",
"output": "670018"
},
{
"input": "],[>>-,].><-.[]>.-+]<<<+]>-,+,.,[+[<+[],]-+>+>+--+]+<,]-+],><>+,>.-,+]-,,",
"output": "965603"
},
{
"input": ">..-",
"output": "36043"
},
{
"input": "-[..[]+>>.>,]>.+]>.]>[.>+,>+",
"output": "860720"
},
{
"input": "]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]",
"output": "266644"
},
{
"input": "[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]",
"output": "47113"
},
{
"input": "[][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][]",
"output": "889131"
},
{
"input": ">>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "208877"
}
] | 1,696,085,957 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 67 | 124 | 0 | a = input()
massiv=[]
massiv+=a
ans = 0
for i in range(len(massiv)):
if massiv[i] == ">":
massiv[i] = "1000"
elif massiv[i] == "<":
massiv[i] = "1001"
elif massiv[i] == "+":
massiv[i] = "1010"
elif massiv[i] == "-":
massiv[i] = "1011"
elif massiv[i] == ".":
massiv[i] = "1100"
elif massiv[i] == ",":
massiv[i] = "1101"
elif massiv[i] == "[":
massiv[i] = "1110"
else:
massiv[i] = "1111"
ans = ''.join(massiv)
print(int(ans, 2)%1000003)
| Title: Unary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Unary is a minimalistic Brainfuck dialect in which programs are written using only one token.
Brainfuck programs use 8 commands: "+", "-", "[", "]", "<", ">", "." and "," (their meaning is not important for the purposes of this problem). Unary programs are created from Brainfuck programs using the following algorithm. First, replace each command with a corresponding binary code, using the following conversion table:
- ">" <=→<= 1000, - "<" <=→<= 1001, - "+" <=→<= 1010, - "-" <=→<= 1011, - "." <=→<= 1100, - "," <=→<= 1101, - "[" <=→<= 1110, - "]" <=→<= 1111.
Next, concatenate the resulting binary codes into one binary number in the same order as in the program. Finally, write this number using unary numeral system — this is the Unary program equivalent to the original Brainfuck one.
You are given a Brainfuck program. Your task is to calculate the size of the equivalent Unary program, and print it modulo 1000003 (106<=+<=3).
Input Specification:
The input will consist of a single line *p* which gives a Brainfuck program. String *p* will contain between 1 and 100 characters, inclusive. Each character of *p* will be "+", "-", "[", "]", "<", ">", "." or ",".
Output Specification:
Output the size of the equivalent Unary program modulo 1000003 (106<=+<=3).
Demo Input:
[',.\n', '++++[>,.<-]\n']
Demo Output:
['220\n', '61425\n']
Note:
To write a number *n* in unary numeral system, one simply has to write 1 *n* times. For example, 5 written in unary system will be 11111.
In the first example replacing Brainfuck commands with binary code will give us 1101 1100. After we concatenate the codes, we'll get 11011100 in binary system, or 220 in decimal. That's exactly the number of tokens in the equivalent Unary program. | ```python
a = input()
massiv=[]
massiv+=a
ans = 0
for i in range(len(massiv)):
if massiv[i] == ">":
massiv[i] = "1000"
elif massiv[i] == "<":
massiv[i] = "1001"
elif massiv[i] == "+":
massiv[i] = "1010"
elif massiv[i] == "-":
massiv[i] = "1011"
elif massiv[i] == ".":
massiv[i] = "1100"
elif massiv[i] == ",":
massiv[i] = "1101"
elif massiv[i] == "[":
massiv[i] = "1110"
else:
massiv[i] = "1111"
ans = ''.join(massiv)
print(int(ans, 2)%1000003)
``` | 3 |
|
776 | A | A Serial Killer | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern. | First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters. | Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order. | [
"ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n",
"icm codeforces\n1\ncodeforces technex\n"
] | [
"ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n",
"icm codeforces\nicm technex\n"
] | In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears. | 500 | [
{
"input": "ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler",
"output": "ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler"
},
{
"input": "icm codeforces\n1\ncodeforces technex",
"output": "icm codeforces\nicm technex"
},
{
"input": "a b\n3\na c\nb d\nd e",
"output": "a b\nc b\nc d\nc e"
},
{
"input": "ze udggmyop\n4\nze szhrbmft\nudggmyop mjorab\nszhrbmft ojdtfnzxj\nojdtfnzxj yjlkg",
"output": "ze udggmyop\nszhrbmft udggmyop\nszhrbmft mjorab\nojdtfnzxj mjorab\nyjlkg mjorab"
},
{
"input": "q s\n10\nq b\nb j\ns g\nj f\nf m\ng c\nc a\nm d\nd z\nz o",
"output": "q s\nb s\nj s\nj g\nf g\nm g\nm c\nm a\nd a\nz a\no a"
},
{
"input": "iii iiiiii\n7\niii iiiiiiiiii\niiiiiiiiii iiii\niiii i\niiiiii iiiiiiii\niiiiiiii iiiiiiiii\ni iiiii\niiiii ii",
"output": "iii iiiiii\niiiiiiiiii iiiiii\niiii iiiiii\ni iiiiii\ni iiiiiiii\ni iiiiiiiii\niiiii iiiiiiiii\nii iiiiiiiii"
},
{
"input": "bwyplnjn zkms\n26\nzkms nzmcsytxh\nnzmcsytxh yujsb\nbwyplnjn gtbzhudpb\ngtbzhudpb hpk\nyujsb xvy\nhpk wrwnfokml\nwrwnfokml ndouuikw\nndouuikw ucgrja\nucgrja tgfmpldz\nxvy nycrfphn\nnycrfphn quvs\nquvs htdy\nhtdy k\ntgfmpldz xtdpkxm\nxtdpkxm suwqxs\nk fv\nsuwqxs qckllwy\nqckllwy diun\nfv lefa\nlefa gdoqjysx\ndiun dhpz\ngdoqjysx bdmqdyt\ndhpz dgz\ndgz v\nbdmqdyt aswy\naswy ydkayhlrnm",
"output": "bwyplnjn zkms\nbwyplnjn nzmcsytxh\nbwyplnjn yujsb\ngtbzhudpb yujsb\nhpk yujsb\nhpk xvy\nwrwnfokml xvy\nndouuikw xvy\nucgrja xvy\ntgfmpldz xvy\ntgfmpldz nycrfphn\ntgfmpldz quvs\ntgfmpldz htdy\ntgfmpldz k\nxtdpkxm k\nsuwqxs k\nsuwqxs fv\nqckllwy fv\ndiun fv\ndiun lefa\ndiun gdoqjysx\ndhpz gdoqjysx\ndhpz bdmqdyt\ndgz bdmqdyt\nv bdmqdyt\nv aswy\nv ydkayhlrnm"
},
{
"input": "wxz hbeqwqp\n7\nhbeqwqp cpieghnszh\ncpieghnszh tlqrpd\ntlqrpd ttwrtio\nttwrtio xapvds\nxapvds zk\nwxz yryk\nzk b",
"output": "wxz hbeqwqp\nwxz cpieghnszh\nwxz tlqrpd\nwxz ttwrtio\nwxz xapvds\nwxz zk\nyryk zk\nyryk b"
},
{
"input": "wced gnsgv\n23\ngnsgv japawpaf\njapawpaf nnvpeu\nnnvpeu a\na ddupputljq\nddupputljq qyhnvbh\nqyhnvbh pqwijl\nwced khuvs\nkhuvs bjkh\npqwijl ysacmboc\nbjkh srf\nsrf jknoz\njknoz hodf\nysacmboc xqtkoyh\nhodf rfp\nxqtkoyh bivgnwqvoe\nbivgnwqvoe nknf\nnknf wuig\nrfp e\ne bqqknq\nwuig sznhhhu\nbqqknq dhrtdld\ndhrtdld n\nsznhhhu bguylf",
"output": "wced gnsgv\nwced japawpaf\nwced nnvpeu\nwced a\nwced ddupputljq\nwced qyhnvbh\nwced pqwijl\nkhuvs pqwijl\nbjkh pqwijl\nbjkh ysacmboc\nsrf ysacmboc\njknoz ysacmboc\nhodf ysacmboc\nhodf xqtkoyh\nrfp xqtkoyh\nrfp bivgnwqvoe\nrfp nknf\nrfp wuig\ne wuig\nbqqknq wuig\nbqqknq sznhhhu\ndhrtdld sznhhhu\nn sznhhhu\nn bguylf"
},
{
"input": "qqqqqqqqqq qqqqqqqq\n3\nqqqqqqqq qqqqqqqqq\nqqqqqqqqq qqqqq\nqqqqq q",
"output": "qqqqqqqqqq qqqqqqqq\nqqqqqqqqqq qqqqqqqqq\nqqqqqqqqqq qqqqq\nqqqqqqqqqq q"
},
{
"input": "wwwww w\n8\nwwwww wwwwwwww\nwwwwwwww wwwwwwwww\nwwwwwwwww wwwwwwwwww\nw www\nwwwwwwwwww wwww\nwwww ww\nwww wwwwww\nwwwwww wwwwwww",
"output": "wwwww w\nwwwwwwww w\nwwwwwwwww w\nwwwwwwwwww w\nwwwwwwwwww www\nwwww www\nww www\nww wwwwww\nww wwwwwww"
},
{
"input": "k d\n17\nk l\nd v\nv z\nl r\nz i\nr s\ns p\np w\nw j\nj h\ni c\nh m\nm q\nc o\no g\nq x\nx n",
"output": "k d\nl d\nl v\nl z\nr z\nr i\ns i\np i\nw i\nj i\nh i\nh c\nm c\nq c\nq o\nq g\nx g\nn g"
}
] | 1,632,396,202 | 2,147,483,647 | Python 3 | OK | TESTS | 57 | 77 | 7,065,600 | initial = input().split()
q = int(input())
vct = []
p = q
while p > 0:
s, t = input().split()
vct.append([s, t])
p -= 1
answer = initial[0] + " " + initial[1]
for i in range(len(vct)):
if initial[0] == vct[i][0]:
initial[0] = vct[i][1]
answer += "\n" + initial[0] + " " + initial[1]
elif initial[1] == vct[i][0]:
initial[1] = vct[i][1]
answer += "\n" + initial[0] + " " + initial[1]
print(answer) | Title: A Serial Killer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input Specification:
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output Specification:
Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Demo Input:
['ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n', 'icm codeforces\n1\ncodeforces technex\n']
Demo Output:
['ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n', 'icm codeforces\nicm technex\n']
Note:
In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears. | ```python
initial = input().split()
q = int(input())
vct = []
p = q
while p > 0:
s, t = input().split()
vct.append([s, t])
p -= 1
answer = initial[0] + " " + initial[1]
for i in range(len(vct)):
if initial[0] == vct[i][0]:
initial[0] = vct[i][1]
answer += "\n" + initial[0] + " " + initial[1]
elif initial[1] == vct[i][0]:
initial[1] = vct[i][1]
answer += "\n" + initial[0] + " " + initial[1]
print(answer)
``` | 3 |
|
987 | B | High School: Become Human | PROGRAMMING | 1,100 | [
"math"
] | null | null | Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare $x^y$ with $y^x$. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare $x^y$ with $y^x$ for Vasya, maybe then other androids will respect him. | On the only line of input there are two integers $x$ and $y$ ($1 \le x, y \le 10^{9}$). | If $x^y < y^x$, then print '<' (without quotes). If $x^y > y^x$, then print '>' (without quotes). If $x^y = y^x$, then print '=' (without quotes). | [
"5 8\n",
"10 3\n",
"6 6\n"
] | [
">\n",
"<\n",
"=\n"
] | In the first example $5^8 = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 390625$, and $8^5 = 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 = 32768$. So you should print '>'.
In the second example $10^3 = 1000 < 3^{10} = 59049$.
In the third example $6^6 = 46656 = 6^6$. | 1,000 | [
{
"input": "5 8",
"output": ">"
},
{
"input": "10 3",
"output": "<"
},
{
"input": "6 6",
"output": "="
},
{
"input": "14 1",
"output": ">"
},
{
"input": "2 4",
"output": "="
},
{
"input": "987654321 123456987",
"output": "<"
},
{
"input": "1 10",
"output": "<"
},
{
"input": "9 1",
"output": ">"
},
{
"input": "1 1",
"output": "="
},
{
"input": "2 2",
"output": "="
},
{
"input": "3 3",
"output": "="
},
{
"input": "4 4",
"output": "="
},
{
"input": "5 5",
"output": "="
},
{
"input": "2 3",
"output": "<"
},
{
"input": "2 5",
"output": ">"
},
{
"input": "3 2",
"output": ">"
},
{
"input": "3 4",
"output": ">"
},
{
"input": "3 5",
"output": ">"
},
{
"input": "4 2",
"output": "="
},
{
"input": "4 3",
"output": "<"
},
{
"input": "4 5",
"output": ">"
},
{
"input": "5 2",
"output": "<"
},
{
"input": "5 3",
"output": "<"
},
{
"input": "5 4",
"output": "<"
},
{
"input": "100 101",
"output": ">"
},
{
"input": "876543 372647",
"output": "<"
},
{
"input": "1000000000 999999999",
"output": "<"
},
{
"input": "1000000000 2",
"output": "<"
},
{
"input": "1000000000 1",
"output": ">"
},
{
"input": "987654321 123456789",
"output": "<"
},
{
"input": "1000000000 1000000000",
"output": "="
},
{
"input": "4359435 4396510",
"output": ">"
},
{
"input": "25936809 25936809",
"output": "="
},
{
"input": "53602896 3",
"output": "<"
},
{
"input": "13208659 1",
"output": ">"
},
{
"input": "620537015 620537016",
"output": ">"
},
{
"input": "56498103 56498102",
"output": "<"
},
{
"input": "4 1000000000",
"output": ">"
},
{
"input": "11 15",
"output": ">"
},
{
"input": "2 6",
"output": ">"
},
{
"input": "1 100",
"output": "<"
},
{
"input": "3 9",
"output": ">"
},
{
"input": "5 25",
"output": ">"
},
{
"input": "17 18",
"output": ">"
},
{
"input": "100 99",
"output": "<"
},
{
"input": "10000035 1000432",
"output": "<"
},
{
"input": "27 3",
"output": "<"
},
{
"input": "15657413 15657414",
"output": ">"
}
] | 1,657,973,761 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 5 | 46 | 0 | from math import pow
x,y = map(int,input().split(" "))
m = pow(x,y)
n = pow(y,x)
if(m>n):
print(">")
elif(m<n):
print("<")
else:
print("=") | Title: High School: Become Human
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.
It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.
One of the popular pranks on Vasya is to force him to compare $x^y$ with $y^x$. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.
Please help Vasya! Write a fast program to compare $x^y$ with $y^x$ for Vasya, maybe then other androids will respect him.
Input Specification:
On the only line of input there are two integers $x$ and $y$ ($1 \le x, y \le 10^{9}$).
Output Specification:
If $x^y < y^x$, then print '<' (without quotes). If $x^y > y^x$, then print '>' (without quotes). If $x^y = y^x$, then print '=' (without quotes).
Demo Input:
['5 8\n', '10 3\n', '6 6\n']
Demo Output:
['>\n', '<\n', '=\n']
Note:
In the first example $5^8 = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 390625$, and $8^5 = 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 = 32768$. So you should print '>'.
In the second example $10^3 = 1000 < 3^{10} = 59049$.
In the third example $6^6 = 46656 = 6^6$. | ```python
from math import pow
x,y = map(int,input().split(" "))
m = pow(x,y)
n = pow(y,x)
if(m>n):
print(">")
elif(m<n):
print("<")
else:
print("=")
``` | -1 |
|
463 | B | Caisa and Pylons | PROGRAMMING | 1,100 | [
"brute force",
"implementation",
"math"
] | null | null | Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game? | The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons. | Print a single number representing the minimum number of dollars paid by Caisa. | [
"5\n3 4 3 2 4\n",
"3\n4 4 4\n"
] | [
"4\n",
"4\n"
] | In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | 1,000 | [
{
"input": "5\n3 4 3 2 4",
"output": "4"
},
{
"input": "3\n4 4 4",
"output": "4"
},
{
"input": "99\n1401 2019 1748 3785 3236 3177 3443 3772 2138 1049 353 908 310 2388 1322 88 2160 2783 435 2248 1471 706 2468 2319 3156 3506 2794 1999 1983 2519 2597 3735 537 344 3519 3772 3872 2961 3895 2010 10 247 3269 671 2986 942 758 1146 77 1545 3745 1547 2250 2565 217 1406 2070 3010 3404 404 1528 2352 138 2065 3047 3656 2188 2919 2616 2083 1280 2977 2681 548 4000 1667 1489 1109 3164 1565 2653 3260 3463 903 1824 3679 2308 245 2689 2063 648 568 766 785 2984 3812 440 1172 2730",
"output": "4000"
},
{
"input": "68\n477 1931 3738 3921 2306 1823 3328 2057 661 3993 2967 3520 171 1739 1525 1817 209 3475 1902 2666 518 3283 3412 3040 3383 2331 1147 1460 1452 1800 1327 2280 82 1416 2200 2388 3238 1879 796 250 1872 114 121 2042 1853 1645 211 2061 1472 2464 726 1989 1746 489 1380 1128 2819 2527 2939 622 678 265 2902 1111 2032 1453 3850 1621",
"output": "3993"
},
{
"input": "30\n30 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": "30"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n69",
"output": "69"
}
] | 1,628,612,351 | 2,147,483,647 | PyPy 3 | OK | TESTS | 49 | 155 | 30,822,400 | if __name__ == '__main__':
n = input()
a = list(map(int, input().split()))
print(max(a)) | Title: Caisa and Pylons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons.
Output Specification:
Print a single number representing the minimum number of dollars paid by Caisa.
Demo Input:
['5\n3 4 3 2 4\n', '3\n4 4 4\n']
Demo Output:
['4\n', '4\n']
Note:
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | ```python
if __name__ == '__main__':
n = input()
a = list(map(int, input().split()))
print(max(a))
``` | 3 |
|
742 | A | Arpa’s hard exam and Mehrdad’s naive cheat | PROGRAMMING | 1,000 | [
"implementation",
"math",
"number theory"
] | null | null | There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat. | The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109). | Print single integer — the last digit of 1378*n*. | [
"1\n",
"2\n"
] | [
"8",
"4"
] | In the first example, last digit of 1378<sup class="upper-index">1</sup> = 1378 is 8.
In the second example, last digit of 1378<sup class="upper-index">2</sup> = 1378·1378 = 1898884 is 4. | 500 | [
{
"input": "1",
"output": "8"
},
{
"input": "2",
"output": "4"
},
{
"input": "1000",
"output": "6"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "6"
},
{
"input": "1000000000",
"output": "6"
},
{
"input": "5",
"output": "8"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "2"
},
{
"input": "1378",
"output": "4"
},
{
"input": "13781378",
"output": "4"
},
{
"input": "51202278",
"output": "4"
},
{
"input": "999999998",
"output": "4"
},
{
"input": "999999997",
"output": "8"
},
{
"input": "12193721",
"output": "8"
},
{
"input": "0",
"output": "1"
},
{
"input": "989898989",
"output": "8"
},
{
"input": "7",
"output": "2"
},
{
"input": "8",
"output": "6"
},
{
"input": "9",
"output": "8"
},
{
"input": "10",
"output": "4"
},
{
"input": "11",
"output": "2"
},
{
"input": "12",
"output": "6"
},
{
"input": "13",
"output": "8"
},
{
"input": "14",
"output": "4"
},
{
"input": "15",
"output": "2"
},
{
"input": "16",
"output": "6"
},
{
"input": "999999996",
"output": "6"
},
{
"input": "999999995",
"output": "2"
},
{
"input": "999999994",
"output": "4"
}
] | 1,658,863,580 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 31 | 0 | import math
n = int(input())
x = 8
sum1 = 0
for i in range(n):
sum1 +=x
print(str(sum1)[-1]) | Title: Arpa’s hard exam and Mehrdad’s naive cheat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
Input Specification:
The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109).
Output Specification:
Print single integer — the last digit of 1378*n*.
Demo Input:
['1\n', '2\n']
Demo Output:
['8', '4']
Note:
In the first example, last digit of 1378<sup class="upper-index">1</sup> = 1378 is 8.
In the second example, last digit of 1378<sup class="upper-index">2</sup> = 1378·1378 = 1898884 is 4. | ```python
import math
n = int(input())
x = 8
sum1 = 0
for i in range(n):
sum1 +=x
print(str(sum1)[-1])
``` | 0 |
|
1,003 | A | Polycarp's Pockets | PROGRAMMING | 800 | [
"implementation"
] | null | null | Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that. | The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins. | Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket. | [
"6\n1 2 4 3 3 2\n",
"1\n100\n"
] | [
"2\n",
"1\n"
] | none | 0 | [
{
"input": "6\n1 2 4 3 3 2",
"output": "2"
},
{
"input": "1\n100",
"output": "1"
},
{
"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": "100"
},
{
"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": "100"
},
{
"input": "100\n59 47 39 47 47 71 47 28 58 47 35 79 58 47 38 47 47 47 47 27 47 43 29 95 47 49 46 71 47 74 79 47 47 32 45 67 47 47 30 37 47 47 16 67 22 76 47 86 84 10 5 47 47 47 47 47 1 51 47 54 47 8 47 47 9 47 47 47 47 28 47 47 26 47 47 47 47 47 47 92 47 47 77 47 47 24 45 47 10 47 47 89 47 27 47 89 47 67 24 71",
"output": "51"
},
{
"input": "100\n45 99 10 27 16 85 39 38 17 32 15 23 67 48 50 97 42 70 62 30 44 81 64 73 34 22 46 5 83 52 58 60 33 74 47 88 18 61 78 53 25 95 94 31 3 75 1 57 20 54 59 9 68 7 77 43 21 87 86 24 4 80 11 49 2 72 36 84 71 8 65 55 79 100 41 14 35 89 66 69 93 37 56 82 90 91 51 19 26 92 6 96 13 98 12 28 76 40 63 29",
"output": "1"
},
{
"input": "100\n45 29 5 2 6 50 22 36 14 15 9 48 46 20 8 37 7 47 12 50 21 38 18 27 33 19 40 10 5 49 38 42 34 37 27 30 35 24 10 3 40 49 41 3 4 44 13 25 28 31 46 36 23 1 1 23 7 22 35 26 21 16 48 42 32 8 11 16 34 11 39 32 47 28 43 41 39 4 14 19 26 45 13 18 15 25 2 44 17 29 17 33 43 6 12 30 9 20 31 24",
"output": "2"
},
{
"input": "50\n7 7 3 3 7 4 5 6 4 3 7 5 6 4 5 4 4 5 6 7 7 7 4 5 5 5 3 7 6 3 4 6 3 6 4 4 5 4 6 6 3 5 6 3 5 3 3 7 7 6",
"output": "10"
},
{
"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 99 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "99"
},
{
"input": "7\n1 2 3 3 3 1 2",
"output": "3"
},
{
"input": "5\n1 2 3 4 5",
"output": "1"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "1"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "1"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "1"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "2"
},
{
"input": "11\n1 2 3 4 5 6 7 8 9 1 1",
"output": "3"
},
{
"input": "12\n1 2 1 1 1 1 1 1 1 1 1 1",
"output": "11"
},
{
"input": "13\n1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "13"
},
{
"input": "14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "14"
},
{
"input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "15"
},
{
"input": "16\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "16"
},
{
"input": "3\n1 1 1",
"output": "3"
},
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "10\n1 1 1 1 2 2 1 1 9 10",
"output": "6"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "56\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",
"output": "56"
},
{
"input": "99\n35 96 73 72 70 83 22 93 98 75 45 32 81 82 45 54 25 7 53 72 29 2 94 19 21 98 34 28 39 99 55 85 44 23 6 47 98 2 33 34 19 57 49 35 67 4 60 4 4 23 55 6 57 66 16 68 34 45 84 79 48 63 4 9 46 88 98 13 19 27 83 12 4 63 57 22 44 77 44 62 28 52 44 64 9 24 55 22 48 4 2 9 80 76 45 1 56 22 92",
"output": "6"
},
{
"input": "10\n1 2 2 3 3 3 4 4 4 4",
"output": "4"
},
{
"input": "99\n97 44 33 56 42 10 61 85 64 26 40 39 82 34 75 9 51 51 39 73 58 38 74 31 13 99 58 1 28 89 76 19 52 7 40 56 12 27 72 72 67 75 62 46 22 55 35 16 18 39 60 63 92 42 85 69 34 61 73 50 57 95 30 4 45 63 76 58 32 35 48 81 10 78 95 79 55 97 21 21 22 94 30 17 78 57 89 93 100 44 16 89 68 55 19 46 42 73 21",
"output": "3"
},
{
"input": "5\n5 5 5 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 5 2 6",
"output": "3"
},
{
"input": "3\n58 59 58",
"output": "2"
},
{
"input": "9\n1 2 3 4 5 6 7 8 8",
"output": "2"
},
{
"input": "97\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",
"output": "97"
},
{
"input": "3\n95 95 4",
"output": "2"
},
{
"input": "3\n2 2 5",
"output": "2"
}
] | 1,654,106,660 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 46 | 0 | a = int(input())
k = input().split()
o = 0
for n in range(a):
j = k.count((k[n]))
if o < j:
o = j
print(o) | Title: Polycarp's Pockets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins.
Output Specification:
Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.
Demo Input:
['6\n1 2 4 3 3 2\n', '1\n100\n']
Demo Output:
['2\n', '1\n']
Note:
none | ```python
a = int(input())
k = input().split()
o = 0
for n in range(a):
j = k.count((k[n]))
if o < j:
o = j
print(o)
``` | 3 |
|
757 | B | Bash's Big Day | PROGRAMMING | 1,400 | [
"greedy",
"math",
"number theory"
] | null | null | Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu's Lab. Since Bash is Professor Zulu's favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.
But Zulu warns him that a group of *k*<=><=1 Pokemon with strengths {*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*} tend to fight among each other if *gcd*(*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*)<==<=1 (see notes for *gcd* definition).
Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?
Note: A Pokemon cannot fight with itself. | The input consists of two lines.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105), the number of Pokemon in the lab.
The next line contains *n* space separated integers, where the *i*-th of them denotes *s**i* (1<=≤<=*s**i*<=≤<=105), the strength of the *i*-th Pokemon. | Print single integer — the maximum number of Pokemons Bash can take. | [
"3\n2 3 4\n",
"5\n2 3 4 6 7\n"
] | [
"2\n",
"3\n"
] | *gcd* (greatest common divisor) of positive integers set {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>} is the maximum positive integer that divides all the integers {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>}.
In the first sample, we can take Pokemons with strengths {2, 4} since *gcd*(2, 4) = 2.
In the second sample, we can take Pokemons with strengths {2, 4, 6}, and there is no larger group with *gcd* ≠ 1. | 1,000 | [
{
"input": "3\n2 3 4",
"output": "2"
},
{
"input": "5\n2 3 4 6 7",
"output": "3"
},
{
"input": "3\n5 6 4",
"output": "2"
},
{
"input": "8\n41 74 4 27 85 39 100 36",
"output": "4"
},
{
"input": "6\n89 20 86 81 62 23",
"output": "3"
},
{
"input": "71\n23 84 98 8 14 4 42 56 83 87 28 22 32 50 5 96 90 1 59 74 77 88 71 38 62 36 85 97 99 6 81 20 49 57 66 9 45 41 29 68 35 19 27 76 78 72 55 25 46 48 26 53 39 31 94 34 63 37 64 16 79 24 82 17 12 3 89 61 80 30 10",
"output": "38"
},
{
"input": "95\n72 38 75 62 87 30 11 65 35 16 73 23 18 48 19 4 22 42 14 60 49 83 59 15 51 27 80 97 37 100 64 81 54 71 52 20 5 98 78 86 26 55 25 57 36 3 8 74 82 21 29 1 76 2 79 61 39 9 89 77 70 63 56 28 92 53 31 45 93 47 67 99 58 12 84 44 32 34 69 40 13 7 66 68 17 85 6 90 33 91 94 24 46 10 50",
"output": "48"
},
{
"input": "44\n39706 21317 26213 55086 10799 31825 29024 6565 96535 11412 14642 91901 41932 24538 81351 53861 63403 34199 82286 32594 29684 42753 16857 73821 71085 36306 70080 11233 21023 8551 85406 95390 92375 52675 77938 46265 74855 5229 5856 66713 65730 24525 84078 20684",
"output": "19"
},
{
"input": "35\n45633 86983 46174 48399 33926 51395 76300 6387 48852 82808 28694 79864 4482 35982 21956 76522 19656 74518 28480 71481 25700 46815 14170 95705 8535 96993 29029 8898 97637 62710 14615 22864 69849 27068 68557",
"output": "20"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n10 7 9 8 3 3 10 7 3 3",
"output": "5"
},
{
"input": "9\n10 10 6 10 9 1 8 3 5",
"output": "5"
},
{
"input": "7\n9 4 2 3 3 9 8",
"output": "4"
},
{
"input": "1\n4",
"output": "1"
},
{
"input": "6\n1623 45906 37856 34727 27156 12598",
"output": "4"
},
{
"input": "30\n83172 59163 67334 83980 5932 8773 77649 41428 62789 28159 17183 10199 41496 59500 14614 10468 54886 64679 42382 57021 50499 95643 77239 61434 16181 30505 59152 55972 18265 70566",
"output": "15"
},
{
"input": "23\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 22 16 2 13 16",
"output": "22"
},
{
"input": "46\n12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 15 1 18 28 20 6 31 16 5 23 21 38 3 11 18 11 3 25 33",
"output": "27"
},
{
"input": "43\n8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8 23 40 33 11 5 21 16 19 15 41 30 28 31 5 32 16 5 38 11 21 34",
"output": "21"
},
{
"input": "25\n58427 26687 48857 46477 7039 25423 58757 48119 38113 40637 22391 48337 4157 10597 8167 19031 64613 70913 69313 18047 17159 77491 13499 70949 24107",
"output": "1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "2\n3 6",
"output": "2"
},
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "5\n3 3 3 3 3",
"output": "5"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "2\n541 541",
"output": "2"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n99989 99989",
"output": "2"
},
{
"input": "3\n3 9 27",
"output": "3"
},
{
"input": "2\n1009 1009",
"output": "2"
},
{
"input": "4\n1 1 1 1",
"output": "1"
},
{
"input": "6\n2 10 20 5 15 25",
"output": "5"
},
{
"input": "3\n3 3 6",
"output": "3"
},
{
"input": "3\n457 457 457",
"output": "3"
},
{
"input": "2\n34 17",
"output": "2"
},
{
"input": "3\n12 24 3",
"output": "3"
},
{
"input": "10\n99991 99991 99991 99991 99991 99991 99991 99991 99991 99991",
"output": "10"
},
{
"input": "2\n1009 2018",
"output": "2"
},
{
"input": "3\n3 3 3",
"output": "3"
},
{
"input": "7\n6 9 12 15 21 27 33",
"output": "7"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "2\n557 557",
"output": "2"
},
{
"input": "3\n1 1 2",
"output": "1"
},
{
"input": "5\n2 2 101 101 101",
"output": "3"
},
{
"input": "2\n122 3721",
"output": "2"
},
{
"input": "2\n49201 98402",
"output": "2"
},
{
"input": "2\n88258 44129",
"output": "2"
},
{
"input": "2\n7919 47514",
"output": "2"
},
{
"input": "5\n1 2 1 1 1",
"output": "1"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "5\n1 1 1 1 2",
"output": "1"
},
{
"input": "3\n10007 20014 30021",
"output": "3"
},
{
"input": "4\n10 10 5 5",
"output": "4"
},
{
"input": "2\n6 9",
"output": "2"
},
{
"input": "3\n6 12 9",
"output": "3"
},
{
"input": "2\n14 7",
"output": "2"
},
{
"input": "2\n199 199",
"output": "2"
},
{
"input": "3\n49999 49999 99998",
"output": "3"
},
{
"input": "5\n1009 1009 1009 1009 1009",
"output": "5"
},
{
"input": "3\n17 17 17",
"output": "3"
},
{
"input": "2\n503 1509",
"output": "2"
},
{
"input": "3\n49999 99998 99998",
"output": "3"
},
{
"input": "2\n28657 28657",
"output": "2"
},
{
"input": "3\n121 22 33",
"output": "3"
},
{
"input": "2\n10079 20158",
"output": "2"
},
{
"input": "2\n6 15",
"output": "2"
},
{
"input": "4\n6 6 6 3",
"output": "4"
},
{
"input": "2\n991 1982",
"output": "2"
},
{
"input": "3\n20362 30543 50905",
"output": "3"
},
{
"input": "2\n443 886",
"output": "2"
},
{
"input": "4\n3 7 7 21",
"output": "3"
},
{
"input": "3\n3 6 9",
"output": "3"
},
{
"input": "2\n217 31",
"output": "2"
},
{
"input": "6\n10 10 10 5 5 5",
"output": "6"
},
{
"input": "4\n49441 25225 9081 4036",
"output": "4"
},
{
"input": "2\n4 9",
"output": "1"
},
{
"input": "2\n18 27",
"output": "2"
},
{
"input": "2\n13 26",
"output": "2"
},
{
"input": "4\n6 6 6 15",
"output": "4"
}
] | 1,633,524,912 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<bits/stdc++.h>
using namespace std;
/*-------------------------------------------------------------------------------------------------------------------------------------*/
#define int long long
#define endl '\n'
#define pi 3.141592653589793
#define all(s) s.begin(),s.end()
#define debug(x) cerr << #x <<" = "; _print(x); cerr << endl;
/*-------------------------------------------------------------------------------------------------------------------------------------*/
const int imax=2147483647;
const int imin=-2147483648;
const int mod=1e9+7;
const int N=1e5+10;
bool prime[N];
int factorials[N];
/*-------------------------------------------------------------------------------------------------------------------------------------*/
typedef priority_queue<int> maxheap;
typedef priority_queue<int,vector<int>,greater<int>> minheap;
/*-------------------------------------------------------------------------------------------------------------------------------------*/
void _print(int t) {cerr << t;}
void _print(unsigned int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(long double t) {cerr << t;}
void _print(double t) {cerr << t;}
template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.first); cerr << ","; _print(p.second); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(unordered_map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
/*-------------------------------------------------------------------------------------------------------------------------------------*/
void IO(){
ios_base::sync_with_stdio(false);cin.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);freopen("uttar.txt", "w", stdout);freopen("error.txt", "w", stderr);
#endif
}
/*-------------------------------------------------------------------------------------------------------------------------------------*/
void seive(){memset(prime,true,sizeof(prime));for(int p=2;p*p<=N;p++){if (prime[p] == true){for(int j=p*p;j<=N;j+=p){prime[j]=false;}}}}
void factorial(){factorials[0]=factorials[1]=1;for(int i=2;i<N;i++){factorials[i]=factorials[i-1]*i;}}
int set_bits(int n){if(n==0){return 0;}return (n&1) + set_bits(n>>1);}
void swap(auto &a, auto &b){a^=b;b^=a;a^=b;}
unordered_map<int,int> Counter(vector<int> &a){unordered_map<int,int> mp;for(auto &i : a){mp[i]++;}return mp;}
void input(vector<int> &arr){for(int &i : arr){cin>>i;}}
/*-------------------------------------------------------------------------------------------------------------------------------------*/
void solve(){
int n;
cin>>n;
unordered_map<int,int> mp;
vector<int> arr(n);
for(int &i : arr){
cin>>i;
mp[i]++;
}
int res = 1;
for(int i = 2; i < 100001;i++){
int sum = 0;
for(int j = i;j < 100001;j+=i){
sum += mp[j];
}
res = max(res, sum);
}
cout<<res<<endl;
}
int32_t main(){
IO();
int t;t=1;
//cin>>t;
while(t--){
solve();
}
return 0;
}
/*-------------------------------------------------------------------------------------------------------------------------------------*/ | Title: Bash's Big Day
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu's Lab. Since Bash is Professor Zulu's favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.
But Zulu warns him that a group of *k*<=><=1 Pokemon with strengths {*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*} tend to fight among each other if *gcd*(*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*)<==<=1 (see notes for *gcd* definition).
Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?
Note: A Pokemon cannot fight with itself.
Input Specification:
The input consists of two lines.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105), the number of Pokemon in the lab.
The next line contains *n* space separated integers, where the *i*-th of them denotes *s**i* (1<=≤<=*s**i*<=≤<=105), the strength of the *i*-th Pokemon.
Output Specification:
Print single integer — the maximum number of Pokemons Bash can take.
Demo Input:
['3\n2 3 4\n', '5\n2 3 4 6 7\n']
Demo Output:
['2\n', '3\n']
Note:
*gcd* (greatest common divisor) of positive integers set {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>} is the maximum positive integer that divides all the integers {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>}.
In the first sample, we can take Pokemons with strengths {2, 4} since *gcd*(2, 4) = 2.
In the second sample, we can take Pokemons with strengths {2, 4, 6}, and there is no larger group with *gcd* ≠ 1. | ```python
#include<bits/stdc++.h>
using namespace std;
/*-------------------------------------------------------------------------------------------------------------------------------------*/
#define int long long
#define endl '\n'
#define pi 3.141592653589793
#define all(s) s.begin(),s.end()
#define debug(x) cerr << #x <<" = "; _print(x); cerr << endl;
/*-------------------------------------------------------------------------------------------------------------------------------------*/
const int imax=2147483647;
const int imin=-2147483648;
const int mod=1e9+7;
const int N=1e5+10;
bool prime[N];
int factorials[N];
/*-------------------------------------------------------------------------------------------------------------------------------------*/
typedef priority_queue<int> maxheap;
typedef priority_queue<int,vector<int>,greater<int>> minheap;
/*-------------------------------------------------------------------------------------------------------------------------------------*/
void _print(int t) {cerr << t;}
void _print(unsigned int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(long double t) {cerr << t;}
void _print(double t) {cerr << t;}
template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.first); cerr << ","; _print(p.second); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(unordered_map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
/*-------------------------------------------------------------------------------------------------------------------------------------*/
void IO(){
ios_base::sync_with_stdio(false);cin.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);freopen("uttar.txt", "w", stdout);freopen("error.txt", "w", stderr);
#endif
}
/*-------------------------------------------------------------------------------------------------------------------------------------*/
void seive(){memset(prime,true,sizeof(prime));for(int p=2;p*p<=N;p++){if (prime[p] == true){for(int j=p*p;j<=N;j+=p){prime[j]=false;}}}}
void factorial(){factorials[0]=factorials[1]=1;for(int i=2;i<N;i++){factorials[i]=factorials[i-1]*i;}}
int set_bits(int n){if(n==0){return 0;}return (n&1) + set_bits(n>>1);}
void swap(auto &a, auto &b){a^=b;b^=a;a^=b;}
unordered_map<int,int> Counter(vector<int> &a){unordered_map<int,int> mp;for(auto &i : a){mp[i]++;}return mp;}
void input(vector<int> &arr){for(int &i : arr){cin>>i;}}
/*-------------------------------------------------------------------------------------------------------------------------------------*/
void solve(){
int n;
cin>>n;
unordered_map<int,int> mp;
vector<int> arr(n);
for(int &i : arr){
cin>>i;
mp[i]++;
}
int res = 1;
for(int i = 2; i < 100001;i++){
int sum = 0;
for(int j = i;j < 100001;j+=i){
sum += mp[j];
}
res = max(res, sum);
}
cout<<res<<endl;
}
int32_t main(){
IO();
int t;t=1;
//cin>>t;
while(t--){
solve();
}
return 0;
}
/*-------------------------------------------------------------------------------------------------------------------------------------*/
``` | -1 |
|
109 | A | Lucky Sum of Digits | PROGRAMMING | 1,000 | [
"brute force",
"implementation"
] | A. Lucky Sum of Digits | 2 | 256 | Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task. | The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number. | Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1. | [
"11\n",
"10\n"
] | [
"47\n",
"-1\n"
] | none | 500 | [
{
"input": "11",
"output": "47"
},
{
"input": "10",
"output": "-1"
},
{
"input": "64",
"output": "4477777777"
},
{
"input": "1",
"output": "-1"
},
{
"input": "4",
"output": "4"
},
{
"input": "7",
"output": "7"
},
{
"input": "12",
"output": "444"
},
{
"input": "1000000",
"output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "47",
"output": "44477777"
},
{
"input": "100",
"output": "4444777777777777"
},
{
"input": "700",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "485",
"output": "44447777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "111",
"output": "444447777777777777"
},
{
"input": "85",
"output": "4477777777777"
},
{
"input": "114",
"output": "444477777777777777"
},
{
"input": "474",
"output": "444777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "74",
"output": "47777777777"
},
{
"input": "1000",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "1024",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "4444",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "45784",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10000",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "9854",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "186",
"output": "477777777777777777777777777"
},
{
"input": "10416",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10417",
"output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "3840",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "100000",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "9876",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "99999",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "777777",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "854759",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "11000",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "18951",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999999",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "888887",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999998",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "40008",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10691",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "-1"
},
{
"input": "9",
"output": "-1"
},
{
"input": "8",
"output": "44"
},
{
"input": "2",
"output": "-1"
},
{
"input": "3",
"output": "-1"
},
{
"input": "999997",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999996",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999990",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999980",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "800000",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "980000",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
}
] | 1,617,890,378 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 280 | 4,505,600 | import sys
coins=[4,7]
sum=int(input())
def minCoins(coins, m, V):
# base case
if (V == 0):
return 0
# Initialize result
res = sys.maxsize
# Try every coin that has smaller value than V
for i in range(0, m):
if (coins[i] <= V):
sub_res = minCoins(coins, m, V - coins[i])
# Check for INT_MAX to avoid overflow and see if
# result can minimized
if (sub_res != sys.maxsize and sub_res + 1 < res):
res = sub_res + 1
return res
a=minCoins(coins,2,sum)
if a<sum:
for i in range(a):
if 4*i+7*(a-1)==sum:
print("4"*i+"7"*(a-1))
break
else:
print(-1) | Title: Lucky Sum of Digits
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.
Output Specification:
Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
Demo Input:
['11\n', '10\n']
Demo Output:
['47\n', '-1\n']
Note:
none | ```python
import sys
coins=[4,7]
sum=int(input())
def minCoins(coins, m, V):
# base case
if (V == 0):
return 0
# Initialize result
res = sys.maxsize
# Try every coin that has smaller value than V
for i in range(0, m):
if (coins[i] <= V):
sub_res = minCoins(coins, m, V - coins[i])
# Check for INT_MAX to avoid overflow and see if
# result can minimized
if (sub_res != sys.maxsize and sub_res + 1 < res):
res = sub_res + 1
return res
a=minCoins(coins,2,sum)
if a<sum:
for i in range(a):
if 4*i+7*(a-1)==sum:
print("4"*i+"7"*(a-1))
break
else:
print(-1)
``` | 0 |
686 | A | Free Ice Cream | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids. | The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place. | Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress. | [
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] | [
"22 1\n",
"3 2\n"
] | Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | 500 | [
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
"input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46",
"output": "46 0"
},
{
"input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100",
"output": "700 0"
},
{
"input": "1 0\n- 526403222",
"output": "0 1"
},
{
"input": "1 897986543\n- 371188251",
"output": "526798292 0"
},
{
"input": "1 0\n+ 1",
"output": "1 0"
},
{
"input": "1 0\n- 1",
"output": "0 1"
},
{
"input": "1 10\n+ 10",
"output": "20 0"
},
{
"input": "1 3\n- 5",
"output": "3 1"
},
{
"input": "1 0\n- 5",
"output": "0 1"
},
{
"input": "1 0\n+ 5",
"output": "5 0"
}
] | 1,669,378,037 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | g = int(input())
c = 0
for i in range (int(input())) :
s = int(input())
if (g + s) < 0 :
c += 1
else :
g += s
print(g,c)
| Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining. | ```python
g = int(input())
c = 0
for i in range (int(input())) :
s = int(input())
if (g + s) < 0 :
c += 1
else :
g += s
print(g,c)
``` | -1 |
|
320 | B | Ping-Pong (Easy Version) | PROGRAMMING | 1,500 | [
"dfs and similar",
"graphs"
] | null | null | In this problem at each moment you have a set of intervals. You can move from interval (*a*,<=*b*) from our set to interval (*c*,<=*d*) from our set if and only if *c*<=<<=*a*<=<<=*d* or *c*<=<<=*b*<=<<=*d*. Also there is a path from interval *I*1 from our set to interval *I*2 from our set if there is a sequence of successive moves starting from *I*1 so that we can reach *I*2.
Your program should handle the queries of the following two types:
1. "1 x y" (*x*<=<<=*y*) — add the new interval (*x*,<=*y*) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals.1. "2 a b" (*a*<=≠<=*b*) — answer the question: is there a path from *a*-th (one-based) added interval to *b*-th (one-based) added interval?
Answer all the queries. Note, that initially you have an empty set of intervals. | The first line of the input contains integer *n* denoting the number of queries, (1<=≤<=*n*<=≤<=100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value.
It's guaranteed that all queries are correct. | For each query of the second type print "YES" or "NO" on a separate line depending on the answer. | [
"5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2\n"
] | [
"NO\nYES\n"
] | none | 1,000 | [
{
"input": "5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2",
"output": "NO\nYES"
},
{
"input": "10\n1 -311 -186\n1 -1070 -341\n1 -1506 -634\n1 688 1698\n2 2 4\n1 70 1908\n2 1 2\n2 2 4\n1 -1053 1327\n2 5 4",
"output": "NO\nNO\nNO\nYES"
},
{
"input": "10\n1 -1365 -865\n1 1244 1834\n2 1 2\n1 -1508 -752\n2 3 2\n2 2 1\n1 -779 595\n1 -1316 877\n2 2 1\n1 -698 1700",
"output": "NO\nNO\nNO\nNO"
},
{
"input": "20\n1 1208 1583\n1 -258 729\n1 -409 1201\n1 194 1938\n1 -958 1575\n1 -1466 1752\n2 1 2\n2 1 2\n2 6 5\n1 -1870 1881\n1 -2002 2749\n1 -2002 2984\n1 -2002 3293\n2 2 4\n2 8 10\n2 9 6\n1 -2002 3572\n1 -2002 4175\n1 -2002 4452\n1 -2002 4605",
"output": "YES\nYES\nYES\nYES\nYES\nNO"
},
{
"input": "9\n1 1 4\n1 5 20\n1 11 30\n1 29 60\n1 59 100\n1 100 200\n2 1 5\n2 1 6\n2 2 5",
"output": "NO\nNO\nYES"
}
] | 1,681,493,226 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 124 | 0 | import sys
input = lambda: sys.stdin.readline().rstrip()
class UnionFind:
def __init__(self, n):
self.parent = list(range(n))
def find(self, a):
acopy = a
while a != self.parent[a]:
a = self.parent[a]
while acopy != a:
self.parent[acopy], acopy = a, self.parent[acopy]
return a
def merge(self, a, b):
self.parent[self.find(b)] = self.find(a)
N = int(input())
uni = UnionFind(N)
A = []
for _ in range(N):
t,x,y = map(int, input().split())
if t==1:
m = len(A)
for i,(a,b) in enumerate(A):
if a<x<b or a<y<b:
uni.merge(i,m)
A.append((x,y))
else:
if uni.find(x-1)==uni.find(y-1):
print('YES')
else:
print('NO')
| Title: Ping-Pong (Easy Version)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem at each moment you have a set of intervals. You can move from interval (*a*,<=*b*) from our set to interval (*c*,<=*d*) from our set if and only if *c*<=<<=*a*<=<<=*d* or *c*<=<<=*b*<=<<=*d*. Also there is a path from interval *I*1 from our set to interval *I*2 from our set if there is a sequence of successive moves starting from *I*1 so that we can reach *I*2.
Your program should handle the queries of the following two types:
1. "1 x y" (*x*<=<<=*y*) — add the new interval (*x*,<=*y*) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals.1. "2 a b" (*a*<=≠<=*b*) — answer the question: is there a path from *a*-th (one-based) added interval to *b*-th (one-based) added interval?
Answer all the queries. Note, that initially you have an empty set of intervals.
Input Specification:
The first line of the input contains integer *n* denoting the number of queries, (1<=≤<=*n*<=≤<=100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value.
It's guaranteed that all queries are correct.
Output Specification:
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
Demo Input:
['5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2\n']
Demo Output:
['NO\nYES\n']
Note:
none | ```python
import sys
input = lambda: sys.stdin.readline().rstrip()
class UnionFind:
def __init__(self, n):
self.parent = list(range(n))
def find(self, a):
acopy = a
while a != self.parent[a]:
a = self.parent[a]
while acopy != a:
self.parent[acopy], acopy = a, self.parent[acopy]
return a
def merge(self, a, b):
self.parent[self.find(b)] = self.find(a)
N = int(input())
uni = UnionFind(N)
A = []
for _ in range(N):
t,x,y = map(int, input().split())
if t==1:
m = len(A)
for i,(a,b) in enumerate(A):
if a<x<b or a<y<b:
uni.merge(i,m)
A.append((x,y))
else:
if uni.find(x-1)==uni.find(y-1):
print('YES')
else:
print('NO')
``` | 0 |
|
1,011 | B | Planning The Expedition | PROGRAMMING | 1,200 | [
"binary search",
"brute force",
"implementation"
] | null | null | Natasha is planning an expedition to Mars for $n$ people. One of the important tasks is to provide food for each participant.
The warehouse has $m$ daily food packages. Each package has some food type $a_i$.
Each participant must eat exactly one food package each day. Due to extreme loads, each participant must eat the same food type throughout the expedition. Different participants may eat different (or the same) types of food.
Formally, for each participant $j$ Natasha should select his food type $b_j$ and each day $j$-th participant will eat one food package of type $b_j$. The values $b_j$ for different participants may be different.
What is the maximum possible number of days the expedition can last, following the requirements above? | The first line contains two integers $n$ and $m$ ($1 \le n \le 100$, $1 \le m \le 100$) — the number of the expedition participants and the number of the daily food packages available.
The second line contains sequence of integers $a_1, a_2, \dots, a_m$ ($1 \le a_i \le 100$), where $a_i$ is the type of $i$-th food package. | Print the single integer — the number of days the expedition can last. If it is not possible to plan the expedition for even one day, print 0. | [
"4 10\n1 5 2 1 1 1 2 5 7 2\n",
"100 1\n1\n",
"2 5\n5 4 3 2 1\n",
"3 9\n42 42 42 42 42 42 42 42 42\n"
] | [
"2\n",
"0\n",
"1\n",
"3\n"
] | In the first example, Natasha can assign type $1$ food to the first participant, the same type $1$ to the second, type $5$ to the third and type $2$ to the fourth. In this case, the expedition can last for $2$ days, since each participant can get two food packages of his food type (there will be used $4$ packages of type $1$, two packages of type $2$ and two packages of type $5$).
In the second example, there are $100$ participants and only $1$ food package. In this case, the expedition can't last even $1$ day. | 1,000 | [
{
"input": "4 10\n1 5 2 1 1 1 2 5 7 2",
"output": "2"
},
{
"input": "100 1\n1",
"output": "0"
},
{
"input": "2 5\n5 4 3 2 1",
"output": "1"
},
{
"input": "3 9\n42 42 42 42 42 42 42 42 42",
"output": "3"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "4 100\n84 99 66 69 86 94 89 96 98 93 93 82 87 93 91 100 69 99 93 81 99 84 75 100 86 88 98 100 84 96 44 70 94 91 85 78 86 79 45 88 91 78 98 94 81 87 93 72 96 88 96 97 96 62 86 72 94 84 80 98 88 90 93 73 73 98 78 50 91 96 97 82 85 90 87 41 97 82 97 77 100 100 92 83 98 81 70 81 74 78 84 79 98 98 55 99 97 99 79 98",
"output": "5"
},
{
"input": "100 100\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": "1"
},
{
"input": "1 100\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": "100"
},
{
"input": "6 100\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 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 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "15"
},
{
"input": "1 1\n59",
"output": "1"
},
{
"input": "1 50\n39 1 46 21 23 28 100 32 63 63 18 15 40 29 34 49 56 74 47 42 96 97 59 62 76 62 69 61 36 21 66 18 92 58 63 85 5 6 77 75 91 66 38 10 66 43 20 74 37 83",
"output": "3"
},
{
"input": "1 100\n83 72 21 55 49 5 61 60 87 21 89 88 3 75 49 81 36 25 50 61 96 19 36 55 48 8 97 69 50 24 23 39 26 25 41 90 69 20 19 62 38 52 60 6 66 31 9 45 36 12 69 94 22 60 91 65 35 58 13 85 33 87 83 11 95 20 20 85 13 21 57 69 17 94 78 37 59 45 60 7 64 51 60 89 91 22 6 58 95 96 51 53 89 22 28 16 27 56 1 54",
"output": "5"
},
{
"input": "50 1\n75",
"output": "0"
},
{
"input": "50 50\n85 20 12 73 52 78 70 95 88 43 31 88 81 41 80 99 16 11 97 11 21 44 2 34 47 38 87 2 32 47 97 93 52 14 35 37 97 48 58 19 52 55 97 72 17 25 16 85 90 58",
"output": "1"
},
{
"input": "50 100\n2 37 74 32 99 75 73 86 67 33 62 30 15 21 51 41 73 75 67 39 90 10 56 74 72 26 38 65 75 55 46 99 34 49 92 82 11 100 15 71 75 12 22 56 47 74 20 98 59 65 14 76 1 40 89 36 43 93 83 73 75 100 50 95 27 10 72 51 25 69 15 3 57 60 84 99 31 44 12 61 69 95 51 31 28 36 57 35 31 52 44 19 79 12 27 27 7 81 68 1",
"output": "1"
},
{
"input": "100 1\n26",
"output": "0"
},
{
"input": "100 50\n8 82 62 11 85 57 5 32 99 92 77 2 61 86 8 88 10 28 83 4 68 79 8 64 56 98 4 88 22 54 30 60 62 79 72 38 17 28 32 16 62 26 56 44 72 33 22 84 77 45",
"output": "0"
},
{
"input": "100 100\n13 88 64 65 78 10 61 97 16 32 76 9 60 1 40 35 90 61 60 85 26 16 38 36 33 95 24 55 82 88 13 9 47 34 94 2 90 74 11 81 46 70 94 11 55 32 19 36 97 16 17 35 38 82 89 16 74 94 97 79 9 94 88 12 28 2 4 25 72 95 49 31 88 82 6 77 70 98 90 57 57 33 38 61 26 75 2 66 22 44 13 35 16 4 33 16 12 66 32 86",
"output": "1"
},
{
"input": "34 64\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",
"output": "1"
},
{
"input": "53 98\n1 1 2 2 2 2 2 1 2 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 1 2 1 1 1 2 1 2 1 1 1 2 2 1 2 1 1 1 2 2 1 2 1 1 2 1 2 2 1 2 2 2 2 2 2 2 2 2 1 1 2 2 1 2 1 2 1 2 1 1 2 2 2 1 1 2 1 2 1 1 1 1 2 2 2 2 2 1 1 2 2 2 1 1",
"output": "1"
},
{
"input": "17 8\n2 5 3 4 3 2 2 2",
"output": "0"
},
{
"input": "24 77\n8 6 10 4 6 6 4 10 9 7 7 5 5 4 6 7 10 6 3 4 6 6 4 9 4 6 2 5 3 4 4 1 4 6 6 8 1 1 6 4 6 2 5 7 7 2 4 4 10 1 10 9 2 3 8 1 10 4 3 9 3 8 3 5 6 3 4 9 5 3 4 1 1 6 1 2 1",
"output": "2"
},
{
"input": "65 74\n7 19 2 38 28 44 34 49 14 13 30 22 11 4 4 12 8 1 40 8 34 31 44 38 21 35 13 7 19 32 37 5 36 26 7 2 15 11 47 45 48 2 49 10 10 42 42 31 50 24 29 34 31 38 39 48 43 47 32 46 10 1 33 21 12 50 13 44 38 11 41 41 10 7",
"output": "1"
},
{
"input": "37 71\n50 93 15 80 82 23 35 90 70 73 55 23 23 6 86 63 38 70 38 52 88 34 25 75 32 19 6 98 31 38 21 8 66 8 59 71 7 80 69 23 17 70 6 40 72 5 48 59 18 1 48 91 17 41 11 27 53 95 87 31 62 94 94 60 38 99 70 50 81 86 44",
"output": "1"
},
{
"input": "35 4\n100 100 100 100",
"output": "0"
},
{
"input": "68 12\n100 100 100 99 99 100 100 100 99 99 99 99",
"output": "0"
},
{
"input": "91 33\n97 100 96 96 97 100 97 97 96 96 99 99 98 97 97 99 99 98 100 96 96 99 100 96 97 100 97 97 99 98 96 98 97",
"output": "0"
},
{
"input": "59 72\n99 96 96 93 94 97 93 96 100 99 93 99 96 92 92 97 93 100 92 100 91 96 99 100 97 98 93 93 96 100 97 99 100 100 91 93 96 91 93 100 95 96 92 97 91 91 99 100 91 95 95 100 92 93 98 93 95 92 95 100 91 95 99 95 92 91 100 93 98 98 96 99",
"output": "1"
},
{
"input": "9 35\n95 54 78 94 56 63 68 73 61 85 73 62 61 94 67 91 66 54 89 69 92 60 91 87 63 62 85 70 62 68 97 88 96 55 62",
"output": "2"
},
{
"input": "99 100\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": "1"
},
{
"input": "1 50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "50"
}
] | 1,591,902,114 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 124 | 614,400 | from collections import Counter
npar, npack = map(int, input().split())
pack_arr = list(map(int, input().split()))
if npar > npack:
print(0)
elif npack//npar == 1:
print(1)
else:
occurences = Counter(pack_arr)
longest_days = 1
for days in range(2, npack//npar + 1):
current_ful = 0
for k in occurences:
current_ful += occurences[k]//days
if current_ful >= npar:
longest_days = days
break
else:
break
print(longest_days)
| Title: Planning The Expedition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is planning an expedition to Mars for $n$ people. One of the important tasks is to provide food for each participant.
The warehouse has $m$ daily food packages. Each package has some food type $a_i$.
Each participant must eat exactly one food package each day. Due to extreme loads, each participant must eat the same food type throughout the expedition. Different participants may eat different (or the same) types of food.
Formally, for each participant $j$ Natasha should select his food type $b_j$ and each day $j$-th participant will eat one food package of type $b_j$. The values $b_j$ for different participants may be different.
What is the maximum possible number of days the expedition can last, following the requirements above?
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n \le 100$, $1 \le m \le 100$) — the number of the expedition participants and the number of the daily food packages available.
The second line contains sequence of integers $a_1, a_2, \dots, a_m$ ($1 \le a_i \le 100$), where $a_i$ is the type of $i$-th food package.
Output Specification:
Print the single integer — the number of days the expedition can last. If it is not possible to plan the expedition for even one day, print 0.
Demo Input:
['4 10\n1 5 2 1 1 1 2 5 7 2\n', '100 1\n1\n', '2 5\n5 4 3 2 1\n', '3 9\n42 42 42 42 42 42 42 42 42\n']
Demo Output:
['2\n', '0\n', '1\n', '3\n']
Note:
In the first example, Natasha can assign type $1$ food to the first participant, the same type $1$ to the second, type $5$ to the third and type $2$ to the fourth. In this case, the expedition can last for $2$ days, since each participant can get two food packages of his food type (there will be used $4$ packages of type $1$, two packages of type $2$ and two packages of type $5$).
In the second example, there are $100$ participants and only $1$ food package. In this case, the expedition can't last even $1$ day. | ```python
from collections import Counter
npar, npack = map(int, input().split())
pack_arr = list(map(int, input().split()))
if npar > npack:
print(0)
elif npack//npar == 1:
print(1)
else:
occurences = Counter(pack_arr)
longest_days = 1
for days in range(2, npack//npar + 1):
current_ful = 0
for k in occurences:
current_ful += occurences[k]//days
if current_ful >= npar:
longest_days = days
break
else:
break
print(longest_days)
``` | 3 |
|
999 | B | Reversing Encryption | PROGRAMMING | 900 | [
"implementation"
] | null | null | A string $s$ of length $n$ can be encrypted by the following algorithm:
- iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$), - for each divisor $d$, reverse the substring $s[1 \dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$).
For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$).
You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique. | The first line of input consists of a single integer $n$ ($1 \le n \le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters. | Print a string $s$ such that the above algorithm results in $t$. | [
"10\nrocesfedoc\n",
"16\nplmaetwoxesisiht\n",
"1\nz\n"
] | [
"codeforces\n",
"thisisexampletwo\n",
"z\n"
] | The first example is described in the problem statement. | 0 | [
{
"input": "10\nrocesfedoc",
"output": "codeforces"
},
{
"input": "16\nplmaetwoxesisiht",
"output": "thisisexampletwo"
},
{
"input": "1\nz",
"output": "z"
},
{
"input": "2\nir",
"output": "ri"
},
{
"input": "3\nilj",
"output": "jli"
},
{
"input": "4\njfyy",
"output": "yyjf"
},
{
"input": "6\nkrdych",
"output": "hcyrkd"
},
{
"input": "60\nfnebsopcvmlaoecpzmakqigyuutueuozjxutlwwiochekmhjgwxsgfbcrpqj",
"output": "jqprcbfgsxwgjhmkehcoiwwltuxjzokamzpalobnfespcvmoecqigyuutueu"
},
{
"input": "64\nhnlzzhrvqnldswxfsrowfhmyzbxtyoxhogudasgywxycyhzgiseerbislcncvnwy",
"output": "ywnvcnclsibreesigzhycyxwygsadugofxwsdlnqzlhnzhrvsrowfhmyzbxtyoxh"
},
{
"input": "97\nqnqrmdhmbubaijtwsecbidqouhlecladwgwcuxbigckrfzasnbfbslukoayhcgquuacygakhxoubibxtqkpyyhzjipylujgrc",
"output": "crgjulypijzhyypkqtxbibuoxhkagycauuqgchyaokulsbfbnsazfrkcgibxucwgwdalcelhuoqdibceswtjiabubmhdmrqnq"
},
{
"input": "100\nedykhvzcntljuuoqghptioetqnfllwekzohiuaxelgecabvsbibgqodqxvyfkbyjwtgbyhvssntinkwsinwsmalusiwnjmtcoovf",
"output": "fvooctmjnwisulamswniswknitnssvhybgtwjybkfyvxqdoqgbqteoitnczvkyedhljuuoqghptnfllwekzohiuaxelgecabvsbi"
},
{
"input": "96\nqtbcksuvxonzbkokhqlgkrvimzqmqnrvqlihrmksldyydacbtckfphenxszcnzhfjmpeykrvshgiboivkvabhrpphgavvprz",
"output": "zrpvvaghpprhbavkviobighsvrkyepmjfhznczsxnehpfkctvrnqmqzmkokbvuctqbksxonzhqlgkrviqlihrmksldyydacb"
},
{
"input": "90\nmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm"
},
{
"input": "89\nwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww"
},
{
"input": "99\nqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq",
"output": "qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq"
},
{
"input": "100\noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo",
"output": "oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo"
},
{
"input": "60\nwwwwwxwwwwwwfhwwhwwwwwwawwwwwwwwwwwwwnwwwwwwwwwwwwwwwwwwwwww",
"output": "wwwwwwwwwwwwwwwwwwwwwwnwwwwwwwwwwhwwwxwwwwwwwwwfhwwwwawwwwww"
},
{
"input": "90\ncccchccccccccccccccccccccccccccwcccccccccgcccccchccccccccccccccccccccccxccccccncccccccuccc",
"output": "cccucccccccnccccccxcccccccccccccccccccccchccccccccccccccccccccccchccccccccccwcccccccccgccc"
},
{
"input": "97\nfwffffffffffffffffffffffffrffffffffffffffzfffffffffffffffftfcfffffffqffffffffffffffffffffffyfffff",
"output": "fffffyffffffffffffffffffffffqfffffffcftffffffffffffffffzffffffffffffffrffffffffffffffffffffffffwf"
},
{
"input": "100\ndjjjjjjjjjjgjjjjjjjjjjjjjjsvjjjjjjjjjjmjjjjjjjjjjjjjajjjjjjajjjjjjrjjjjjjjjjjjjrjjtjjjjjjjjjjjjjojjj",
"output": "jjjojjjjjjjjjjjjjtjjrjjjjjjjjjjjjrjjjjjjajjjjjjajjjjjjjjjjjjjjdjjjgjjjjjjjjjsvjjjjjjjjjjmjjjjjjjjjjj"
}
] | 1,621,101,336 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 62 | 0 | n = int(input())
s = list(input())
for d in range(1,n+1):
s[0:d] = reversed(s[0:d]) if n%d == 0 else s[0:d]
print("".join(s)) | Title: Reversing Encryption
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string $s$ of length $n$ can be encrypted by the following algorithm:
- iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$), - for each divisor $d$, reverse the substring $s[1 \dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$).
For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$).
You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique.
Input Specification:
The first line of input consists of a single integer $n$ ($1 \le n \le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters.
Output Specification:
Print a string $s$ such that the above algorithm results in $t$.
Demo Input:
['10\nrocesfedoc\n', '16\nplmaetwoxesisiht\n', '1\nz\n']
Demo Output:
['codeforces\n', 'thisisexampletwo\n', 'z\n']
Note:
The first example is described in the problem statement. | ```python
n = int(input())
s = list(input())
for d in range(1,n+1):
s[0:d] = reversed(s[0:d]) if n%d == 0 else s[0:d]
print("".join(s))
``` | 3 |
|
851 | B | Arpa and an exam about geometry | PROGRAMMING | 1,400 | [
"geometry",
"math"
] | null | null | Arpa is taking a geometry exam. Here is the last problem of the exam.
You are given three points *a*,<=*b*,<=*c*.
Find a point and an angle such that if we rotate the page around the point by the angle, the new position of *a* is the same as the old position of *b*, and the new position of *b* is the same as the old position of *c*.
Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not. | The only line contains six integers *a**x*,<=*a**y*,<=*b**x*,<=*b**y*,<=*c**x*,<=*c**y* (|*a**x*|,<=|*a**y*|,<=|*b**x*|,<=|*b**y*|,<=|*c**x*|,<=|*c**y*|<=≤<=109). It's guaranteed that the points are distinct. | Print "Yes" if the problem has a solution, "No" otherwise.
You can print each letter in any case (upper or lower). | [
"0 1 1 1 1 0\n",
"1 1 0 0 1000 1000\n"
] | [
"Yes\n",
"No\n"
] | In the first sample test, rotate the page around (0.5, 0.5) by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9d845923f4d356a48d8ede337db0303821311f0c.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test, you can't find any solution. | 1,000 | [
{
"input": "0 1 1 1 1 0",
"output": "Yes"
},
{
"input": "1 1 0 0 1000 1000",
"output": "No"
},
{
"input": "1 0 2 0 3 0",
"output": "No"
},
{
"input": "3 4 0 0 4 3",
"output": "Yes"
},
{
"input": "-1000000000 1 0 0 1000000000 1",
"output": "Yes"
},
{
"input": "49152 0 0 0 0 81920",
"output": "No"
},
{
"input": "1 -1 4 4 2 -3",
"output": "No"
},
{
"input": "-2 -2 1 4 -2 0",
"output": "No"
},
{
"input": "5 0 4 -2 0 1",
"output": "No"
},
{
"input": "-4 -3 2 -1 -3 4",
"output": "No"
},
{
"input": "-3 -3 5 2 3 -1",
"output": "No"
},
{
"input": "-1000000000 -1000000000 0 0 1000000000 999999999",
"output": "No"
},
{
"input": "-1000000000 -1000000000 0 0 1000000000 1000000000",
"output": "No"
},
{
"input": "-357531221 381512519 -761132895 -224448284 328888775 -237692564",
"output": "No"
},
{
"input": "264193194 -448876521 736684426 -633906160 -328597212 -47935734",
"output": "No"
},
{
"input": "419578772 -125025887 169314071 89851312 961404059 21419450",
"output": "No"
},
{
"input": "-607353321 -620687860 248029390 477864359 728255275 -264646027",
"output": "No"
},
{
"input": "299948862 -648908808 338174789 841279400 -850322448 350263551",
"output": "No"
},
{
"input": "48517753 416240699 7672672 272460100 -917845051 199790781",
"output": "No"
},
{
"input": "-947393823 -495674431 211535284 -877153626 -522763219 -778236665",
"output": "No"
},
{
"input": "-685673792 -488079395 909733355 385950193 -705890324 256550506",
"output": "No"
},
{
"input": "-326038504 547872194 49630307 713863100 303770000 -556852524",
"output": "No"
},
{
"input": "-706921242 -758563024 -588592101 -443440080 858751713 238854303",
"output": "No"
},
{
"input": "-1000000000 -1000000000 0 1000000000 1000000000 -1000000000",
"output": "Yes"
},
{
"input": "1000000000 1000000000 0 -1000000000 -1000000000 1000000000",
"output": "Yes"
},
{
"input": "-999999999 -1000000000 0 0 1000000000 999999999",
"output": "Yes"
},
{
"input": "-1000000000 -999999999 0 0 1000000000 999999999",
"output": "No"
},
{
"input": "-1 -1000000000 0 1000000000 1 -1000000000",
"output": "Yes"
},
{
"input": "0 1000000000 1 0 0 -1000000000",
"output": "Yes"
},
{
"input": "0 1000000000 0 0 0 -1000000000",
"output": "No"
},
{
"input": "0 1 1 2 2 3",
"output": "No"
},
{
"input": "999999999 1000000000 0 0 -1000000000 -999999999",
"output": "Yes"
},
{
"input": "0 0 1 1 2 0",
"output": "Yes"
},
{
"input": "0 0 1 1 2 2",
"output": "No"
},
{
"input": "1 1 2 2 3 3",
"output": "No"
},
{
"input": "0 2 0 3 0 4",
"output": "No"
},
{
"input": "1 1 1 2 1 3",
"output": "No"
},
{
"input": "0 0 3 4 3 9",
"output": "Yes"
},
{
"input": "589824 196608 262144 196608 0 0",
"output": "Yes"
},
{
"input": "0 0 1000000000 1 1000000000 -999999999",
"output": "No"
},
{
"input": "0 0 2 45 0 90",
"output": "Yes"
},
{
"input": "0 0 0 2 0 1",
"output": "No"
},
{
"input": "0 2 4 5 4 0",
"output": "Yes"
},
{
"input": "0 0 2 0 4 0",
"output": "No"
},
{
"input": "1 1 3 3 5 5",
"output": "No"
},
{
"input": "1 1 2 2 3 1",
"output": "Yes"
}
] | 1,504,538,101 | 2,400 | Python 3 | RUNTIME_ERROR | TESTS | 29 | 62 | 307,200 | #B
from math import asin
from math import sqrt
from math import fabs
def solution():
ax, ay, bx, by, cx, cy = map(int, input().split())
k = 0
if ax - bx != 0:
k = (ay - by)/(ax - bx)
b = by - k*bx
if cy == k*cx + b:
print('No')
return
d = 2*(ax*(by - cy) + bx*(cy - ay) + cx*(ay - by))
x = ((ax**2 + ay**2)*(by - cy) + (bx**2 + by**2)*(cy - ay) + (cx**2 + cy**2)*(ay - by))/d
y = ((ax**2 + ay**2)*(cx - bx) + (bx**2 + by**2)*(ax - cx) + (cx**2 + cy**2)*(bx - ax))/d
a = sqrt((bx - ax)**2 + (by - ay)**2)
b = sqrt((cx - bx)**2 + (cy - by)**2)
r = sqrt((ax - x)**2 + (ay - y)**2)
if asin(fabs(a)/(2*r)) != asin(fabs(b)/(2*r)):
print('No')
return
print('Yes')
solution() | Title: Arpa and an exam about geometry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arpa is taking a geometry exam. Here is the last problem of the exam.
You are given three points *a*,<=*b*,<=*c*.
Find a point and an angle such that if we rotate the page around the point by the angle, the new position of *a* is the same as the old position of *b*, and the new position of *b* is the same as the old position of *c*.
Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.
Input Specification:
The only line contains six integers *a**x*,<=*a**y*,<=*b**x*,<=*b**y*,<=*c**x*,<=*c**y* (|*a**x*|,<=|*a**y*|,<=|*b**x*|,<=|*b**y*|,<=|*c**x*|,<=|*c**y*|<=≤<=109). It's guaranteed that the points are distinct.
Output Specification:
Print "Yes" if the problem has a solution, "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['0 1 1 1 1 0\n', '1 1 0 0 1000 1000\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test, rotate the page around (0.5, 0.5) by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9d845923f4d356a48d8ede337db0303821311f0c.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test, you can't find any solution. | ```python
#B
from math import asin
from math import sqrt
from math import fabs
def solution():
ax, ay, bx, by, cx, cy = map(int, input().split())
k = 0
if ax - bx != 0:
k = (ay - by)/(ax - bx)
b = by - k*bx
if cy == k*cx + b:
print('No')
return
d = 2*(ax*(by - cy) + bx*(cy - ay) + cx*(ay - by))
x = ((ax**2 + ay**2)*(by - cy) + (bx**2 + by**2)*(cy - ay) + (cx**2 + cy**2)*(ay - by))/d
y = ((ax**2 + ay**2)*(cx - bx) + (bx**2 + by**2)*(ax - cx) + (cx**2 + cy**2)*(bx - ax))/d
a = sqrt((bx - ax)**2 + (by - ay)**2)
b = sqrt((cx - bx)**2 + (cy - by)**2)
r = sqrt((ax - x)**2 + (ay - y)**2)
if asin(fabs(a)/(2*r)) != asin(fabs(b)/(2*r)):
print('No')
return
print('Yes')
solution()
``` | -1 |
|
787 | A | The Monster | PROGRAMMING | 1,200 | [
"brute force",
"math",
"number theory"
] | null | null | A monster is chasing after Rick and Morty on another planet. They're so frightened that sometimes they scream. More accurately, Rick screams at times *b*,<=*b*<=+<=*a*,<=*b*<=+<=2*a*,<=*b*<=+<=3*a*,<=... and Morty screams at times *d*,<=*d*<=+<=*c*,<=*d*<=+<=2*c*,<=*d*<=+<=3*c*,<=....
The Monster will catch them if at any point they scream at the same time, so it wants to know when it will catch them (the first time they scream at the same time) or that they will never scream at the same time. | The first line of input contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100).
The second line contains two integers *c* and *d* (1<=≤<=*c*,<=*d*<=≤<=100). | Print the first time Rick and Morty will scream at the same time, or <=-<=1 if they will never scream at the same time. | [
"20 2\n9 19\n",
"2 1\n16 12\n"
] | [
"82\n",
"-1\n"
] | In the first sample testcase, Rick's 5th scream and Morty's 8th time are at time 82.
In the second sample testcase, all Rick's screams will be at odd times and Morty's will be at even times, so they will never scream at the same time. | 500 | [
{
"input": "20 2\n9 19",
"output": "82"
},
{
"input": "2 1\n16 12",
"output": "-1"
},
{
"input": "39 52\n88 78",
"output": "1222"
},
{
"input": "59 96\n34 48",
"output": "1748"
},
{
"input": "87 37\n91 29",
"output": "211"
},
{
"input": "11 81\n49 7",
"output": "301"
},
{
"input": "39 21\n95 89",
"output": "3414"
},
{
"input": "59 70\n48 54",
"output": "1014"
},
{
"input": "87 22\n98 32",
"output": "718"
},
{
"input": "15 63\n51 13",
"output": "-1"
},
{
"input": "39 7\n97 91",
"output": "1255"
},
{
"input": "18 18\n71 71",
"output": "1278"
},
{
"input": "46 71\n16 49",
"output": "209"
},
{
"input": "70 11\n74 27",
"output": "2321"
},
{
"input": "94 55\n20 96",
"output": "-1"
},
{
"input": "18 4\n77 78",
"output": "1156"
},
{
"input": "46 44\n23 55",
"output": "-1"
},
{
"input": "74 88\n77 37",
"output": "1346"
},
{
"input": "94 37\n34 7",
"output": "789"
},
{
"input": "22 81\n80 88",
"output": "-1"
},
{
"input": "46 30\n34 62",
"output": "674"
},
{
"input": "40 4\n81 40",
"output": "364"
},
{
"input": "69 48\n39 9",
"output": "48"
},
{
"input": "89 93\n84 87",
"output": "5967"
},
{
"input": "17 45\n42 65",
"output": "317"
},
{
"input": "41 85\n95 46",
"output": "331"
},
{
"input": "69 30\n41 16",
"output": "1410"
},
{
"input": "93 74\n99 93",
"output": "-1"
},
{
"input": "17 19\n44 75",
"output": "427"
},
{
"input": "45 63\n98 53",
"output": "3483"
},
{
"input": "69 11\n48 34",
"output": "-1"
},
{
"input": "55 94\n3 96",
"output": "204"
},
{
"input": "100 100\n100 100",
"output": "100"
},
{
"input": "1 1\n1 1",
"output": "1"
},
{
"input": "1 1\n1 100",
"output": "100"
},
{
"input": "1 100\n100 1",
"output": "101"
},
{
"input": "98 1\n99 100",
"output": "9703"
},
{
"input": "98 1\n99 2",
"output": "9605"
},
{
"input": "97 2\n99 100",
"output": "4852"
},
{
"input": "3 3\n3 1",
"output": "-1"
},
{
"input": "3 2\n7 2",
"output": "2"
},
{
"input": "2 3\n2 5",
"output": "5"
},
{
"input": "2 3\n2 3",
"output": "3"
},
{
"input": "100 3\n100 5",
"output": "-1"
},
{
"input": "6 10\n12 14",
"output": "-1"
},
{
"input": "4 2\n4 4",
"output": "-1"
},
{
"input": "2 3\n2 2",
"output": "-1"
},
{
"input": "2 3\n4 99",
"output": "99"
},
{
"input": "1 5\n1 5",
"output": "5"
},
{
"input": "1 100\n3 1",
"output": "100"
},
{
"input": "2 2\n2 1",
"output": "-1"
},
{
"input": "2 10\n6 20",
"output": "20"
},
{
"input": "2 2\n2 10",
"output": "10"
},
{
"input": "3 7\n3 6",
"output": "-1"
},
{
"input": "1 100\n1 100",
"output": "100"
},
{
"input": "7 25\n39 85",
"output": "319"
},
{
"input": "84 82\n38 6",
"output": "82"
},
{
"input": "7 7\n7 14",
"output": "14"
}
] | 1,490,470,625 | 2,147,483,647 | Python 3 | OK | TESTS | 58 | 514 | 4,608,000 | #!/usr/bin/python3
a,b = [int(x) for x in input().split()]
c,d = [int(x) for x in input().split()]
i = 0
j = 0
flag = 0
for i in range(1000000):
s1 = b + a*i
if (s1 >= d and 0 == (s1-d)%c):
print(s1)
flag = 1
break
if (not flag):
print("-1")
| Title: The Monster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A monster is chasing after Rick and Morty on another planet. They're so frightened that sometimes they scream. More accurately, Rick screams at times *b*,<=*b*<=+<=*a*,<=*b*<=+<=2*a*,<=*b*<=+<=3*a*,<=... and Morty screams at times *d*,<=*d*<=+<=*c*,<=*d*<=+<=2*c*,<=*d*<=+<=3*c*,<=....
The Monster will catch them if at any point they scream at the same time, so it wants to know when it will catch them (the first time they scream at the same time) or that they will never scream at the same time.
Input Specification:
The first line of input contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100).
The second line contains two integers *c* and *d* (1<=≤<=*c*,<=*d*<=≤<=100).
Output Specification:
Print the first time Rick and Morty will scream at the same time, or <=-<=1 if they will never scream at the same time.
Demo Input:
['20 2\n9 19\n', '2 1\n16 12\n']
Demo Output:
['82\n', '-1\n']
Note:
In the first sample testcase, Rick's 5th scream and Morty's 8th time are at time 82.
In the second sample testcase, all Rick's screams will be at odd times and Morty's will be at even times, so they will never scream at the same time. | ```python
#!/usr/bin/python3
a,b = [int(x) for x in input().split()]
c,d = [int(x) for x in input().split()]
i = 0
j = 0
flag = 0
for i in range(1000000):
s1 = b + a*i
if (s1 >= d and 0 == (s1-d)%c):
print(s1)
flag = 1
break
if (not flag):
print("-1")
``` | 3 |
|
181 | A | Series of Crimes | PROGRAMMING | 800 | [
"brute force",
"geometry",
"implementation"
] | null | null | The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed. | The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements. | Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right. | [
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] | [
"1 1\n",
"2 3\n"
] | none | 500 | [
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n.......*..",
"output": "2 1"
},
{
"input": "10 3\n*..\n...\n...\n...\n...\n...\n...\n...\n...\n**.",
"output": "1 2"
},
{
"input": "100 2\n*.\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "2 100\n...................................................................................................*\n*..................................................................................................*",
"output": "1 1"
},
{
"input": "2 2\n*.\n**",
"output": "1 2"
},
{
"input": "2 2\n**\n.*",
"output": "2 1"
},
{
"input": "2 3\n*.*\n..*",
"output": "2 1"
}
] | 1,615,527,479 | 2,147,483,647 | PyPy 3 | OK | TESTS | 36 | 218 | 0 | n,m = input().split()
table = []
row = []
for i in range(int(n)):
a=str(input())
table.append(a)
if a.find('*')>=0:
row.append(i)
for i in range(int(m)):
if table[row[0]][i]=='*' and table[row[1]][i]=='*':
continue
elif table[row[0]][i]=='*' and table[row[1]][i]!='*':
print("{} {}".format(row[1]+1,i+1))
elif table[row[0]][i]!='*' and table[row[1]][i]=='*':
print("{} {}".format(row[0]+1,i+1))
| Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none | ```python
n,m = input().split()
table = []
row = []
for i in range(int(n)):
a=str(input())
table.append(a)
if a.find('*')>=0:
row.append(i)
for i in range(int(m)):
if table[row[0]][i]=='*' and table[row[1]][i]=='*':
continue
elif table[row[0]][i]=='*' and table[row[1]][i]!='*':
print("{} {}".format(row[1]+1,i+1))
elif table[row[0]][i]!='*' and table[row[1]][i]=='*':
print("{} {}".format(row[0]+1,i+1))
``` | 3 |
|
165 | A | Supercentral Point | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | [
"8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n",
"5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n"
] | [
"2\n",
"1\n"
] | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | 500 | [
{
"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3",
"output": "2"
},
{
"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0",
"output": "1"
},
{
"input": "9\n-565 -752\n-184 723\n-184 -752\n-184 1\n950 723\n-565 723\n950 -752\n950 1\n-565 1",
"output": "1"
},
{
"input": "25\n-651 897\n916 897\n-651 -808\n-748 301\n-734 414\n-651 -973\n-734 897\n916 -550\n-758 414\n916 180\n-758 -808\n-758 -973\n125 -550\n125 -973\n125 301\n916 414\n-748 -808\n-651 301\n-734 301\n-307 897\n-651 -550\n-651 414\n125 -808\n-748 -550\n916 -808",
"output": "7"
},
{
"input": "1\n487 550",
"output": "0"
},
{
"input": "10\n990 -396\n990 736\n990 646\n990 -102\n990 -570\n990 155\n990 528\n990 489\n990 268\n990 676",
"output": "0"
},
{
"input": "30\n507 836\n525 836\n-779 196\n507 -814\n525 -814\n525 42\n525 196\n525 -136\n-779 311\n507 -360\n525 300\n507 578\n507 311\n-779 836\n507 300\n525 -360\n525 311\n-779 -360\n-779 578\n-779 300\n507 42\n525 578\n-779 379\n507 196\n525 379\n507 379\n-779 -814\n-779 42\n-779 -136\n507 -136",
"output": "8"
},
{
"input": "25\n890 -756\n890 -188\n-37 -756\n-37 853\n523 998\n-261 853\n-351 853\n-351 -188\n523 -756\n-261 -188\n-37 998\n523 -212\n-351 998\n-37 -188\n-351 -756\n-37 -212\n890 998\n890 -212\n523 853\n-351 -212\n-261 -212\n-261 998\n-261 -756\n890 853\n523 -188",
"output": "9"
},
{
"input": "21\n-813 -11\n486 254\n685 254\n-708 254\n-55 -11\n-671 -191\n486 -11\n-671 -11\n685 -11\n685 -191\n486 -191\n-55 254\n-708 -11\n-813 254\n-708 -191\n41 -11\n-671 254\n-813 -191\n41 254\n-55 -191\n41 -191",
"output": "5"
},
{
"input": "4\n1 0\n2 0\n1 1\n1 -1",
"output": "0"
}
] | 1,679,281,938 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 92 | 0 | #45
n = int(input())
ins = [[int(x) for x in input().split()] for i in range(n)]
v = [set([x[i] for x in ins]) for i in range(2) ]
Maps = [{x : [] for x in v[i]} for i in range(2)]
superCentralPoints = 0
for MapIndex in range(2):
for elemnt in ins:
Maps[MapIndex][elemnt[MapIndex]].append(elemnt)
for point in ins:
hasAdjacents = [False for i in range(4)]
for MapIndex in range(2):
for otherPoint in Maps[MapIndex][point[MapIndex]]:
if point[1 - MapIndex] > otherPoint[1 - MapIndex]:
hasAdjacents[MapIndex * 2] = True
elif point[1 - MapIndex] < otherPoint[1 - MapIndex]:
hasAdjacents[MapIndex * 2 + 1] = True
if False not in hasAdjacents[MapIndex : MapIndex + 2]:
break
if False not in hasAdjacents:
superCentralPoints += 1
print(superCentralPoints) | Title: Supercentral Point
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*). Let's define neighbors for some fixed point from the given set (*x*,<=*y*):
- point (*x*',<=*y*') is (*x*,<=*y*)'s right neighbor, if *x*'<=><=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s left neighbor, if *x*'<=<<=*x* and *y*'<==<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s lower neighbor, if *x*'<==<=*x* and *y*'<=<<=*y* - point (*x*',<=*y*') is (*x*,<=*y*)'s upper neighbor, if *x*'<==<=*x* and *y*'<=><=*y*
We'll consider point (*x*,<=*y*) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.
Input Specification:
The first input line contains the only integer *n* (1<=≤<=*n*<=≤<=200) — the number of points in the given set. Next *n* lines contain the coordinates of the points written as "*x* *y*" (without the quotes) (|*x*|,<=|*y*|<=≤<=1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.
Output Specification:
Print the only number — the number of supercentral points of the given set.
Demo Input:
['8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3\n', '5\n0 0\n0 1\n1 0\n0 -1\n-1 0\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | ```python
#45
n = int(input())
ins = [[int(x) for x in input().split()] for i in range(n)]
v = [set([x[i] for x in ins]) for i in range(2) ]
Maps = [{x : [] for x in v[i]} for i in range(2)]
superCentralPoints = 0
for MapIndex in range(2):
for elemnt in ins:
Maps[MapIndex][elemnt[MapIndex]].append(elemnt)
for point in ins:
hasAdjacents = [False for i in range(4)]
for MapIndex in range(2):
for otherPoint in Maps[MapIndex][point[MapIndex]]:
if point[1 - MapIndex] > otherPoint[1 - MapIndex]:
hasAdjacents[MapIndex * 2] = True
elif point[1 - MapIndex] < otherPoint[1 - MapIndex]:
hasAdjacents[MapIndex * 2 + 1] = True
if False not in hasAdjacents[MapIndex : MapIndex + 2]:
break
if False not in hasAdjacents:
superCentralPoints += 1
print(superCentralPoints)
``` | 0 |
|
263 | A | Beautiful Matrix | PROGRAMMING | 800 | [
"implementation"
] | null | null | You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful. | The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one. | Print a single integer — the minimum number of moves needed to make the matrix beautiful. | [
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] | [
"3\n",
"1\n"
] | none | 500 | [
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,697,992,864 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 25 | 154 | 0 | ans = 0
for i in range(5):
s = input().split()
for j in range(5):
if s[j] == '1':
ans = abs(i - 2) + abs(j - 2)
print(ans) | Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
ans = 0
for i in range(5):
s = input().split()
for j in range(5):
if s[j] == '1':
ans = abs(i - 2) + abs(j - 2)
print(ans)
``` | 3 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | As we all know Barney's job is "PLEASE" and he has not much to do at work. That's why he started playing "cups and key". In this game there are three identical cups arranged in a line from left to right. Initially key to Barney's heart is under the middle cup.
Then at one turn Barney swaps the cup in the middle with any of other two cups randomly (he choses each with equal probability), so the chosen cup becomes the middle one. Game lasts *n* turns and Barney independently choses a cup to swap with the middle one within each turn, and the key always remains in the cup it was at the start.
After *n*-th turn Barney asks a girl to guess which cup contains the key. The girl points to the middle one but Barney was distracted while making turns and doesn't know if the key is under the middle cup. That's why he asked you to tell him the probability that girl guessed right.
Number *n* of game turns can be extremely large, that's why Barney did not give it to you. Instead he gave you an array *a*1,<=*a*2,<=...,<=*a**k* such that
in other words, *n* is multiplication of all elements of the given array.
Because of precision difficulties, Barney asked you to tell him the answer as an irreducible fraction. In other words you need to find it as a fraction *p*<=/<=*q* such that , where is the greatest common divisor. Since *p* and *q* can be extremely large, you only need to find the remainders of dividing each of them by 109<=+<=7.
Please note that we want of *p* and *q* to be 1, not of their remainders after dividing by 109<=+<=7. | The first line of input contains a single integer *k* (1<=≤<=*k*<=≤<=105) — the number of elements in array Barney gave you.
The second line contains *k* integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=1018) — the elements of the array. | In the only line of output print a single string *x*<=/<=*y* where *x* is the remainder of dividing *p* by 109<=+<=7 and *y* is the remainder of dividing *q* by 109<=+<=7. | [
"1\n2\n",
"3\n1 1 1\n"
] | [
"1/2\n",
"0/1\n"
] | none | 0 | [
{
"input": "1\n2",
"output": "1/2"
},
{
"input": "3\n1 1 1",
"output": "0/1"
},
{
"input": "1\n983155795040951739",
"output": "145599903/436799710"
},
{
"input": "2\n467131402341701583 956277077729692725",
"output": "63467752/190403257"
},
{
"input": "10\n217673221404542171 806579927281665969 500754531396239406 214319484250163112 328494187336342674 427465830578952934 951554014286436941 664022909283791499 653206814724654845 66704816231807388",
"output": "896298678/688896019"
},
{
"input": "8\n137264686188377169 524477139880847337 939966121107073137 244138018261712937 158070587508987781 35608416591331673 378899027510195451 81986819972451999",
"output": "993002178/979006521"
},
{
"input": "9\n174496219779575399 193634487267697117 972518022554199573 695317701399937273 464007855398119159 881020180696239657 296973121744507377 544232692627163469 751214074246742731",
"output": "149736910/449210731"
},
{
"input": "12\n254904759290707697 475737283258450340 533306428548398547 442127134828578937 779740159015946254 272877594683860919 93000149670491971 349640818793278778 498293278222136720 551099726729989816 149940343283925029 989425634209891686",
"output": "674872752/24618241"
},
{
"input": "1\n1",
"output": "0/1"
},
{
"input": "1\n1000000000000000000",
"output": "453246046/359738130"
},
{
"input": "1\n3",
"output": "1/4"
},
{
"input": "1\n1000000006",
"output": "500000004/500000004"
},
{
"input": "2\n500000004 1000000006",
"output": "500000004/500000004"
},
{
"input": "1\n500000004",
"output": "666666672/1"
},
{
"input": "2\n500000004 500000004",
"output": "666666672/1"
},
{
"input": "1\n500000003",
"output": "833333339/500000004"
},
{
"input": "2\n500000003 500000004",
"output": "500000004/500000004"
},
{
"input": "2\n500000003 500000003",
"output": "833333339/500000004"
},
{
"input": "1\n1000000005",
"output": "750000005/250000002"
},
{
"input": "2\n1000000005 500000004",
"output": "416666670/250000002"
}
] | 1,468,862,361 | 6,561 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 1,228,800 | from fractions import Fraction
k=int(input(""))
L=[int(i) for i in input("").split()]
N=1
for i in L:
N*=i
N=float(N)
def P(N):
answer=float(1.00/3.00)*( (-1.00)**N*2.00**(1-N)+1.00)
return Fraction.from_float(answer).limit_denominator()
answer=P(N)
print(str(answer.numerator)+"/"+str(answer.denominator)) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As we all know Barney's job is "PLEASE" and he has not much to do at work. That's why he started playing "cups and key". In this game there are three identical cups arranged in a line from left to right. Initially key to Barney's heart is under the middle cup.
Then at one turn Barney swaps the cup in the middle with any of other two cups randomly (he choses each with equal probability), so the chosen cup becomes the middle one. Game lasts *n* turns and Barney independently choses a cup to swap with the middle one within each turn, and the key always remains in the cup it was at the start.
After *n*-th turn Barney asks a girl to guess which cup contains the key. The girl points to the middle one but Barney was distracted while making turns and doesn't know if the key is under the middle cup. That's why he asked you to tell him the probability that girl guessed right.
Number *n* of game turns can be extremely large, that's why Barney did not give it to you. Instead he gave you an array *a*1,<=*a*2,<=...,<=*a**k* such that
in other words, *n* is multiplication of all elements of the given array.
Because of precision difficulties, Barney asked you to tell him the answer as an irreducible fraction. In other words you need to find it as a fraction *p*<=/<=*q* such that , where is the greatest common divisor. Since *p* and *q* can be extremely large, you only need to find the remainders of dividing each of them by 109<=+<=7.
Please note that we want of *p* and *q* to be 1, not of their remainders after dividing by 109<=+<=7.
Input Specification:
The first line of input contains a single integer *k* (1<=≤<=*k*<=≤<=105) — the number of elements in array Barney gave you.
The second line contains *k* integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=1018) — the elements of the array.
Output Specification:
In the only line of output print a single string *x*<=/<=*y* where *x* is the remainder of dividing *p* by 109<=+<=7 and *y* is the remainder of dividing *q* by 109<=+<=7.
Demo Input:
['1\n2\n', '3\n1 1 1\n']
Demo Output:
['1/2\n', '0/1\n']
Note:
none | ```python
from fractions import Fraction
k=int(input(""))
L=[int(i) for i in input("").split()]
N=1
for i in L:
N*=i
N=float(N)
def P(N):
answer=float(1.00/3.00)*( (-1.00)**N*2.00**(1-N)+1.00)
return Fraction.from_float(answer).limit_denominator()
answer=P(N)
print(str(answer.numerator)+"/"+str(answer.denominator))
``` | 0 |
|
907 | A | Masha and Bears | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars. | You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3. | Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes). | [
"50 30 10 10\n",
"100 50 10 21\n"
] | [
"50\n30\n10\n",
"-1\n"
] | In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20. | 500 | [
{
"input": "50 30 10 10",
"output": "50\n30\n10"
},
{
"input": "100 50 10 21",
"output": "-1"
},
{
"input": "100 50 19 10",
"output": "100\n50\n19"
},
{
"input": "99 50 25 49",
"output": "100\n99\n49"
},
{
"input": "3 2 1 1",
"output": "4\n3\n1"
},
{
"input": "100 99 98 100",
"output": "-1"
},
{
"input": "100 40 30 40",
"output": "-1"
},
{
"input": "100 50 19 25",
"output": "100\n51\n25"
},
{
"input": "100 50 19 30",
"output": "100\n61\n30"
},
{
"input": "49 48 25 49",
"output": "-1"
},
{
"input": "48 47 23 46",
"output": "94\n93\n46"
},
{
"input": "37 23 16 20",
"output": "42\n41\n20"
},
{
"input": "98 2 1 1",
"output": "98\n3\n1"
},
{
"input": "99 22 13 14",
"output": "99\n29\n14"
},
{
"input": "97 95 3 2",
"output": "97\n95\n3"
},
{
"input": "27 3 2 3",
"output": "-1"
},
{
"input": "13 7 6 2",
"output": "-1"
},
{
"input": "19 17 11 6",
"output": "19\n17\n11"
},
{
"input": "15 6 4 5",
"output": "15\n11\n5"
},
{
"input": "21 3 1 3",
"output": "-1"
},
{
"input": "100 99 98 97",
"output": "196\n195\n98"
},
{
"input": "99 32 13 5",
"output": "-1"
},
{
"input": "5 4 3 1",
"output": "-1"
},
{
"input": "50 20 4 8",
"output": "50\n20\n8"
},
{
"input": "60 30 10 1",
"output": "-1"
},
{
"input": "100 50 10 1",
"output": "-1"
},
{
"input": "40 30 20 10",
"output": "40\n30\n20"
},
{
"input": "50 49 5 10",
"output": "50\n49\n10"
},
{
"input": "99 98 50 97",
"output": "196\n195\n97"
},
{
"input": "50 30 10 5",
"output": "50\n30\n10"
},
{
"input": "4 3 2 1",
"output": "4\n3\n2"
},
{
"input": "100 50 3 1",
"output": "-1"
}
] | 1,595,276,402 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 93 | 6,656,000 | list1 = input().split()
if list1[3] <= list1[2] < list1[1] < list1[0]:
for i in range(len(list1[:-1])):
print(list1[i])
else:
print(-1)
| Title: Masha and Bears
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input Specification:
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
Output Specification:
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Demo Input:
['50 30 10 10\n', '100 50 10 21\n']
Demo Output:
['50\n30\n10\n', '-1\n']
Note:
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20. | ```python
list1 = input().split()
if list1[3] <= list1[2] < list1[1] < list1[0]:
for i in range(len(list1[:-1])):
print(list1[i])
else:
print(-1)
``` | 0 |
|
780 | A | Andryusha and Socks | PROGRAMMING | 800 | [
"implementation"
] | null | null | Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time? | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair. | Print single integer — the maximum number of socks that were on the table at the same time. | [
"1\n1 1\n",
"3\n2 1 1 3 2 3\n"
] | [
"1\n",
"2\n"
] | In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe. | 500 | [
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n2 1 1 3 2 3",
"output": "2"
},
{
"input": "5\n5 1 3 2 4 3 1 2 4 5",
"output": "5"
},
{
"input": "10\n4 2 6 3 4 8 7 1 1 5 2 10 6 8 3 5 10 9 9 7",
"output": "6"
},
{
"input": "50\n30 47 31 38 37 50 36 43 9 23 2 2 15 31 14 49 9 16 6 44 27 14 5 6 3 47 25 26 1 35 3 15 24 19 8 46 49 41 4 26 40 28 42 11 34 35 46 18 7 28 18 40 19 42 4 41 38 48 50 12 29 39 33 17 25 22 22 21 36 45 27 30 20 7 13 29 39 44 21 8 37 45 34 1 20 10 11 17 33 12 43 13 10 16 48 24 32 5 23 32",
"output": "25"
},
{
"input": "50\n1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 46 46 47 47 48 48 49 49 50 50",
"output": "1"
},
{
"input": "50\n50 50 49 49 48 48 47 47 46 46 45 45 44 44 43 43 42 42 41 41 40 40 39 39 38 38 37 37 36 36 35 35 34 34 33 33 32 32 31 31 30 30 29 29 28 28 27 27 26 26 25 25 24 24 23 23 22 22 21 21 20 20 19 19 18 18 17 17 16 16 15 15 14 14 13 13 12 12 11 11 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1",
"output": "1"
},
{
"input": "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 1 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",
"output": "50"
},
{
"input": "50\n50 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 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": "50"
},
{
"input": "10\n2 9 4 1 6 7 10 3 1 5 8 6 2 3 10 7 4 8 5 9",
"output": "9"
}
] | 1,610,520,566 | 2,147,483,647 | Python 3 | OK | TESTS | 56 | 265 | 13,824,000 | n=int(input())
l=list(map(int,input().split()))
d={}
c=0
m=0
for i in range(2*n):
if(l[i] not in d):
c+=1
d[l[i]]=1
else:
m=max(m,c)
c-=1
d.pop(l[i])
print(m)
| Title: Andryusha and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair.
Output Specification:
Print single integer — the maximum number of socks that were on the table at the same time.
Demo Input:
['1\n1 1\n', '3\n2 1 1 3 2 3\n']
Demo Output:
['1\n', '2\n']
Note:
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe. | ```python
n=int(input())
l=list(map(int,input().split()))
d={}
c=0
m=0
for i in range(2*n):
if(l[i] not in d):
c+=1
d[l[i]]=1
else:
m=max(m,c)
c-=1
d.pop(l[i])
print(m)
``` | 3 |
|
670 | B | Game of Robots | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | In late autumn evening *n* robots gathered in the cheerful company of friends. Each robot has a unique identifier — an integer from 1 to 109.
At some moment, robots decided to play the game "Snowball". Below there are the rules of this game. First, all robots stand in a row. Then the first robot says his identifier. After that the second robot says the identifier of the first robot and then says his own identifier. Then the third robot says the identifier of the first robot, then says the identifier of the second robot and after that says his own. This process continues from left to right until the *n*-th robot says his identifier.
Your task is to determine the *k*-th identifier to be pronounced. | The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*k*<=≤<=*min*(2·109,<=*n*·(*n*<=+<=1)<=/<=2).
The second line contains the sequence *id*1,<=*id*2,<=...,<=*id**n* (1<=≤<=*id**i*<=≤<=109) — identifiers of roborts. It is guaranteed that all identifiers are different. | Print the *k*-th pronounced identifier (assume that the numeration starts from 1). | [
"2 2\n1 2\n",
"4 5\n10 4 18 3\n"
] | [
"1\n",
"4\n"
] | In the first sample identifiers of robots will be pronounced in the following order: 1, 1, 2. As *k* = 2, the answer equals to 1.
In the second test case identifiers of robots will be pronounced in the following order: 10, 10, 4, 10, 4, 18, 10, 4, 18, 3. As *k* = 5, the answer equals to 4. | 750 | [
{
"input": "2 2\n1 2",
"output": "1"
},
{
"input": "4 5\n10 4 18 3",
"output": "4"
},
{
"input": "1 1\n4",
"output": "4"
},
{
"input": "2 1\n5 1",
"output": "5"
},
{
"input": "2 2\n1 4",
"output": "1"
},
{
"input": "2 3\n6 7",
"output": "7"
},
{
"input": "3 1\n4 5 6",
"output": "4"
},
{
"input": "3 2\n4 5 6",
"output": "4"
},
{
"input": "3 3\n4 5 6",
"output": "5"
},
{
"input": "3 4\n4 5 6",
"output": "4"
},
{
"input": "3 5\n4 5 6",
"output": "5"
},
{
"input": "3 6\n4 5 6",
"output": "6"
},
{
"input": "4 1\n5 1000000000 999999999 12",
"output": "5"
},
{
"input": "4 2\n5 1000000000 999999999 12",
"output": "5"
},
{
"input": "4 3\n5 1000000000 999999999 12",
"output": "1000000000"
},
{
"input": "4 4\n5 1000000000 999999999 12",
"output": "5"
},
{
"input": "4 5\n5 1000000000 999999999 12",
"output": "1000000000"
},
{
"input": "4 6\n5 1000000000 999999999 12",
"output": "999999999"
},
{
"input": "4 7\n5 1000000000 999999999 12",
"output": "5"
},
{
"input": "4 8\n5 1000000000 999999999 12",
"output": "1000000000"
},
{
"input": "4 9\n5 1000000000 999999999 12",
"output": "999999999"
},
{
"input": "4 10\n5 1000000000 999999999 12",
"output": "12"
}
] | 1,624,363,765 | 2,147,483,647 | Python 3 | OK | TESTS | 143 | 108 | 7,680,000 | n, k = map(int, input().split())
robos = list(input().split())
for i in range(n):
if k <= i + 1:
print(robos[k - 1])
break
else:
k -= i + 1
| Title: Game of Robots
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In late autumn evening *n* robots gathered in the cheerful company of friends. Each robot has a unique identifier — an integer from 1 to 109.
At some moment, robots decided to play the game "Snowball". Below there are the rules of this game. First, all robots stand in a row. Then the first robot says his identifier. After that the second robot says the identifier of the first robot and then says his own identifier. Then the third robot says the identifier of the first robot, then says the identifier of the second robot and after that says his own. This process continues from left to right until the *n*-th robot says his identifier.
Your task is to determine the *k*-th identifier to be pronounced.
Input Specification:
The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*k*<=≤<=*min*(2·109,<=*n*·(*n*<=+<=1)<=/<=2).
The second line contains the sequence *id*1,<=*id*2,<=...,<=*id**n* (1<=≤<=*id**i*<=≤<=109) — identifiers of roborts. It is guaranteed that all identifiers are different.
Output Specification:
Print the *k*-th pronounced identifier (assume that the numeration starts from 1).
Demo Input:
['2 2\n1 2\n', '4 5\n10 4 18 3\n']
Demo Output:
['1\n', '4\n']
Note:
In the first sample identifiers of robots will be pronounced in the following order: 1, 1, 2. As *k* = 2, the answer equals to 1.
In the second test case identifiers of robots will be pronounced in the following order: 10, 10, 4, 10, 4, 18, 10, 4, 18, 3. As *k* = 5, the answer equals to 4. | ```python
n, k = map(int, input().split())
robos = list(input().split())
for i in range(n):
if k <= i + 1:
print(robos[k - 1])
break
else:
k -= i + 1
``` | 3 |
|
483 | A | Counterexample | PROGRAMMING | 1,100 | [
"brute force",
"implementation",
"math",
"number theory"
] | null | null | Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one.
Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime.
You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*.
More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime. | The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50). | Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.
If the counterexample does not exist, print the single number -1. | [
"2 4\n",
"10 11\n",
"900000000000000009 900000000000000029\n"
] | [
"2 3 4\n",
"-1\n",
"900000000000000009 900000000000000010 900000000000000021\n"
] | In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.
In the second sample you cannot form a group of three distinct integers, so the answer is -1.
In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three. | 500 | [
{
"input": "2 4",
"output": "2 3 4"
},
{
"input": "10 11",
"output": "-1"
},
{
"input": "900000000000000009 900000000000000029",
"output": "900000000000000009 900000000000000010 900000000000000021"
},
{
"input": "640097987171091791 640097987171091835",
"output": "640097987171091792 640097987171091793 640097987171091794"
},
{
"input": "19534350415104721 19534350415104725",
"output": "19534350415104722 19534350415104723 19534350415104724"
},
{
"input": "933700505788726243 933700505788726280",
"output": "933700505788726244 933700505788726245 933700505788726246"
},
{
"input": "1 3",
"output": "-1"
},
{
"input": "1 4",
"output": "2 3 4"
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "266540997167959130 266540997167959164",
"output": "266540997167959130 266540997167959131 266540997167959132"
},
{
"input": "267367244641009850 267367244641009899",
"output": "267367244641009850 267367244641009851 267367244641009852"
},
{
"input": "268193483524125978 268193483524125993",
"output": "268193483524125978 268193483524125979 268193483524125980"
},
{
"input": "269019726702209402 269019726702209432",
"output": "269019726702209402 269019726702209403 269019726702209404"
},
{
"input": "269845965585325530 269845965585325576",
"output": "269845965585325530 269845965585325531 269845965585325532"
},
{
"input": "270672213058376250 270672213058376260",
"output": "270672213058376250 270672213058376251 270672213058376252"
},
{
"input": "271498451941492378 271498451941492378",
"output": "-1"
},
{
"input": "272324690824608506 272324690824608523",
"output": "272324690824608506 272324690824608507 272324690824608508"
},
{
"input": "273150934002691930 273150934002691962",
"output": "273150934002691930 273150934002691931 273150934002691932"
},
{
"input": "996517375802030516 996517375802030524",
"output": "996517375802030516 996517375802030517 996517375802030518"
},
{
"input": "997343614685146644 997343614685146694",
"output": "997343614685146644 997343614685146645 997343614685146646"
},
{
"input": "998169857863230068 998169857863230083",
"output": "998169857863230068 998169857863230069 998169857863230070"
},
{
"input": "998996101041313492 998996101041313522",
"output": "998996101041313492 998996101041313493 998996101041313494"
},
{
"input": "999822344219396916 999822344219396961",
"output": "999822344219396916 999822344219396917 999822344219396918"
},
{
"input": "648583102513043 648583102513053",
"output": "648583102513044 648583102513045 648583102513046"
},
{
"input": "266540997167959130 266540997167959131",
"output": "-1"
},
{
"input": "267367244641009850 267367244641009850",
"output": "-1"
},
{
"input": "268193483524125978 268193483524125979",
"output": "-1"
},
{
"input": "269019726702209402 269019726702209402",
"output": "-1"
},
{
"input": "269845965585325530 269845965585325530",
"output": "-1"
},
{
"input": "270672213058376250 270672213058376254",
"output": "270672213058376250 270672213058376251 270672213058376252"
},
{
"input": "271498451941492378 271498451941492379",
"output": "-1"
},
{
"input": "272324690824608506 272324690824608508",
"output": "272324690824608506 272324690824608507 272324690824608508"
},
{
"input": "273150934002691930 273150934002691931",
"output": "-1"
},
{
"input": "996517375802030516 996517375802030518",
"output": "996517375802030516 996517375802030517 996517375802030518"
},
{
"input": "997343614685146644 997343614685146644",
"output": "-1"
},
{
"input": "2147483647 2147483649",
"output": "-1"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "1 7",
"output": "2 3 4"
},
{
"input": "9 12",
"output": "9 11 12"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "11 13",
"output": "-1"
},
{
"input": "2 2",
"output": "-1"
}
] | 1,470,227,197 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 62 | 0 | import sys
l, r = map(int, sys.stdin.readline().split())
if l % 2 is not 0:
l += 1
if r - l < 2:
print(-1)
else:
print(l, l+1, l+2)
| Title: Counterexample
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one.
Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime.
You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*.
More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime.
Input Specification:
The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50).
Output Specification:
Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.
If the counterexample does not exist, print the single number -1.
Demo Input:
['2 4\n', '10 11\n', '900000000000000009 900000000000000029\n']
Demo Output:
['2 3 4\n', '-1\n', '900000000000000009 900000000000000010 900000000000000021\n']
Note:
In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.
In the second sample you cannot form a group of three distinct integers, so the answer is -1.
In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three. | ```python
import sys
l, r = map(int, sys.stdin.readline().split())
if l % 2 is not 0:
l += 1
if r - l < 2:
print(-1)
else:
print(l, l+1, l+2)
``` | 3 |
|
799 | A | Carrot Cakes | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | In some game by Playrix it takes *t* minutes for an oven to bake *k* carrot cakes, all cakes are ready at the same moment *t* minutes after they started baking. Arkady needs at least *n* cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take *d* minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven.
Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get *n* cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable. | The only line contains four integers *n*, *t*, *k*, *d* (1<=≤<=*n*,<=*t*,<=*k*,<=*d*<=≤<=1<=000) — the number of cakes needed, the time needed for one oven to bake *k* cakes, the number of cakes baked at the same time, the time needed to build the second oven. | If it is reasonable to build the second oven, print "YES". Otherwise print "NO". | [
"8 6 4 5\n",
"8 6 4 6\n",
"10 3 11 4\n",
"4 2 1 4\n"
] | [
"YES\n",
"NO\n",
"NO\n",
"YES\n"
] | In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6 minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven.
In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven.
In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven. | 500 | [
{
"input": "8 6 4 5",
"output": "YES"
},
{
"input": "8 6 4 6",
"output": "NO"
},
{
"input": "10 3 11 4",
"output": "NO"
},
{
"input": "4 2 1 4",
"output": "YES"
},
{
"input": "28 17 16 26",
"output": "NO"
},
{
"input": "60 69 9 438",
"output": "NO"
},
{
"input": "599 97 54 992",
"output": "YES"
},
{
"input": "11 22 18 17",
"output": "NO"
},
{
"input": "1 13 22 11",
"output": "NO"
},
{
"input": "1 1 1 1",
"output": "NO"
},
{
"input": "3 1 1 1",
"output": "YES"
},
{
"input": "1000 1000 1000 1000",
"output": "NO"
},
{
"input": "1000 1000 1 1",
"output": "YES"
},
{
"input": "1000 1000 1 400",
"output": "YES"
},
{
"input": "1000 1000 1 1000",
"output": "YES"
},
{
"input": "1000 1000 1 999",
"output": "YES"
},
{
"input": "53 11 3 166",
"output": "YES"
},
{
"input": "313 2 3 385",
"output": "NO"
},
{
"input": "214 9 9 412",
"output": "NO"
},
{
"input": "349 9 5 268",
"output": "YES"
},
{
"input": "611 16 8 153",
"output": "YES"
},
{
"input": "877 13 3 191",
"output": "YES"
},
{
"input": "340 9 9 10",
"output": "YES"
},
{
"input": "31 8 2 205",
"output": "NO"
},
{
"input": "519 3 2 148",
"output": "YES"
},
{
"input": "882 2 21 219",
"output": "NO"
},
{
"input": "982 13 5 198",
"output": "YES"
},
{
"input": "428 13 6 272",
"output": "YES"
},
{
"input": "436 16 14 26",
"output": "YES"
},
{
"input": "628 10 9 386",
"output": "YES"
},
{
"input": "77 33 18 31",
"output": "YES"
},
{
"input": "527 36 4 8",
"output": "YES"
},
{
"input": "128 18 2 169",
"output": "YES"
},
{
"input": "904 4 2 288",
"output": "YES"
},
{
"input": "986 4 3 25",
"output": "YES"
},
{
"input": "134 8 22 162",
"output": "NO"
},
{
"input": "942 42 3 69",
"output": "YES"
},
{
"input": "894 4 9 4",
"output": "YES"
},
{
"input": "953 8 10 312",
"output": "YES"
},
{
"input": "43 8 1 121",
"output": "YES"
},
{
"input": "12 13 19 273",
"output": "NO"
},
{
"input": "204 45 10 871",
"output": "YES"
},
{
"input": "342 69 50 425",
"output": "NO"
},
{
"input": "982 93 99 875",
"output": "NO"
},
{
"input": "283 21 39 132",
"output": "YES"
},
{
"input": "1000 45 83 686",
"output": "NO"
},
{
"input": "246 69 36 432",
"output": "NO"
},
{
"input": "607 93 76 689",
"output": "NO"
},
{
"input": "503 21 24 435",
"output": "NO"
},
{
"input": "1000 45 65 989",
"output": "NO"
},
{
"input": "30 21 2 250",
"output": "YES"
},
{
"input": "1000 49 50 995",
"output": "NO"
},
{
"input": "383 69 95 253",
"output": "YES"
},
{
"input": "393 98 35 999",
"output": "YES"
},
{
"input": "1000 22 79 552",
"output": "NO"
},
{
"input": "268 294 268 154",
"output": "NO"
},
{
"input": "963 465 706 146",
"output": "YES"
},
{
"input": "304 635 304 257",
"output": "NO"
},
{
"input": "4 2 1 6",
"output": "NO"
},
{
"input": "1 51 10 50",
"output": "NO"
},
{
"input": "5 5 4 4",
"output": "YES"
},
{
"input": "3 2 1 1",
"output": "YES"
},
{
"input": "3 4 3 3",
"output": "NO"
},
{
"input": "7 3 4 1",
"output": "YES"
},
{
"input": "101 10 1 1000",
"output": "NO"
},
{
"input": "5 1 1 1",
"output": "YES"
},
{
"input": "5 10 5 5",
"output": "NO"
},
{
"input": "19 1 7 1",
"output": "YES"
},
{
"input": "763 572 745 262",
"output": "YES"
},
{
"input": "1 2 1 1",
"output": "NO"
},
{
"input": "5 1 1 3",
"output": "YES"
},
{
"input": "170 725 479 359",
"output": "NO"
},
{
"input": "6 2 1 7",
"output": "YES"
},
{
"input": "6 2 5 1",
"output": "YES"
},
{
"input": "1 2 2 1",
"output": "NO"
},
{
"input": "24 2 8 3",
"output": "YES"
},
{
"input": "7 3 3 3",
"output": "YES"
},
{
"input": "5 2 2 2",
"output": "YES"
},
{
"input": "3 2 1 2",
"output": "YES"
},
{
"input": "1000 2 200 8",
"output": "NO"
},
{
"input": "3 100 2 100",
"output": "NO"
},
{
"input": "2 999 1 1000",
"output": "NO"
},
{
"input": "2 1 1 1",
"output": "NO"
},
{
"input": "2 3 5 1",
"output": "NO"
},
{
"input": "100 1 5 1",
"output": "YES"
},
{
"input": "7 2 3 3",
"output": "YES"
},
{
"input": "4 1 1 3",
"output": "NO"
},
{
"input": "3 2 2 1",
"output": "YES"
},
{
"input": "1 1 1 2",
"output": "NO"
},
{
"input": "91 8 7 13",
"output": "YES"
},
{
"input": "3 1 2 1",
"output": "NO"
},
{
"input": "5 3 2 3",
"output": "YES"
},
{
"input": "9 6 6 3",
"output": "YES"
}
] | 1,678,614,126 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | def solve():
(n, t, k, d) = map(int ,input().split())
T1 = t/k*n
T2 = d + t/k*n/2
return (T1 > T2)
t = 1
for _ in range(t):
print("YES" if solve() else "NO") | Title: Carrot Cakes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In some game by Playrix it takes *t* minutes for an oven to bake *k* carrot cakes, all cakes are ready at the same moment *t* minutes after they started baking. Arkady needs at least *n* cakes to complete a task, but he currently don't have any. However, he has infinitely many ingredients and one oven. Moreover, Arkady can build one more similar oven to make the process faster, it would take *d* minutes to build the oven. While the new oven is being built, only old one can bake cakes, after the new oven is built, both ovens bake simultaneously. Arkady can't build more than one oven.
Determine if it is reasonable to build the second oven, i.e. will it decrease the minimum time needed to get *n* cakes or not. If the time needed with the second oven is the same as with one oven, then it is unreasonable.
Input Specification:
The only line contains four integers *n*, *t*, *k*, *d* (1<=≤<=*n*,<=*t*,<=*k*,<=*d*<=≤<=1<=000) — the number of cakes needed, the time needed for one oven to bake *k* cakes, the number of cakes baked at the same time, the time needed to build the second oven.
Output Specification:
If it is reasonable to build the second oven, print "YES". Otherwise print "NO".
Demo Input:
['8 6 4 5\n', '8 6 4 6\n', '10 3 11 4\n', '4 2 1 4\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n', 'YES\n']
Note:
In the first example it is possible to get 8 cakes in 12 minutes using one oven. The second oven can be built in 5 minutes, so after 6 minutes the first oven bakes 4 cakes, the second oven bakes 4 more ovens after 11 minutes. Thus, it is reasonable to build the second oven.
In the second example it doesn't matter whether we build the second oven or not, thus it takes 12 minutes to bake 8 cakes in both cases. Thus, it is unreasonable to build the second oven.
In the third example the first oven bakes 11 cakes in 3 minutes, that is more than needed 10. It is unreasonable to build the second oven, because its building takes more time that baking the needed number of cakes using the only oven. | ```python
def solve():
(n, t, k, d) = map(int ,input().split())
T1 = t/k*n
T2 = d + t/k*n/2
return (T1 > T2)
t = 1
for _ in range(t):
print("YES" if solve() else "NO")
``` | 0 |
|
771 | B | Bear and Different Names | PROGRAMMING | 1,500 | [
"constructive algorithms",
"greedy"
] | null | null | In the army, it isn't easy to form a group of soldiers that will be effective on the battlefield. The communication is crucial and thus no two soldiers should share a name (what would happen if they got an order that Bob is a scouter, if there are two Bobs?).
A group of soldiers is effective if and only if their names are different. For example, a group (John, Bob, Limak) would be effective, while groups (Gary, Bob, Gary) and (Alice, Alice) wouldn't.
You are a spy in the enemy's camp. You noticed *n* soldiers standing in a row, numbered 1 through *n*. The general wants to choose a group of *k* consecutive soldiers. For every *k* consecutive soldiers, the general wrote down whether they would be an effective group or not.
You managed to steal the general's notes, with *n*<=-<=*k*<=+<=1 strings *s*1,<=*s*2,<=...,<=*s**n*<=-<=*k*<=+<=1, each either "YES" or "NO".
- The string *s*1 describes a group of soldiers 1 through *k* ("YES" if the group is effective, and "NO" otherwise). - The string *s*2 describes a group of soldiers 2 through *k*<=+<=1. - And so on, till the string *s**n*<=-<=*k*<=+<=1 that describes a group of soldiers *n*<=-<=*k*<=+<=1 through *n*.
Your task is to find possible names of *n* soldiers. Names should match the stolen notes. Each name should be a string that consists of between 1 and 10 English letters, inclusive. The first letter should be uppercase, and all other letters should be lowercase. Names don't have to be existing names — it's allowed to print "Xyzzzdj" or "T" for example.
Find and print any solution. It can be proved that there always exists at least one solution. | The first line of the input contains two integers *n* and *k* (2<=≤<=*k*<=≤<=*n*<=≤<=50) — the number of soldiers and the size of a group respectively.
The second line contains *n*<=-<=*k*<=+<=1 strings *s*1,<=*s*2,<=...,<=*s**n*<=-<=*k*<=+<=1. The string *s**i* is "YES" if the group of soldiers *i* through *i*<=+<=*k*<=-<=1 is effective, and "NO" otherwise. | Find any solution satisfying all given conditions. In one line print *n* space-separated strings, denoting possible names of soldiers in the order. The first letter of each name should be uppercase, while the other letters should be lowercase. Each name should contain English letters only and has length from 1 to 10.
If there are multiple valid solutions, print any of them. | [
"8 3\nNO NO YES YES YES NO\n",
"9 8\nYES NO\n",
"3 2\nNO NO\n"
] | [
"Adam Bob Bob Cpqepqwer Limak Adam Bob Adam",
"R Q Ccccccccc Ccocc Ccc So Strong Samples Ccc",
"Na Na Na"
] | In the first sample, there are 8 soldiers. For every 3 consecutive ones we know whether they would be an effective group. Let's analyze the provided sample output:
- First three soldiers (i.e. Adam, Bob, Bob) wouldn't be an effective group because there are two Bobs. Indeed, the string *s*<sub class="lower-index">1</sub> is "NO". - Soldiers 2 through 4 (Bob, Bob, Cpqepqwer) wouldn't be effective either, and the string *s*<sub class="lower-index">2</sub> is "NO". - Soldiers 3 through 5 (Bob, Cpqepqwer, Limak) would be effective, and the string *s*<sub class="lower-index">3</sub> is "YES". - ..., - Soldiers 6 through 8 (Adam, Bob, Adam) wouldn't be effective, and the string *s*<sub class="lower-index">6</sub> is "NO". | 500 | [
{
"input": "8 3\nNO NO YES YES YES NO",
"output": "Ab Ac Ab Ac Af Ag Ah Ag "
},
{
"input": "9 8\nYES NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Ac "
},
{
"input": "3 2\nNO NO",
"output": "Ab Ab Ab "
},
{
"input": "2 2\nYES",
"output": "Ab Ac "
},
{
"input": "2 2\nNO",
"output": "Ab Ab "
},
{
"input": "7 2\nYES NO YES YES NO YES",
"output": "Ab Ac Ac Ae Af Af Ah "
},
{
"input": "18 7\nYES YES YES YES YES YES YES NO NO NO NO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ai Aj Ak Al Am "
},
{
"input": "50 3\nNO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO YES NO",
"output": "Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Ab Ac Bx Ac "
},
{
"input": "19 15\nNO YES YES YES NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ab Aq Ar As Af "
},
{
"input": "3 2\nNO NO",
"output": "Ab Ab Ab "
},
{
"input": "3 2\nNO YES",
"output": "Ab Ab Ad "
},
{
"input": "3 2\nYES NO",
"output": "Ab Ac Ac "
},
{
"input": "3 2\nYES YES",
"output": "Ab Ac Ad "
},
{
"input": "26 17\nNO YES YES YES NO YES NO YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ab As At Au Af Aw Ah Ay Az Ba "
},
{
"input": "12 2\nYES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am "
},
{
"input": "16 2\nNO NO NO NO NO NO NO NO NO NO NO NO NO NO NO",
"output": "Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab Ab "
},
{
"input": "42 20\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq "
},
{
"input": "37 14\nNO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al "
},
{
"input": "29 10\nYES NO YES NO YES NO YES YES YES YES YES NO NO NO NO NO YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Ac Am Ae Ao Ag Aq Ar As At Au Am Ae Ao Ag Aq Ba Bb Bc Bd "
},
{
"input": "37 3\nYES NO YES NO YES NO YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES NO NO YES NO NO YES YES YES YES NO",
"output": "Ab Ac Ad Ac Af Ac Ah Ac Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Ba Bb Be Bb Be Bh Bi Bj Bk Bj "
},
{
"input": "44 11\nNO NO YES NO YES NO YES YES YES YES YES YES YES YES YES YES YES YES YES NO YES YES YES YES YES NO NO YES NO NO YES YES YES NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Ab Ac An Ae Ap Ag Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Au Bf Bg Bh Bi Bj Ba Bb Bm Bd Au Bp Bq Br Bi "
},
{
"input": "50 49\nNO YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Ab By "
},
{
"input": "50 49\nYES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "50 49\nNO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Ab Ac "
},
{
"input": "50 49\nYES NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx Ac "
},
{
"input": "46 42\nNO YES YES YES NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Ab Br Bs Bt Af "
},
{
"input": "45 26\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt "
},
{
"input": "45 26\nNO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au "
},
{
"input": "50 3\nNO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES",
"output": "Ab Ac Ab Ae Ab Ag Ab Ai Ab Ak Ab Am Ab Ao Ab Aq Ab As Ab Au Ab Aw Ab Ay Ab Ba Ab Bc Ab Be Ab Bg Ab Bi Ab Bk Ab Bm Ab Bo Ab Bq Ab Bs Ab Bu Ab Bw Ab By "
},
{
"input": "50 2\nNO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO",
"output": "Ab Ab Ad Ad Af Af Ah Ah Aj Aj Al Al An An Ap Ap Ar Ar At At Av Av Ax Ax Az Az Bb Bb Bd Bd Bf Bf Bh Bh Bj Bj Bl Bl Bn Bn Bp Bp Br Br Bt Bt Bv Bv Bx Bx "
},
{
"input": "50 3\nNO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES YES YES YES YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES",
"output": "Ab Ac Ab Ae Ab Ag Ab Ai Ab Ak Ab Am Ab Ao Ab Aq Ab As Ab Au Ab Aw Ab Ay Ab Ba Ab Bc Bd Be Bf Bg Bf Bi Bf Bk Bf Bm Bf Bo Bf Bq Bf Bs Bf Bu Bf Bw Bf By "
},
{
"input": "49 2\nNO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO NO NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES",
"output": "Ab Ab Ad Ad Af Af Ah Ah Aj Aj Al Al An An Ap Ap Ar Ar At At Av Av Ax Ax Ax Ax Bb Bb Bd Bd Bf Bf Bh Bh Bj Bj Bl Bl Bn Bn Bp Bp Br Br Bt Bt Bv Bv Bx "
},
{
"input": "35 22\nNO NO NO NO NO NO NO NO NO NO NO NO NO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao "
},
{
"input": "46 41\nYES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu "
},
{
"input": "12 4\nYES YES NO NO NO NO NO YES YES",
"output": "Ab Ac Ad Ae Af Ad Ae Af Ad Ae Al Am "
},
{
"input": "50 2\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "50 4\nYES YES YES YES YES NO YES YES YES YES NO NO YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES NO YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Ag Ak Al Am An Al Am Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bc Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "34 5\nYES YES YES YES YES NO YES YES YES YES NO NO YES YES YES NO NO YES NO YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ag Al Am An Ao Al Am Ar As At Am Ar Aw At Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi "
},
{
"input": "50 43\nYES NO YES NO YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Ac Bt Ae Bv Bw Bx By "
},
{
"input": "38 30\nNO NO YES NO YES NO NO NO NO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Ab Ac Bg Ae Bi Ag Ah Ai Aj "
},
{
"input": "50 50\nNO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx Ab "
},
{
"input": "50 50\nYES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "5 3\nYES NO YES",
"output": "Ab Ac Ad Ac Af "
},
{
"input": "30 2\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be "
},
{
"input": "50 50\nYES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "27 27\nYES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb "
},
{
"input": "28 2\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc "
},
{
"input": "50 2\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "8 3\nYES NO YES NO YES NO",
"output": "Ab Ac Ad Ac Af Ac Ah Ac "
},
{
"input": "42 30\nNO YES YES NO NO YES NO YES NO YES NO NO YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Ab Bf Bg Ae Af Bj Ah Bl Aj Bn Al Am Bq "
},
{
"input": "50 49\nYES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "50 3\nYES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx By "
},
{
"input": "7 5\nYES NO YES",
"output": "Ab Ac Ad Ae Af Ac Ah "
},
{
"input": "8 4\nNO YES NO YES NO",
"output": "Ab Ac Ad Ab Af Ad Ah Af "
},
{
"input": "50 50\nNO",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Bx Ab "
},
{
"input": "50 48\nYES NO YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ap Aq Ar As At Au Av Aw Ax Ay Az Ba Bb Bc Bd Be Bf Bg Bh Bi Bj Bk Bl Bm Bn Bo Bp Bq Br Bs Bt Bu Bv Bw Ac By "
},
{
"input": "29 14\nYES NO YES NO NO YES YES NO NO YES YES NO NO YES YES YES",
"output": "Ab Ac Ad Ae Af Ag Ah Ai Aj Ak Al Am An Ao Ac Aq Ae Af At Au Ai Aj Ax Ay Am An Bb Bc Bd "
},
{
"input": "10 3\nNO YES NO YES NO YES NO YES",
"output": "Ab Ac Ab Ae Ab Ag Ab Ai Ab Ak "
},
{
"input": "10 5\nYES NO YES NO YES NO",
"output": "Ab Ac Ad Ae Af Ac Ah Ae Aj Ac "
}
] | 1,590,383,023 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 108 | 307,200 | n, k = map(int, input().split())
s = input().split()
words = []
for i in range(26):
words.append(chr(65 + i))
for i in range(26):
words.append(chr(65 + i) + chr(i + 97))
m = n - k + 1
ans = [0] * n
for i in range(m):
if s[i] == "YES":
for j in range(i, i + k):
if ans[j] == 0:
for word in words:
if word not in ans[i:i+k]:
ans[j] = word
break
last = "A"
for i in range(n):
if ans[i] == 0:
ans[i] = last
last = ans[i]
print(*ans) | Title: Bear and Different Names
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the army, it isn't easy to form a group of soldiers that will be effective on the battlefield. The communication is crucial and thus no two soldiers should share a name (what would happen if they got an order that Bob is a scouter, if there are two Bobs?).
A group of soldiers is effective if and only if their names are different. For example, a group (John, Bob, Limak) would be effective, while groups (Gary, Bob, Gary) and (Alice, Alice) wouldn't.
You are a spy in the enemy's camp. You noticed *n* soldiers standing in a row, numbered 1 through *n*. The general wants to choose a group of *k* consecutive soldiers. For every *k* consecutive soldiers, the general wrote down whether they would be an effective group or not.
You managed to steal the general's notes, with *n*<=-<=*k*<=+<=1 strings *s*1,<=*s*2,<=...,<=*s**n*<=-<=*k*<=+<=1, each either "YES" or "NO".
- The string *s*1 describes a group of soldiers 1 through *k* ("YES" if the group is effective, and "NO" otherwise). - The string *s*2 describes a group of soldiers 2 through *k*<=+<=1. - And so on, till the string *s**n*<=-<=*k*<=+<=1 that describes a group of soldiers *n*<=-<=*k*<=+<=1 through *n*.
Your task is to find possible names of *n* soldiers. Names should match the stolen notes. Each name should be a string that consists of between 1 and 10 English letters, inclusive. The first letter should be uppercase, and all other letters should be lowercase. Names don't have to be existing names — it's allowed to print "Xyzzzdj" or "T" for example.
Find and print any solution. It can be proved that there always exists at least one solution.
Input Specification:
The first line of the input contains two integers *n* and *k* (2<=≤<=*k*<=≤<=*n*<=≤<=50) — the number of soldiers and the size of a group respectively.
The second line contains *n*<=-<=*k*<=+<=1 strings *s*1,<=*s*2,<=...,<=*s**n*<=-<=*k*<=+<=1. The string *s**i* is "YES" if the group of soldiers *i* through *i*<=+<=*k*<=-<=1 is effective, and "NO" otherwise.
Output Specification:
Find any solution satisfying all given conditions. In one line print *n* space-separated strings, denoting possible names of soldiers in the order. The first letter of each name should be uppercase, while the other letters should be lowercase. Each name should contain English letters only and has length from 1 to 10.
If there are multiple valid solutions, print any of them.
Demo Input:
['8 3\nNO NO YES YES YES NO\n', '9 8\nYES NO\n', '3 2\nNO NO\n']
Demo Output:
['Adam Bob Bob Cpqepqwer Limak Adam Bob Adam', 'R Q Ccccccccc Ccocc Ccc So Strong Samples Ccc', 'Na Na Na']
Note:
In the first sample, there are 8 soldiers. For every 3 consecutive ones we know whether they would be an effective group. Let's analyze the provided sample output:
- First three soldiers (i.e. Adam, Bob, Bob) wouldn't be an effective group because there are two Bobs. Indeed, the string *s*<sub class="lower-index">1</sub> is "NO". - Soldiers 2 through 4 (Bob, Bob, Cpqepqwer) wouldn't be effective either, and the string *s*<sub class="lower-index">2</sub> is "NO". - Soldiers 3 through 5 (Bob, Cpqepqwer, Limak) would be effective, and the string *s*<sub class="lower-index">3</sub> is "YES". - ..., - Soldiers 6 through 8 (Adam, Bob, Adam) wouldn't be effective, and the string *s*<sub class="lower-index">6</sub> is "NO". | ```python
n, k = map(int, input().split())
s = input().split()
words = []
for i in range(26):
words.append(chr(65 + i))
for i in range(26):
words.append(chr(65 + i) + chr(i + 97))
m = n - k + 1
ans = [0] * n
for i in range(m):
if s[i] == "YES":
for j in range(i, i + k):
if ans[j] == 0:
for word in words:
if word not in ans[i:i+k]:
ans[j] = word
break
last = "A"
for i in range(n):
if ans[i] == 0:
ans[i] = last
last = ans[i]
print(*ans)
``` | 0 |
|
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"
}
] | 1,682,803,155 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 61 | 124 | 0 | from sys import stdin, setrecursionlimit, stdout
def solve():
n = int(stdin.readline())
mas = [int(s) for s in stdin.readline().split()]
k = 0
ans = [0]*3
for i in range(n):
ans[k] += mas[i]
k = (k + 1) % 3
if ans[0] > ans[1] and ans[0] > ans[2]:
print('chest')
elif ans[1] > ans[0] and ans[1] > ans[2]:
print('biceps')
else:
print('back')
if __name__ == '__main__':
solve()
| 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
from sys import stdin, setrecursionlimit, stdout
def solve():
n = int(stdin.readline())
mas = [int(s) for s in stdin.readline().split()]
k = 0
ans = [0]*3
for i in range(n):
ans[k] += mas[i]
k = (k + 1) % 3
if ans[0] > ans[1] and ans[0] > ans[2]:
print('chest')
elif ans[1] > ans[0] and ans[1] > ans[2]:
print('biceps')
else:
print('back')
if __name__ == '__main__':
solve()
``` | 3 |
|
869 | B | The Eternal Immortality | PROGRAMMING | 1,100 | [
"math"
] | null | null | Even if the world is full of counterfeits, I still regard it as wonderful.
Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.
The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1.
Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer.
As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge. | The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018). | Output one line containing a single decimal digit — the last digit of the value that interests Koyomi. | [
"2 4\n",
"0 10\n",
"107 109\n"
] | [
"2\n",
"0\n",
"2\n"
] | In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2;
In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0;
In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2. | 1,000 | [
{
"input": "2 4",
"output": "2"
},
{
"input": "0 10",
"output": "0"
},
{
"input": "107 109",
"output": "2"
},
{
"input": "10 13",
"output": "6"
},
{
"input": "998244355 998244359",
"output": "4"
},
{
"input": "999999999000000000 1000000000000000000",
"output": "0"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "3 15",
"output": "0"
},
{
"input": "24 26",
"output": "0"
},
{
"input": "14 60",
"output": "0"
},
{
"input": "11 79",
"output": "0"
},
{
"input": "1230 1232",
"output": "2"
},
{
"input": "2633 2634",
"output": "4"
},
{
"input": "535 536",
"output": "6"
},
{
"input": "344319135 396746843",
"output": "0"
},
{
"input": "696667767 696667767",
"output": "1"
},
{
"input": "419530302 610096911",
"output": "0"
},
{
"input": "238965115 821731161",
"output": "0"
},
{
"input": "414626436 728903812",
"output": "0"
},
{
"input": "274410639 293308324",
"output": "0"
},
{
"input": "650636673091305697 650636673091305702",
"output": "0"
},
{
"input": "651240548333620923 651240548333620924",
"output": "4"
},
{
"input": "500000000000000000 1000000000000000000",
"output": "0"
},
{
"input": "999999999999999999 1000000000000000000",
"output": "0"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1"
},
{
"input": "0 4",
"output": "4"
},
{
"input": "50000000062000007 50000000062000011",
"output": "0"
},
{
"input": "0 0",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "0 2",
"output": "2"
},
{
"input": "10000000000012 10000000000015",
"output": "0"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "12 23",
"output": "0"
},
{
"input": "0 11",
"output": "0"
},
{
"input": "11111234567890 11111234567898",
"output": "0"
},
{
"input": "0 3",
"output": "6"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999999999999999997 999999999999999999",
"output": "2"
},
{
"input": "4 5",
"output": "5"
},
{
"input": "0 1",
"output": "1"
},
{
"input": "101 1002",
"output": "0"
},
{
"input": "0 100000000000000001",
"output": "0"
},
{
"input": "99999999999999997 99999999999999999",
"output": "2"
},
{
"input": "14 15",
"output": "5"
},
{
"input": "8 19",
"output": "0"
},
{
"input": "12 22",
"output": "0"
},
{
"input": "999999999999996 999999999999999",
"output": "4"
},
{
"input": "1 3",
"output": "6"
},
{
"input": "124 125",
"output": "5"
},
{
"input": "11 32",
"output": "0"
},
{
"input": "0 5",
"output": "0"
},
{
"input": "0 999999",
"output": "0"
},
{
"input": "151151151515 151151151526",
"output": "0"
},
{
"input": "6 107",
"output": "0"
},
{
"input": "5 16",
"output": "0"
},
{
"input": "7 16",
"output": "0"
},
{
"input": "6 19",
"output": "0"
},
{
"input": "11113111111111 13111111111111",
"output": "0"
},
{
"input": "1 1000",
"output": "0"
},
{
"input": "24 25",
"output": "5"
},
{
"input": "0 100000000000",
"output": "0"
},
{
"input": "1 22",
"output": "0"
},
{
"input": "999999999999999996 999999999999999999",
"output": "4"
}
] | 1,507,965,724 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 77 | 0 | l = input().split(" ")
n =int(l[0])
m = int(l[1])
res=1
if m-n>=10:
print(0)
else:
while m>n:
res = ((m%10)*res)%10
m-=1
print(res) | Title: The Eternal Immortality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Even if the world is full of counterfeits, I still regard it as wonderful.
Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.
The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1.
Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer.
As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.
Input Specification:
The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018).
Output Specification:
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.
Demo Input:
['2 4\n', '0 10\n', '107 109\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2;
In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0;
In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2. | ```python
l = input().split(" ")
n =int(l[0])
m = int(l[1])
res=1
if m-n>=10:
print(0)
else:
while m>n:
res = ((m%10)*res)%10
m-=1
print(res)
``` | 3 |
|
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,609,077,189 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 93 | 0 | text = input()
chat = 'hello'
index = 0
for i in range(len(text)):
if text[i] == chat[index]:
index += 1
elif i == 0:
continue
elif text[i] != text[i-1]:
index = 0
if(index == len(chat)): break
print('YES' if(index == len(chat)) else 'NO') | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
text = input()
chat = 'hello'
index = 0
for i in range(len(text)):
if text[i] == chat[index]:
index += 1
elif i == 0:
continue
elif text[i] != text[i-1]:
index = 0
if(index == len(chat)): break
print('YES' if(index == len(chat)) else 'NO')
``` | 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"
}
] | 1,694,389,127 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 92 | 0 | n, m = map(int, input().split())
count = 0
stop = m*n
while stop >1:
stop -= 2
count += 1
print(count) | 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
n, m = map(int, input().split())
count = 0
stop = m*n
while stop >1:
stop -= 2
count += 1
print(count)
``` | 3.977 |
349 | B | Color the Fence | PROGRAMMING | 1,700 | [
"data structures",
"dp",
"greedy",
"implementation"
] | null | null | Igor has fallen in love with Tanya. Now Igor wants to show his feelings and write a number on the fence opposite to Tanya's house. Igor thinks that the larger the number is, the more chance to win Tanya's heart he has.
Unfortunately, Igor could only get *v* liters of paint. He did the math and concluded that digit *d* requires *a**d* liters of paint. Besides, Igor heard that Tanya doesn't like zeroes. That's why Igor won't use them in his number.
Help Igor find the maximum number he can write on the fence. | The first line contains a positive integer *v* (0<=≤<=*v*<=≤<=106). The second line contains nine positive integers *a*1,<=*a*2,<=...,<=*a*9 (1<=≤<=*a**i*<=≤<=105). | Print the maximum number Igor can write on the fence. If he has too little paint for any digit (so, he cannot write anything), print -1. | [
"5\n5 4 3 2 1 2 3 4 5\n",
"2\n9 11 1 12 5 8 9 10 6\n",
"0\n1 1 1 1 1 1 1 1 1\n"
] | [
"55555\n",
"33\n",
"-1\n"
] | none | 1,000 | [
{
"input": "5\n5 4 3 2 1 2 3 4 5",
"output": "55555"
},
{
"input": "2\n9 11 1 12 5 8 9 10 6",
"output": "33"
},
{
"input": "0\n1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "50\n5 3 10 2 2 4 3 6 5",
"output": "5555555555555555555555555"
},
{
"input": "22\n405 343 489 474 385 23 100 94 276",
"output": "-1"
},
{
"input": "62800\n867 936 2 888 474 530 287 822 220",
"output": "3333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333..."
},
{
"input": "27\n836 637 966 929 82 678 213 465 688",
"output": "-1"
},
{
"input": "1000000\n100000 100000 100000 100000 100000 100000 100000 100000 100000",
"output": "9999999999"
},
{
"input": "898207\n99745 99746 99748 99752 99760 99776 99808 99872 100000",
"output": "987654321"
},
{
"input": "80910\n64537 83748 97081 82722 12334 3056 9491 59130 28478",
"output": "66666666666666666666666666"
},
{
"input": "120081\n11268 36403 73200 12674 83919 74218 74172 91581 68432",
"output": "4444411111"
},
{
"input": "839851\n29926 55862 57907 51153 56350 86145 1909 22622 89861",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "751233\n69761 51826 91095 73642 98995 93262 377 38818 97480",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "306978\n95955 99204 81786 41258 96065 46946 64532 36297 70808",
"output": "88888888"
},
{
"input": "366313\n18486 12701 92334 95391 61480 14118 20465 69784 13592",
"output": "9999999999922222222222222222"
},
{
"input": "320671\n95788 46450 97582 95928 47742 15508 10466 10301 38822",
"output": "8888888888888888888888888888888"
},
{
"input": "913928\n80373 47589 53204 68236 44060 97485 82241 44149 59825",
"output": "99888888888888855555"
},
{
"input": "630384\n19652 11530 20316 3161 87360 64207 74067 77894 81452",
"output": "4444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444"
},
{
"input": "95\n22076 12056 63350 12443 43123 585 52908 18372 96799",
"output": "-1"
},
{
"input": "271380\n19135 80309 23783 48534 98990 37278 85258 67602 40288",
"output": "11111111111111"
},
{
"input": "80085\n56973 29725 30219 17439 53162 6051 41388 35555 39392",
"output": "6666666666666"
},
{
"input": "201332\n20008 22829 30296 1967 32154 67760 11437 90972 79865",
"output": "444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444"
},
{
"input": "3402\n64151 98148 81468 82342 48823 93464 5989 58868 77138",
"output": "-1"
},
{
"input": "432544\n95724 98294 23292 24174 57778 95072 81898 50019 86824",
"output": "444444444444444333"
},
{
"input": "1000000\n1 1 1 1 1 1 1 1 1",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999..."
},
{
"input": "1000000\n2 2 2 2 2 2 2 2 2",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999..."
},
{
"input": "1000000\n2 3 2 2 3 2 2 3 2",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999..."
},
{
"input": "999999\n2 3 2 2 3 2 2 3 3",
"output": "9777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "153\n85 91 28 53 29 30 92 36 89",
"output": "86653"
},
{
"input": "26531\n64 93 48 49 86 57 93 60 96",
"output": "8864433333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333..."
},
{
"input": "17186\n50 90 76 51 91 54 71 90 73",
"output": "9666411111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "11213\n51 82 49 50 99 52 69 96 85",
"output": "964433333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333"
},
{
"input": "20075\n57 42 99 45 56 80 76 71 63",
"output": "954422222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222"
},
{
"input": "21069\n31 19 49 30 28 43 21 25 28",
"output": "9872222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222..."
},
{
"input": "4822\n35 36 21 13 34 36 14 16 20",
"output": "9877444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444"
}
] | 1,690,881,022 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 92 | 0 | def main():
v = int(input())
aseq = read_ints()
amin_idx = -1
amin = float('inf')
for i in range(9, 0, -1):
if aseq[i-1] < amin:
amin = aseq[i-1]
amin_idx = i-1
m = v // amin
if m == 0:
print(-1)
return
res = [amin_idx+1] * m
rem = v - m * amin
for i in range(m):
if rem == 0:
break
for d in range(9, 0, -1):
if rem + amin >= aseq[d-1] and d > amin_idx + 1:
res[i] = d
rem += amin
rem -= aseq[d-1]
break
print(' '.join(map(str, res)))
def input(): return next(test).strip()
def read_ints(): return [int(c) for c in input().split()]
def print_lines(lst): print('\n'.join(map(str, lst)))
if __name__ == "__main__":
import sys
from os import environ as env
if 'COMPUTERNAME' in env and 'L2A6HRI' in env['COMPUTERNAME']:
sys.stdout = open('out.txt', 'w')
sys.stdin = open('in.txt', 'r')
test = iter(sys.stdin.readlines())
main()
| Title: Color the Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Igor has fallen in love with Tanya. Now Igor wants to show his feelings and write a number on the fence opposite to Tanya's house. Igor thinks that the larger the number is, the more chance to win Tanya's heart he has.
Unfortunately, Igor could only get *v* liters of paint. He did the math and concluded that digit *d* requires *a**d* liters of paint. Besides, Igor heard that Tanya doesn't like zeroes. That's why Igor won't use them in his number.
Help Igor find the maximum number he can write on the fence.
Input Specification:
The first line contains a positive integer *v* (0<=≤<=*v*<=≤<=106). The second line contains nine positive integers *a*1,<=*a*2,<=...,<=*a*9 (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Print the maximum number Igor can write on the fence. If he has too little paint for any digit (so, he cannot write anything), print -1.
Demo Input:
['5\n5 4 3 2 1 2 3 4 5\n', '2\n9 11 1 12 5 8 9 10 6\n', '0\n1 1 1 1 1 1 1 1 1\n']
Demo Output:
['55555\n', '33\n', '-1\n']
Note:
none | ```python
def main():
v = int(input())
aseq = read_ints()
amin_idx = -1
amin = float('inf')
for i in range(9, 0, -1):
if aseq[i-1] < amin:
amin = aseq[i-1]
amin_idx = i-1
m = v // amin
if m == 0:
print(-1)
return
res = [amin_idx+1] * m
rem = v - m * amin
for i in range(m):
if rem == 0:
break
for d in range(9, 0, -1):
if rem + amin >= aseq[d-1] and d > amin_idx + 1:
res[i] = d
rem += amin
rem -= aseq[d-1]
break
print(' '.join(map(str, res)))
def input(): return next(test).strip()
def read_ints(): return [int(c) for c in input().split()]
def print_lines(lst): print('\n'.join(map(str, lst)))
if __name__ == "__main__":
import sys
from os import environ as env
if 'COMPUTERNAME' in env and 'L2A6HRI' in env['COMPUTERNAME']:
sys.stdout = open('out.txt', 'w')
sys.stdin = open('in.txt', 'r')
test = iter(sys.stdin.readlines())
main()
``` | 0 |
|
558 | B | Amr and The Large Array | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Amr has got a large array of size *n*. Amr doesn't like large arrays so he intends to make it smaller.
Amr doesn't care about anything in the array except the beauty of it. The beauty of the array is defined to be the maximum number of times that some number occurs in this array. He wants to choose the smallest subsegment of this array such that the beauty of it will be the same as the original array.
Help Amr by choosing the smallest subsegment possible. | The first line contains one number *n* (1<=≤<=*n*<=≤<=105), the size of the array.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=106), representing elements of the array. | Output two integers *l*,<=*r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), the beginning and the end of the subsegment chosen respectively.
If there are several possible answers you may output any of them. | [
"5\n1 1 2 2 1\n",
"5\n1 2 2 3 1\n",
"6\n1 2 2 1 1 2\n"
] | [
"1 5",
"2 3",
"1 5"
] | A subsegment *B* of an array *A* from *l* to *r* is an array of size *r* - *l* + 1 where *B*<sub class="lower-index">*i*</sub> = *A*<sub class="lower-index">*l* + *i* - 1</sub> for all 1 ≤ *i* ≤ *r* - *l* + 1 | 1,000 | [
{
"input": "5\n1 1 2 2 1",
"output": "1 5"
},
{
"input": "5\n1 2 2 3 1",
"output": "2 3"
},
{
"input": "6\n1 2 2 1 1 2",
"output": "1 5"
},
{
"input": "10\n1 1000000 2 1000000 3 2 1000000 1 2 1",
"output": "2 7"
},
{
"input": "10\n1 2 3 4 5 5 1 2 3 4",
"output": "5 6"
},
{
"input": "10\n1 1 2 1 1 2 2 1 2 3",
"output": "1 8"
},
{
"input": "10\n5 4 3 2 1 1 2 3 4 5",
"output": "5 6"
},
{
"input": "10\n1 10 100 1000 10000 1 10 100 1000 10000",
"output": "1 6"
},
{
"input": "10\n9 9 9 9 8 9 8 8 8 8",
"output": "1 6"
},
{
"input": "10\n1 11 111 1111 1 11 11 1 1111 1111",
"output": "2 7"
}
] | 1,589,882,264 | 2,147,483,647 | PyPy 3 | OK | TESTS | 49 | 218 | 15,564,800 | from sys import stdin
def main():
n = int(stdin.readline())
ar = list(map(int, stdin.readline().split()))
c = [0] * (10 ** 6 + 1)
mf = 0
for elm in ar:
c[elm] += 1
if c[elm] > mf:
mf = c[elm]
start = 0
end = 0
lk = [0] * (10 ** 6 + 1)
f = [0] * (n + 1)
f[0] = n
cf = 0
ml = n + 1
il = -1
ir = -1
while end < n:
while end < n and cf < mf:
f[lk[ar[end]]] -= 1
lk[ar[end]] += 1
f[lk[ar[end]]] += 1
if lk[ar[end]] > cf:
cf = lk[ar[end]]
end += 1
if cf == mf:
if end - start < ml:
ml = end - start
il = start + 1
ir = end
while start < end and cf == mf:
f[lk[ar[start]]] -= 1
lk[ar[start]] -= 1
f[lk[ar[start]]] += 1
start += 1
if f[cf] == 0:
cf = cf - 1
else:
if end - start < ml:
ml = end - start
il = start + 1
ir = end
print(il, ir)
if __name__ == "__main__":
main()
| Title: Amr and The Large Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Amr has got a large array of size *n*. Amr doesn't like large arrays so he intends to make it smaller.
Amr doesn't care about anything in the array except the beauty of it. The beauty of the array is defined to be the maximum number of times that some number occurs in this array. He wants to choose the smallest subsegment of this array such that the beauty of it will be the same as the original array.
Help Amr by choosing the smallest subsegment possible.
Input Specification:
The first line contains one number *n* (1<=≤<=*n*<=≤<=105), the size of the array.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=106), representing elements of the array.
Output Specification:
Output two integers *l*,<=*r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), the beginning and the end of the subsegment chosen respectively.
If there are several possible answers you may output any of them.
Demo Input:
['5\n1 1 2 2 1\n', '5\n1 2 2 3 1\n', '6\n1 2 2 1 1 2\n']
Demo Output:
['1 5', '2 3', '1 5']
Note:
A subsegment *B* of an array *A* from *l* to *r* is an array of size *r* - *l* + 1 where *B*<sub class="lower-index">*i*</sub> = *A*<sub class="lower-index">*l* + *i* - 1</sub> for all 1 ≤ *i* ≤ *r* - *l* + 1 | ```python
from sys import stdin
def main():
n = int(stdin.readline())
ar = list(map(int, stdin.readline().split()))
c = [0] * (10 ** 6 + 1)
mf = 0
for elm in ar:
c[elm] += 1
if c[elm] > mf:
mf = c[elm]
start = 0
end = 0
lk = [0] * (10 ** 6 + 1)
f = [0] * (n + 1)
f[0] = n
cf = 0
ml = n + 1
il = -1
ir = -1
while end < n:
while end < n and cf < mf:
f[lk[ar[end]]] -= 1
lk[ar[end]] += 1
f[lk[ar[end]]] += 1
if lk[ar[end]] > cf:
cf = lk[ar[end]]
end += 1
if cf == mf:
if end - start < ml:
ml = end - start
il = start + 1
ir = end
while start < end and cf == mf:
f[lk[ar[start]]] -= 1
lk[ar[start]] -= 1
f[lk[ar[start]]] += 1
start += 1
if f[cf] == 0:
cf = cf - 1
else:
if end - start < ml:
ml = end - start
il = start + 1
ir = end
print(il, ir)
if __name__ == "__main__":
main()
``` | 3 |
|
205 | A | Little Elephant and Rozdil | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities. | Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes). | [
"2\n7 4\n",
"7\n7 4 47 100 4 9 12\n"
] | [
"2\n",
"Still Rozdil\n"
] | In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil". | 500 | [
{
"input": "2\n7 4",
"output": "2"
},
{
"input": "7\n7 4 47 100 4 9 12",
"output": "Still Rozdil"
},
{
"input": "1\n47",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "Still Rozdil"
},
{
"input": "7\n7 6 5 4 3 2 1",
"output": "7"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "Still Rozdil"
},
{
"input": "4\n1000000000 100000000 1000000 1000000",
"output": "Still Rozdil"
},
{
"input": "20\n7 1 1 2 1 1 8 7 7 8 4 3 7 10 5 3 10 5 10 6",
"output": "Still Rozdil"
},
{
"input": "20\n3 3 6 9 8 2 4 1 7 3 2 9 7 7 9 7 2 6 2 7",
"output": "8"
},
{
"input": "47\n35 79 84 56 67 95 80 34 77 68 14 55 95 32 40 89 58 79 96 66 50 79 35 86 31 74 91 35 22 72 84 38 11 59 73 51 65 11 11 62 30 12 32 71 69 15 11",
"output": "Still Rozdil"
},
{
"input": "47\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 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "32"
},
{
"input": "25\n73 56 65 70 57 57 49 50 65 68 54 48 66 55 55 66 60 70 47 62 63 70 62 70 63",
"output": "19"
},
{
"input": "100\n922 938 103 689 526 118 314 496 373 47 224 544 495 656 804 754 269 356 354 536 817 597 787 178 516 659 790 993 786 169 866 779 997 350 674 437 219 546 564 893 402 449 751 628 238 686 392 951 716 439 332 563 215 146 746 311 866 773 548 675 731 183 890 802 488 944 473 25 662 559 732 534 777 144 502 593 839 692 84 337 679 718 1005 663 709 940 86 469 310 960 902 233 606 761 116 231 662 903 68 622",
"output": "68"
},
{
"input": "5\n5 5 2 3 1",
"output": "5"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "3\n7 7 4",
"output": "3"
},
{
"input": "3\n2 2 1",
"output": "3"
},
{
"input": "3\n4 4 3",
"output": "3"
},
{
"input": "2\n10000001 10000002",
"output": "1"
},
{
"input": "4\n3 3 2 1",
"output": "4"
},
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "2\n1000000000 100000000",
"output": "2"
},
{
"input": "1\n99999999",
"output": "1"
},
{
"input": "3\n5 5 3",
"output": "3"
},
{
"input": "3\n7 7 5",
"output": "3"
},
{
"input": "3\n3 3 1",
"output": "3"
},
{
"input": "4\n10 10 10 1",
"output": "4"
}
] | 1,615,979,358 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 186 | 8,499,200 | n = int(input())
l = list(map(int,input().split()))
k = min(l)
if l.count(k) > 1:
print("Still Rozdil")
else:
print(l.index(k)+1)
| Title: Little Elephant and Rozdil
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities.
Output Specification:
Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes).
Demo Input:
['2\n7 4\n', '7\n7 4 47 100 4 9 12\n']
Demo Output:
['2\n', 'Still Rozdil\n']
Note:
In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil". | ```python
n = int(input())
l = list(map(int,input().split()))
k = min(l)
if l.count(k) > 1:
print("Still Rozdil")
else:
print(l.index(k)+1)
``` | 3 |
|
606 | A | Magic Spheres | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Carl is a beginner magician. He has *a* blue, *b* violet and *c* orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least *x* blue, *y* violet and *z* orange spheres. Can he get them (possible, in multiple actions)? | The first line of the input contains three integers *a*, *b* and *c* (0<=≤<=*a*,<=*b*,<=*c*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that he needs to get. | If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No". | [
"4 4 0\n2 1 2\n",
"5 6 1\n2 7 2\n",
"3 3 3\n2 2 2\n"
] | [
"Yes\n",
"No\n",
"Yes\n"
] | In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs. | 500 | [
{
"input": "4 4 0\n2 1 2",
"output": "Yes"
},
{
"input": "5 6 1\n2 7 2",
"output": "No"
},
{
"input": "3 3 3\n2 2 2",
"output": "Yes"
},
{
"input": "0 0 0\n0 0 0",
"output": "Yes"
},
{
"input": "0 0 0\n0 0 1",
"output": "No"
},
{
"input": "0 1 0\n0 0 0",
"output": "Yes"
},
{
"input": "1 0 0\n1 0 0",
"output": "Yes"
},
{
"input": "2 2 1\n1 1 2",
"output": "No"
},
{
"input": "1 3 1\n2 1 1",
"output": "Yes"
},
{
"input": "1000000 1000000 1000000\n1000000 1000000 1000000",
"output": "Yes"
},
{
"input": "1000000 500000 500000\n0 750000 750000",
"output": "Yes"
},
{
"input": "500000 1000000 500000\n750001 0 750000",
"output": "No"
},
{
"input": "499999 500000 1000000\n750000 750000 0",
"output": "No"
},
{
"input": "500000 500000 0\n0 0 500000",
"output": "Yes"
},
{
"input": "0 500001 499999\n500000 0 0",
"output": "No"
},
{
"input": "1000000 500000 1000000\n500000 1000000 500000",
"output": "Yes"
},
{
"input": "1000000 1000000 499999\n500000 500000 1000000",
"output": "No"
},
{
"input": "500000 1000000 1000000\n1000000 500001 500000",
"output": "No"
},
{
"input": "1000000 500000 500000\n0 1000000 500000",
"output": "Yes"
},
{
"input": "500000 500000 1000000\n500001 1000000 0",
"output": "No"
},
{
"input": "500000 999999 500000\n1000000 0 500000",
"output": "No"
},
{
"input": "4 0 3\n2 2 1",
"output": "Yes"
},
{
"input": "0 2 4\n2 0 2",
"output": "Yes"
},
{
"input": "3 1 0\n1 1 1",
"output": "Yes"
},
{
"input": "4 4 1\n1 3 2",
"output": "Yes"
},
{
"input": "1 2 4\n2 1 3",
"output": "No"
},
{
"input": "1 1 0\n0 0 1",
"output": "No"
},
{
"input": "4 0 0\n0 1 1",
"output": "Yes"
},
{
"input": "0 3 0\n1 0 1",
"output": "No"
},
{
"input": "0 0 3\n1 0 1",
"output": "Yes"
},
{
"input": "1 12 1\n4 0 4",
"output": "Yes"
},
{
"input": "4 0 4\n1 2 1",
"output": "Yes"
},
{
"input": "4 4 0\n1 1 3",
"output": "No"
},
{
"input": "0 9 0\n2 2 2",
"output": "No"
},
{
"input": "0 10 0\n2 2 2",
"output": "Yes"
},
{
"input": "9 0 9\n0 8 0",
"output": "Yes"
},
{
"input": "0 9 9\n9 0 0",
"output": "No"
},
{
"input": "9 10 0\n0 0 9",
"output": "Yes"
},
{
"input": "10 0 9\n0 10 0",
"output": "No"
},
{
"input": "0 10 10\n10 0 0",
"output": "Yes"
},
{
"input": "10 10 0\n0 0 11",
"output": "No"
},
{
"input": "307075 152060 414033\n381653 222949 123101",
"output": "No"
},
{
"input": "569950 228830 153718\n162186 357079 229352",
"output": "No"
},
{
"input": "149416 303568 749016\n238307 493997 190377",
"output": "No"
},
{
"input": "438332 298094 225324\n194220 400244 245231",
"output": "No"
},
{
"input": "293792 300060 511272\n400687 382150 133304",
"output": "No"
},
{
"input": "295449 518151 368838\n382897 137148 471892",
"output": "No"
},
{
"input": "191789 291147 691092\n324321 416045 176232",
"output": "Yes"
},
{
"input": "286845 704749 266526\n392296 104421 461239",
"output": "Yes"
},
{
"input": "135522 188282 377041\n245719 212473 108265",
"output": "Yes"
},
{
"input": "404239 359124 133292\n180069 184791 332544",
"output": "No"
},
{
"input": "191906 624432 244408\n340002 367217 205432",
"output": "No"
},
{
"input": "275980 429361 101824\n274288 302579 166062",
"output": "No"
},
{
"input": "136092 364927 395302\n149173 343146 390922",
"output": "No"
},
{
"input": "613852 334661 146012\n363786 326286 275233",
"output": "No"
},
{
"input": "348369 104625 525203\n285621 215396 366411",
"output": "No"
},
{
"input": "225307 153572 114545\n154753 153282 149967",
"output": "Yes"
},
{
"input": "438576 124465 629784\n375118 276028 390116",
"output": "Yes"
},
{
"input": "447521 327510 158732\n395759 178458 259139",
"output": "Yes"
},
{
"input": "8 5 5\n5 5 5",
"output": "Yes"
},
{
"input": "100 100 100\n1 1 1",
"output": "Yes"
},
{
"input": "100 100 100\n0 0 0",
"output": "Yes"
},
{
"input": "3 2 3\n2 3 2",
"output": "No"
},
{
"input": "5 4 3\n2 2 2",
"output": "Yes"
},
{
"input": "14 9 8\n12 5 10",
"output": "Yes"
},
{
"input": "10 10 10\n1 1 1",
"output": "Yes"
},
{
"input": "6 3 3\n3 3 3",
"output": "Yes"
},
{
"input": "10 0 4\n2 4 2",
"output": "Yes"
},
{
"input": "100 100 100\n2 2 2",
"output": "Yes"
},
{
"input": "4 6 0\n2 1 2",
"output": "Yes"
},
{
"input": "4 6 3\n4 2 3",
"output": "Yes"
},
{
"input": "5 5 5\n1 1 1",
"output": "Yes"
},
{
"input": "41 17 34\n0 19 24",
"output": "Yes"
},
{
"input": "8 8 8\n3 3 3",
"output": "Yes"
},
{
"input": "7 7 1\n1 1 2",
"output": "Yes"
},
{
"input": "6 6 0\n2 2 2",
"output": "Yes"
},
{
"input": "5 5 5\n2 2 2",
"output": "Yes"
},
{
"input": "400 400 400\n1 1 1",
"output": "Yes"
},
{
"input": "4 4 4\n2 2 2",
"output": "Yes"
}
] | 1,668,144,816 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 79 | 62 | 0 | import sys
input = sys.stdin.readline
# for _ in range(int(input())):
a,b,c = map(int,input().split(" "))
x,y,z = map(int,input().split(" "))
diff = [a-x,b-y,c-z]
ans = 0
for i in range(3):
if diff[i] > 0:
diff[i] //= 2
ans += diff[i]
if ans >= 0:
print("Yes")
else:
print("No") | Title: Magic Spheres
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Carl is a beginner magician. He has *a* blue, *b* violet and *c* orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least *x* blue, *y* violet and *z* orange spheres. Can he get them (possible, in multiple actions)?
Input Specification:
The first line of the input contains three integers *a*, *b* and *c* (0<=≤<=*a*,<=*b*,<=*c*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=1<=000<=000) — the number of blue, violet and orange spheres that he needs to get.
Output Specification:
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Demo Input:
['4 4 0\n2 1 2\n', '5 6 1\n2 7 2\n', '3 3 3\n2 2 2\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n']
Note:
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs. | ```python
import sys
input = sys.stdin.readline
# for _ in range(int(input())):
a,b,c = map(int,input().split(" "))
x,y,z = map(int,input().split(" "))
diff = [a-x,b-y,c-z]
ans = 0
for i in range(3):
if diff[i] > 0:
diff[i] //= 2
ans += diff[i]
if ans >= 0:
print("Yes")
else:
print("No")
``` | 3 |
|
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,694,587,862 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 9 | 62 | 0 | if __name__ == '__main__':
s=str(input())
t=str(input())
n=len(s)
cnt=0
for i in range(n):
if s[i]==t[(n-1)-i]:
cnt=cnt+1
if cnt==n:
print("YES")
else:
print("NO")
| Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
if __name__ == '__main__':
s=str(input())
t=str(input())
n=len(s)
cnt=0
for i in range(n):
if s[i]==t[(n-1)-i]:
cnt=cnt+1
if cnt==n:
print("YES")
else:
print("NO")
``` | -1 |
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"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"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"
}
] | 1,695,200,207 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | result = 0
for i in range(int(input)):
result -= eval(input().replace(" ", "-")
print(result) | 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
result = 0
for i in range(int(input)):
result -= eval(input().replace(" ", "-")
print(result)
``` | -1 |
|
268 | B | Buttons | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Manao is trying to open a rather challenging lock. The lock has *n* buttons on it and to open it, you should press the buttons in a certain order to open the lock. When you push some button, it either stays pressed into the lock (that means that you've guessed correctly and pushed the button that goes next in the sequence), or all pressed buttons return to the initial position. When all buttons are pressed into the lock at once, the lock opens.
Consider an example with three buttons. Let's say that the opening sequence is: {2, 3, 1}. If you first press buttons 1 or 3, the buttons unpress immediately. If you first press button 2, it stays pressed. If you press 1 after 2, all buttons unpress. If you press 3 after 2, buttons 3 and 2 stay pressed. As soon as you've got two pressed buttons, you only need to press button 1 to open the lock.
Manao doesn't know the opening sequence. But he is really smart and he is going to act in the optimal way. Calculate the number of times he's got to push a button in order to open the lock in the worst-case scenario. | A single line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of buttons the lock has. | In a single line print the number of times Manao has to push a button in the worst-case scenario. | [
"2\n",
"3\n"
] | [
"3\n",
"7\n"
] | Consider the first test sample. Manao can fail his first push and push the wrong button. In this case he will already be able to guess the right one with his second push. And his third push will push the second right button. Thus, in the worst-case scenario he will only need 3 pushes. | 1,000 | [
{
"input": "2",
"output": "3"
},
{
"input": "3",
"output": "7"
},
{
"input": "4",
"output": "14"
},
{
"input": "1",
"output": "1"
},
{
"input": "10",
"output": "175"
},
{
"input": "2000",
"output": "1333335000"
},
{
"input": "1747",
"output": "888644743"
},
{
"input": "889",
"output": "117099969"
},
{
"input": "1999",
"output": "1331335999"
},
{
"input": "914",
"output": "127259419"
},
{
"input": "996",
"output": "164675486"
},
{
"input": "17",
"output": "833"
},
{
"input": "50",
"output": "20875"
},
{
"input": "91",
"output": "125671"
},
{
"input": "92",
"output": "129858"
},
{
"input": "256",
"output": "2796416"
},
{
"input": "512",
"output": "22370048"
},
{
"input": "666",
"output": "49235271"
},
{
"input": "667",
"output": "49457383"
},
{
"input": "314",
"output": "5160119"
},
{
"input": "1241",
"output": "318541121"
},
{
"input": "1500",
"output": "562501250"
},
{
"input": "1837",
"output": "1033182073"
},
{
"input": "1000",
"output": "166667500"
}
] | 1,672,266,496 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 92 | 0 | n = int(input());print(int(n*(n-1)*0.5 + (n-2)**2 + n))
| Title: Buttons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao is trying to open a rather challenging lock. The lock has *n* buttons on it and to open it, you should press the buttons in a certain order to open the lock. When you push some button, it either stays pressed into the lock (that means that you've guessed correctly and pushed the button that goes next in the sequence), or all pressed buttons return to the initial position. When all buttons are pressed into the lock at once, the lock opens.
Consider an example with three buttons. Let's say that the opening sequence is: {2, 3, 1}. If you first press buttons 1 or 3, the buttons unpress immediately. If you first press button 2, it stays pressed. If you press 1 after 2, all buttons unpress. If you press 3 after 2, buttons 3 and 2 stay pressed. As soon as you've got two pressed buttons, you only need to press button 1 to open the lock.
Manao doesn't know the opening sequence. But he is really smart and he is going to act in the optimal way. Calculate the number of times he's got to push a button in order to open the lock in the worst-case scenario.
Input Specification:
A single line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of buttons the lock has.
Output Specification:
In a single line print the number of times Manao has to push a button in the worst-case scenario.
Demo Input:
['2\n', '3\n']
Demo Output:
['3\n', '7\n']
Note:
Consider the first test sample. Manao can fail his first push and push the wrong button. In this case he will already be able to guess the right one with his second push. And his third push will push the second right button. Thus, in the worst-case scenario he will only need 3 pushes. | ```python
n = int(input());print(int(n*(n-1)*0.5 + (n-2)**2 + n))
``` | 0 |
|
796 | A | Buying A House | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.
The girl lives in house *m* of a village. There are *n* houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house *n*. The village is also well-structured: house *i* and house *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.
You will be given *n* integers *a*1,<=*a*2,<=...,<=*a**n* that denote the availability and the prices of the houses. If house *i* is occupied, and therefore cannot be bought, then *a**i* equals 0. Otherwise, house *i* can be bought, and *a**i* represents the money required to buy it, in dollars.
As Zane has only *k* dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love. | The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — denoting the availability and the prices of the houses.
It is guaranteed that *a**m*<==<=0 and that it is possible to purchase some house with no more than *k* dollars. | Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy. | [
"5 1 20\n0 27 32 21 19\n",
"7 3 50\n62 0 0 0 99 33 22\n",
"10 5 100\n1 0 1 0 0 0 0 0 1 1\n"
] | [
"40",
"30",
"20"
] | In the first sample, with *k* = 20 dollars, Zane can buy only house 5. The distance from house *m* = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.
In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house *m* = 3 and house 6 are only 30 meters away, while house *m* = 3 and house 7 are 40 meters away. | 500 | [
{
"input": "5 1 20\n0 27 32 21 19",
"output": "40"
},
{
"input": "7 3 50\n62 0 0 0 99 33 22",
"output": "30"
},
{
"input": "10 5 100\n1 0 1 0 0 0 0 0 1 1",
"output": "20"
},
{
"input": "5 3 1\n1 1 0 0 1",
"output": "10"
},
{
"input": "5 5 5\n1 0 5 6 0",
"output": "20"
},
{
"input": "15 10 50\n20 0 49 50 50 50 50 50 50 0 50 50 49 0 20",
"output": "10"
},
{
"input": "7 5 1\n0 100 2 2 0 2 1",
"output": "20"
},
{
"input": "100 50 100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 0 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "10"
},
{
"input": "100 50 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 0 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "490"
},
{
"input": "100 77 50\n50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 0 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0",
"output": "10"
},
{
"input": "100 1 1\n0 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0",
"output": "980"
},
{
"input": "100 1 100\n0 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "10"
},
{
"input": "100 10 99\n0 0 0 0 0 0 0 0 0 0 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 98",
"output": "890"
},
{
"input": "7 4 5\n1 0 6 0 5 6 0",
"output": "10"
},
{
"input": "7 4 5\n1 6 5 0 0 6 0",
"output": "10"
},
{
"input": "100 42 59\n50 50 50 50 50 50 50 50 50 50 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 60 60 60 60 60 60 60 60 0 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 0",
"output": "90"
},
{
"input": "2 1 100\n0 1",
"output": "10"
},
{
"input": "2 2 100\n1 0",
"output": "10"
},
{
"input": "10 1 88\n0 95 0 0 0 0 0 94 0 85",
"output": "90"
},
{
"input": "10 2 14\n2 0 1 26 77 39 41 100 13 32",
"output": "10"
},
{
"input": "10 3 11\n0 0 0 0 0 62 0 52 1 35",
"output": "60"
},
{
"input": "20 12 44\n27 40 58 69 53 38 31 39 75 95 8 0 28 81 77 90 38 61 21 88",
"output": "10"
},
{
"input": "30 29 10\n59 79 34 12 100 6 1 58 18 73 54 11 37 46 89 90 80 85 73 45 64 5 31 0 89 19 0 74 0 82",
"output": "70"
},
{
"input": "40 22 1\n7 95 44 53 0 0 19 93 0 68 65 0 24 91 10 58 17 0 71 0 100 0 94 90 79 73 0 73 4 61 54 81 7 13 21 84 5 41 0 1",
"output": "180"
},
{
"input": "40 22 99\n60 0 100 0 0 100 100 0 0 0 0 100 100 0 0 100 100 0 100 100 100 0 100 100 100 0 100 100 0 0 100 100 100 0 0 100 0 100 0 0",
"output": "210"
},
{
"input": "50 10 82\n56 54 0 0 0 0 88 93 0 0 83 93 0 0 91 89 0 30 62 52 24 84 80 8 38 13 92 78 16 87 23 30 71 55 16 63 15 99 4 93 24 6 3 35 4 42 73 27 86 37",
"output": "80"
},
{
"input": "63 49 22\n18 3 97 52 75 2 12 24 58 75 80 97 22 10 79 51 30 60 68 99 75 2 35 3 97 88 9 7 18 5 0 0 0 91 0 91 56 36 76 0 0 0 52 27 35 0 51 72 0 96 57 0 0 0 0 92 55 28 0 30 0 78 77",
"output": "190"
},
{
"input": "74 38 51\n53 36 55 42 64 5 87 9 0 16 86 78 9 22 19 1 25 72 1 0 0 0 79 0 0 0 77 58 70 0 0 100 64 0 99 59 0 0 0 0 65 74 0 96 0 58 89 93 61 88 0 0 82 89 0 0 49 24 7 77 89 87 94 61 100 31 93 70 39 49 39 14 20 84",
"output": "190"
},
{
"input": "89 22 11\n36 0 68 89 0 85 72 0 38 56 0 44 0 94 0 28 71 0 0 18 0 0 0 89 0 0 0 75 0 0 0 32 66 0 0 0 0 0 0 48 63 0 64 58 0 23 48 0 0 52 93 61 57 0 18 0 0 34 62 17 0 41 0 0 53 59 44 0 0 51 40 0 0 100 100 54 0 88 0 5 45 56 57 67 24 16 88 86 15",
"output": "580"
},
{
"input": "97 44 100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 51 19",
"output": "520"
},
{
"input": "100 1 1\n0 0 0 0 10 54 84 6 17 94 65 82 34 0 61 46 42 0 2 16 56 0 100 0 82 0 0 0 89 78 96 56 0 0 0 0 0 0 0 0 77 70 0 96 67 0 0 32 44 1 72 50 14 11 24 61 100 64 19 5 67 69 44 82 93 22 67 93 22 61 53 64 79 41 84 48 43 97 7 24 8 49 23 16 72 52 97 29 69 47 29 49 64 91 4 73 17 18 51 67",
"output": "490"
},
{
"input": "100 1 50\n0 0 0 60 0 0 54 0 80 0 0 0 97 0 68 97 84 0 0 93 0 0 0 0 68 0 0 62 0 0 55 68 65 87 0 69 0 0 0 0 0 52 61 100 0 71 0 82 88 78 0 81 0 95 0 57 0 67 0 0 0 55 86 0 60 72 0 0 73 0 83 0 0 60 64 0 56 0 0 77 84 0 58 63 84 0 0 67 0 16 3 88 0 98 31 52 40 35 85 23",
"output": "890"
},
{
"input": "100 1 100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 91 70 14",
"output": "970"
},
{
"input": "100 1 29\n0 0 0 0 64 0 89 97 0 0 0 59 0 67 62 0 59 0 0 80 0 0 0 0 0 97 0 57 0 64 32 0 44 0 0 48 0 47 38 0 42 0 0 0 0 0 0 46 74 0 86 33 33 0 44 0 79 0 0 0 0 91 59 0 59 65 55 0 0 58 33 95 0 97 76 0 81 0 41 0 38 81 80 0 85 0 31 0 0 92 0 0 45 96 0 85 91 87 0 10",
"output": "990"
},
{
"input": "100 50 20\n3 0 32 0 48 32 64 0 54 26 0 0 0 0 0 28 0 0 54 0 0 45 49 0 38 74 0 0 39 42 62 48 75 96 89 42 0 44 0 0 30 21 76 0 50 0 79 0 0 0 0 99 0 84 62 0 0 0 0 53 80 0 28 0 0 53 0 0 38 0 62 0 0 62 0 0 88 0 44 32 0 81 35 45 49 0 69 73 38 27 72 0 96 72 69 0 0 22 76 10",
"output": "490"
},
{
"input": "100 50 20\n49 0 56 0 87 25 40 0 50 0 0 97 0 0 36 29 0 0 0 0 0 73 29 71 44 0 0 0 91 92 69 0 0 60 81 49 48 38 0 87 0 82 0 32 0 82 46 39 0 0 29 0 0 29 0 79 47 0 0 0 0 0 49 0 24 33 70 0 63 45 97 90 0 0 29 53 55 0 84 0 0 100 26 0 88 0 0 0 0 81 70 0 30 80 0 75 59 98 0 2",
"output": "500"
},
{
"input": "100 2 2\n0 0 43 90 47 5 2 97 52 69 21 48 64 10 34 97 97 74 8 19 68 56 55 24 47 38 43 73 72 72 60 60 51 36 33 44 100 45 13 54 72 52 0 15 3 6 50 8 88 4 78 26 40 27 30 63 67 83 61 91 33 97 54 20 92 27 89 35 10 7 84 50 11 95 74 88 24 44 74 100 18 56 34 91 41 34 51 51 11 91 89 54 19 100 83 89 10 17 76 20",
"output": "50"
},
{
"input": "100 100 34\n5 73 0 0 44 0 0 0 79 55 0 0 0 0 0 0 0 0 83 67 75 0 0 0 0 59 0 74 0 0 47 98 0 0 72 41 0 55 87 0 0 78 84 0 0 39 0 79 72 95 0 0 0 0 0 85 53 84 0 0 0 0 37 75 0 66 0 0 0 0 61 0 70 0 37 60 42 78 92 52 0 0 0 55 77 57 0 63 37 0 0 0 96 70 0 94 97 0 0 0",
"output": "990"
},
{
"input": "100 100 100\n43 79 21 87 84 14 28 69 92 16 3 71 79 37 48 37 72 58 12 72 62 49 37 17 60 54 41 99 15 72 40 89 76 1 99 87 14 56 63 48 69 37 96 64 7 14 1 73 85 33 98 70 97 71 96 28 49 71 56 2 67 22 100 2 98 100 62 77 92 76 98 98 47 26 22 47 50 56 9 16 72 47 5 62 29 78 81 1 0 63 32 65 87 3 40 53 8 80 93 0",
"output": "10"
},
{
"input": "100 38 1\n3 59 12 81 33 95 0 41 36 17 63 76 42 77 85 56 3 96 55 41 24 87 18 9 0 37 0 61 69 0 0 0 67 0 0 0 0 0 0 18 0 0 47 56 74 0 0 80 0 42 0 1 60 59 62 9 19 87 92 48 58 30 98 51 99 10 42 94 51 53 50 89 24 5 52 82 50 39 98 8 95 4 57 21 10 0 44 32 19 14 64 34 79 76 17 3 15 22 71 51",
"output": "140"
},
{
"input": "100 72 1\n56 98 8 27 9 23 16 76 56 1 34 43 96 73 75 49 62 20 18 23 51 55 30 84 4 20 89 40 75 16 69 35 1 0 16 0 80 0 41 17 0 0 76 23 0 92 0 34 0 91 82 54 0 0 0 63 85 59 98 24 29 0 8 77 26 0 34 95 39 0 0 0 74 0 0 0 0 12 0 92 0 0 55 95 66 30 0 0 29 98 0 0 0 47 0 0 80 0 0 4",
"output": "390"
},
{
"input": "100 66 1\n38 50 64 91 37 44 74 21 14 41 80 90 26 51 78 85 80 86 44 14 49 75 93 48 78 89 23 72 35 22 14 48 100 71 62 22 7 95 80 66 32 20 17 47 79 30 41 52 15 62 67 71 1 6 0 9 0 0 0 11 0 0 24 0 31 0 77 0 51 0 0 0 0 0 0 77 0 36 44 19 90 45 6 25 100 87 93 30 4 97 36 88 33 50 26 71 97 71 51 68",
"output": "130"
},
{
"input": "100 55 1\n0 33 45 83 56 96 58 24 45 30 38 60 39 69 21 87 59 21 72 73 27 46 61 61 11 97 77 5 39 3 3 35 76 37 53 84 24 75 9 48 31 90 100 84 74 81 83 83 42 23 29 94 18 1 0 53 52 99 86 37 94 54 28 75 28 80 17 14 98 68 76 20 32 23 42 31 57 79 60 14 18 27 1 98 32 3 96 25 15 38 2 6 3 28 59 54 63 2 43 59",
"output": "10"
},
{
"input": "100 55 1\n24 52 41 6 55 11 58 25 63 12 70 39 23 28 72 17 96 85 7 84 21 13 34 37 97 43 36 32 15 30 58 5 14 71 40 70 9 92 44 73 31 58 96 90 19 35 29 91 25 36 48 95 61 78 0 1 99 61 81 88 42 53 61 57 42 55 74 45 41 92 99 30 20 25 89 50 37 4 17 24 6 65 15 44 40 2 38 43 7 90 38 59 75 87 96 28 12 67 24 32",
"output": "10"
},
{
"input": "100 21 1\n62 5 97 80 81 28 83 0 26 0 0 0 0 23 0 0 90 0 0 0 0 0 0 0 0 54 71 8 0 0 42 0 73 0 17 0 1 31 71 78 58 72 84 39 54 59 13 29 16 41 71 35 88 55 70 50 33 100 100 60 52 90 7 66 44 55 51 42 90 17 86 44 46 8 52 74 8 22 2 92 34 37 58 98 70 74 19 91 74 25 4 38 71 68 50 68 63 14 60 98",
"output": "160"
},
{
"input": "5 2 20\n27 0 32 21 19",
"output": "30"
},
{
"input": "6 4 10\n10 0 0 0 0 10",
"output": "20"
},
{
"input": "8 7 100\n1 0 0 0 0 0 0 1",
"output": "10"
},
{
"input": "5 3 20\n1 21 0 0 1",
"output": "20"
},
{
"input": "4 3 1\n0 0 0 1",
"output": "10"
},
{
"input": "5 2 3\n4 0 5 6 1",
"output": "30"
},
{
"input": "5 3 87\n88 89 0 1 90",
"output": "10"
},
{
"input": "5 3 20\n15 30 0 15 35",
"output": "10"
},
{
"input": "6 3 50\n0 0 0 1 2 0",
"output": "10"
},
{
"input": "6 4 9\n100 9 10 0 0 9",
"output": "20"
},
{
"input": "5 4 20\n0 20 0 0 20",
"output": "10"
},
{
"input": "6 3 3\n1 5 0 2 2 0",
"output": "10"
},
{
"input": "5 4 100\n0 1 0 0 1",
"output": "10"
}
] | 1,602,425,054 | 2,147,483,647 | Python 3 | OK | TESTS | 58 | 109 | 0 | __author__ = "Daniel Mitre"
n, m, k = map(int, input().split())
prices = map(int, input().split())
def dist_to_girl(pos: int) -> int:
return abs(pos - m)
buy = float('inf')
for idx, price in zip(range(1, n + 1), prices):
if (price > 0) and (price <= k) and (dist_to_girl(idx) < dist_to_girl(buy)):
buy = idx
print(dist_to_girl(buy) * 10)
| Title: Buying A House
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.
The girl lives in house *m* of a village. There are *n* houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house *n*. The village is also well-structured: house *i* and house *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.
You will be given *n* integers *a*1,<=*a*2,<=...,<=*a**n* that denote the availability and the prices of the houses. If house *i* is occupied, and therefore cannot be bought, then *a**i* equals 0. Otherwise, house *i* can be bought, and *a**i* represents the money required to buy it, in dollars.
As Zane has only *k* dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love.
Input Specification:
The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — denoting the availability and the prices of the houses.
It is guaranteed that *a**m*<==<=0 and that it is possible to purchase some house with no more than *k* dollars.
Output Specification:
Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy.
Demo Input:
['5 1 20\n0 27 32 21 19\n', '7 3 50\n62 0 0 0 99 33 22\n', '10 5 100\n1 0 1 0 0 0 0 0 1 1\n']
Demo Output:
['40', '30', '20']
Note:
In the first sample, with *k* = 20 dollars, Zane can buy only house 5. The distance from house *m* = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.
In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house *m* = 3 and house 6 are only 30 meters away, while house *m* = 3 and house 7 are 40 meters away. | ```python
__author__ = "Daniel Mitre"
n, m, k = map(int, input().split())
prices = map(int, input().split())
def dist_to_girl(pos: int) -> int:
return abs(pos - m)
buy = float('inf')
for idx, price in zip(range(1, n + 1), prices):
if (price > 0) and (price <= k) and (dist_to_girl(idx) < dist_to_girl(buy)):
buy = idx
print(dist_to_girl(buy) * 10)
``` | 3 |
|
603 | A | Alternative Thinking | PROGRAMMING | 1,600 | [
"dp",
"greedy",
"math"
] | null | null | Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have. | The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000).
The following line contains a binary string of length *n* representing Kevin's results on the USAICO. | Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring. | [
"8\n10000011\n",
"2\n01\n"
] | [
"5\n",
"2\n"
] | In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score. | 500 | [
{
"input": "8\n10000011",
"output": "5"
},
{
"input": "2\n01",
"output": "2"
},
{
"input": "5\n10101",
"output": "5"
},
{
"input": "75\n010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "75"
},
{
"input": "11\n00000000000",
"output": "3"
},
{
"input": "56\n10101011010101010101010101010101010101011010101010101010",
"output": "56"
},
{
"input": "50\n01011010110101010101010101010101010101010101010100",
"output": "49"
},
{
"input": "7\n0110100",
"output": "7"
},
{
"input": "8\n11011111",
"output": "5"
},
{
"input": "6\n000000",
"output": "3"
},
{
"input": "5\n01000",
"output": "5"
},
{
"input": "59\n10101010101010101010101010101010101010101010101010101010101",
"output": "59"
},
{
"input": "88\n1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "88"
},
{
"input": "93\n010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "93"
},
{
"input": "70\n0101010101010101010101010101010101010101010101010101010101010101010101",
"output": "70"
},
{
"input": "78\n010101010101010101010101010101101010101010101010101010101010101010101010101010",
"output": "78"
},
{
"input": "83\n10101010101010101010101010101010101010101010101010110101010101010101010101010101010",
"output": "83"
},
{
"input": "87\n101010101010101010101010101010101010101010101010101010101010101010101010101010010101010",
"output": "87"
},
{
"input": "65\n01010101101010101010101010101010101010101010101010101010101010101",
"output": "65"
},
{
"input": "69\n010101010101010101101010101010101010101010101010101010101010101010101",
"output": "69"
},
{
"input": "74\n01010101010101010101010101010101010101010101010101010101010101000101010101",
"output": "74"
},
{
"input": "77\n01010101010101001010101010101010100101010101010101010101010101010101010101010",
"output": "77"
},
{
"input": "60\n101010110101010101010101010110101010101010101010101010101010",
"output": "60"
},
{
"input": "89\n01010101010101010101010101010101010101010101010101010101101010101010101010100101010101010",
"output": "89"
},
{
"input": "68\n01010101010101010101010101010101010100101010100101010101010100101010",
"output": "67"
},
{
"input": "73\n0101010101010101010101010101010101010101010111011010101010101010101010101",
"output": "72"
},
{
"input": "55\n1010101010101010010101010101101010101010101010100101010",
"output": "54"
},
{
"input": "85\n1010101010101010101010101010010101010101010101101010101010101010101011010101010101010",
"output": "84"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1111111111",
"output": "3"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "2\n00",
"output": "2"
},
{
"input": "3\n000",
"output": "3"
},
{
"input": "3\n001",
"output": "3"
},
{
"input": "3\n010",
"output": "3"
},
{
"input": "3\n011",
"output": "3"
},
{
"input": "3\n100",
"output": "3"
},
{
"input": "3\n101",
"output": "3"
},
{
"input": "3\n110",
"output": "3"
},
{
"input": "3\n111",
"output": "3"
},
{
"input": "4\n0000",
"output": "3"
},
{
"input": "4\n0001",
"output": "4"
},
{
"input": "4\n0010",
"output": "4"
},
{
"input": "4\n0011",
"output": "4"
},
{
"input": "4\n0100",
"output": "4"
},
{
"input": "4\n0101",
"output": "4"
},
{
"input": "4\n0110",
"output": "4"
},
{
"input": "4\n0111",
"output": "4"
},
{
"input": "4\n1000",
"output": "4"
},
{
"input": "4\n1001",
"output": "4"
},
{
"input": "4\n1010",
"output": "4"
},
{
"input": "4\n1011",
"output": "4"
},
{
"input": "4\n1100",
"output": "4"
},
{
"input": "4\n1101",
"output": "4"
},
{
"input": "4\n1110",
"output": "4"
},
{
"input": "4\n1111",
"output": "3"
},
{
"input": "5\n00000",
"output": "3"
},
{
"input": "5\n00001",
"output": "4"
},
{
"input": "5\n00010",
"output": "5"
},
{
"input": "5\n00011",
"output": "4"
},
{
"input": "5\n00100",
"output": "5"
},
{
"input": "5\n00101",
"output": "5"
},
{
"input": "5\n00110",
"output": "5"
},
{
"input": "5\n00111",
"output": "4"
},
{
"input": "5\n01000",
"output": "5"
},
{
"input": "5\n01001",
"output": "5"
},
{
"input": "5\n01010",
"output": "5"
},
{
"input": "5\n01011",
"output": "5"
},
{
"input": "5\n01100",
"output": "5"
},
{
"input": "5\n01101",
"output": "5"
},
{
"input": "5\n01110",
"output": "5"
},
{
"input": "5\n01111",
"output": "4"
},
{
"input": "5\n10000",
"output": "4"
},
{
"input": "5\n10001",
"output": "5"
},
{
"input": "5\n10010",
"output": "5"
},
{
"input": "5\n10100",
"output": "5"
},
{
"input": "5\n10101",
"output": "5"
},
{
"input": "5\n10110",
"output": "5"
},
{
"input": "5\n10111",
"output": "5"
},
{
"input": "5\n11000",
"output": "4"
},
{
"input": "5\n11001",
"output": "5"
},
{
"input": "5\n11010",
"output": "5"
},
{
"input": "5\n11011",
"output": "5"
},
{
"input": "5\n11100",
"output": "4"
},
{
"input": "5\n11101",
"output": "5"
},
{
"input": "5\n11110",
"output": "4"
},
{
"input": "5\n11111",
"output": "3"
}
] | 1,492,179,121 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 46 | 4,608,000 | #! /bin/python
n = int(input())
tab = str(input())
d = 1
tmp = 1
changes = 0
for i in range(1, n):
if tab[i] == tab[i - 1]:
tmp += 1
d = max(tmp, d)
else:
changes += 1
tmp = 1
if d >= 3:
changes += 3
elif d == 1:
changes -= 1
print(changes)
| Title: Alternative Thinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length *n*. Each character of Kevin's string represents Kevin's score on one of the *n* questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0,<=1,<=0,<=1}, {1,<=0,<=1}, and {1,<=0,<=1,<=0} are alternating sequences, while {1,<=0,<=0} and {0,<=1,<=0,<=1,<=1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input Specification:
The first line contains the number of questions on the olympiad *n* (1<=≤<=*n*<=≤<=100<=000).
The following line contains a binary string of length *n* representing Kevin's results on the USAICO.
Output Specification:
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Demo Input:
['8\n10000011\n', '2\n01\n']
Demo Output:
['5\n', '2\n']
Note:
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score. | ```python
#! /bin/python
n = int(input())
tab = str(input())
d = 1
tmp = 1
changes = 0
for i in range(1, n):
if tab[i] == tab[i - 1]:
tmp += 1
d = max(tmp, d)
else:
changes += 1
tmp = 1
if d >= 3:
changes += 3
elif d == 1:
changes -= 1
print(changes)
``` | 0 |
|
879 | A | Borya's Diagnosis | PROGRAMMING | 900 | [
"implementation"
] | null | null | It seems that Borya is seriously sick. He is going visit *n* doctors to find out the exact diagnosis. Each of the doctors needs the information about all previous visits, so Borya has to visit them in the prescribed order (i.e. Borya should first visit doctor 1, then doctor 2, then doctor 3 and so on). Borya will get the information about his health from the last doctor.
Doctors have a strange working schedule. The doctor *i* goes to work on the *s**i*-th day and works every *d**i* day. So, he works on days *s**i*,<=*s**i*<=+<=*d**i*,<=*s**i*<=+<=2*d**i*,<=....
The doctor's appointment takes quite a long time, so Borya can not see more than one doctor per day. What is the minimum time he needs to visit all doctors? | First line contains an integer *n* — number of doctors (1<=≤<=*n*<=≤<=1000).
Next *n* lines contain two numbers *s**i* and *d**i* (1<=≤<=*s**i*,<=*d**i*<=≤<=1000). | Output a single integer — the minimum day at which Borya can visit the last doctor. | [
"3\n2 2\n1 2\n2 2\n",
"2\n10 1\n6 5\n"
] | [
"4\n",
"11\n"
] | In the first sample case, Borya can visit all doctors on days 2, 3 and 4.
In the second sample case, Borya can visit all doctors on days 10 and 11. | 500 | [
{
"input": "3\n2 2\n1 2\n2 2",
"output": "4"
},
{
"input": "2\n10 1\n6 5",
"output": "11"
},
{
"input": "3\n6 10\n3 3\n8 2",
"output": "10"
},
{
"input": "4\n4 8\n10 10\n4 2\n8 2",
"output": "14"
},
{
"input": "5\n7 1\n5 1\n6 1\n1 6\n6 8",
"output": "14"
},
{
"input": "6\n1 3\n2 5\n4 7\n7 5\n6 8\n8 8",
"output": "16"
},
{
"input": "10\n4 10\n8 7\n6 5\n2 1\n2 3\n8 8\n2 4\n2 2\n6 7\n7 9",
"output": "34"
},
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "1\n1000 1000",
"output": "1000"
},
{
"input": "42\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2",
"output": "83"
},
{
"input": "2\n5 5\n5 1",
"output": "6"
},
{
"input": "2\n5 5\n5 5",
"output": "10"
},
{
"input": "2\n1 1\n1 1",
"output": "2"
},
{
"input": "2\n1 6\n7 1",
"output": "7"
},
{
"input": "2\n4 3\n4 5",
"output": "9"
},
{
"input": "2\n1 2\n1 3",
"output": "4"
},
{
"input": "3\n2 3\n5 1\n2 1",
"output": "6"
},
{
"input": "2\n2 1\n6 3",
"output": "6"
},
{
"input": "3\n10 1\n4 4\n12 1",
"output": "13"
},
{
"input": "2\n2 2\n10 2",
"output": "10"
},
{
"input": "2\n1 1\n1000 2",
"output": "1000"
},
{
"input": "14\n1000 1\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1 1",
"output": "13001"
},
{
"input": "2\n2 4\n2 1",
"output": "3"
},
{
"input": "3\n1 100\n100 3\n200 1",
"output": "200"
},
{
"input": "7\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1 1",
"output": "6001"
},
{
"input": "2\n5 5\n15 5",
"output": "15"
},
{
"input": "2\n2 2\n2 4",
"output": "6"
},
{
"input": "2\n1 1\n10 1",
"output": "10"
},
{
"input": "2\n10 1\n100 1",
"output": "100"
},
{
"input": "15\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1 1",
"output": "14001"
},
{
"input": "3\n2 1\n1 3\n4 7",
"output": "11"
},
{
"input": "2\n5 5\n100 5",
"output": "100"
},
{
"input": "2\n1 10\n2 30",
"output": "2"
},
{
"input": "4\n2 2\n2 2\n2 2\n2 2",
"output": "8"
},
{
"input": "1\n10 10",
"output": "10"
}
] | 1,659,459,794 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | total = 0
for _ in range(int(input())):
s, d = map(int, input().split())
if total < s:
total = s
elif s <= total:
i = 1
while (s+ d*i) <= total:
i +=1
total = (s+d*i)
print(total) | Title: Borya's Diagnosis
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It seems that Borya is seriously sick. He is going visit *n* doctors to find out the exact diagnosis. Each of the doctors needs the information about all previous visits, so Borya has to visit them in the prescribed order (i.e. Borya should first visit doctor 1, then doctor 2, then doctor 3 and so on). Borya will get the information about his health from the last doctor.
Doctors have a strange working schedule. The doctor *i* goes to work on the *s**i*-th day and works every *d**i* day. So, he works on days *s**i*,<=*s**i*<=+<=*d**i*,<=*s**i*<=+<=2*d**i*,<=....
The doctor's appointment takes quite a long time, so Borya can not see more than one doctor per day. What is the minimum time he needs to visit all doctors?
Input Specification:
First line contains an integer *n* — number of doctors (1<=≤<=*n*<=≤<=1000).
Next *n* lines contain two numbers *s**i* and *d**i* (1<=≤<=*s**i*,<=*d**i*<=≤<=1000).
Output Specification:
Output a single integer — the minimum day at which Borya can visit the last doctor.
Demo Input:
['3\n2 2\n1 2\n2 2\n', '2\n10 1\n6 5\n']
Demo Output:
['4\n', '11\n']
Note:
In the first sample case, Borya can visit all doctors on days 2, 3 and 4.
In the second sample case, Borya can visit all doctors on days 10 and 11. | ```python
total = 0
for _ in range(int(input())):
s, d = map(int, input().split())
if total < s:
total = s
elif s <= total:
i = 1
while (s+ d*i) <= total:
i +=1
total = (s+d*i)
print(total)
``` | 0 |
|
78 | A | Haiku | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Haiku | 2 | 256 | Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not. | The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification. | Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes). | [
"on codeforces \nbeta round is running\n a rustling of keys \n",
"how many gallons\nof edo s rain did you drink\n cuckoo\n"
] | [
"YES",
"NO"
] | none | 500 | [
{
"input": "on codeforces \nbeta round is running\n a rustling of keys ",
"output": "YES"
},
{
"input": "how many gallons\nof edo s rain did you drink\n cuckoo",
"output": "NO"
},
{
"input": " hatsu shigure\n saru mo komino wo\nhoshige nari",
"output": "YES"
},
{
"input": "o vetus stagnum\n rana de ripa salit\n ac sonant aquae",
"output": "NO"
},
{
"input": " furuike ya\nkawazu tobikomu\nmizu no oto ",
"output": "YES"
},
{
"input": " noch da leich\na stamperl zum aufwaerma\n da pfarrer kimmt a ",
"output": "NO"
},
{
"input": " sommerfuglene \n hvorfor bruge mange ord\n et kan gore det",
"output": "YES"
},
{
"input": " ab der mittagszeit\n ist es etwas schattiger\n ein wolkenhimmel",
"output": "NO"
},
{
"input": "tornando a vederli\ni fiori di ciliegio la sera\nson divenuti frutti",
"output": "NO"
},
{
"input": "kutaburete\nyado karu koro ya\nfuji no hana",
"output": "YES"
},
{
"input": " beginnings of poetry\n the rice planting songs \n of the interior",
"output": "NO"
},
{
"input": " door zomerregens\n zijn de kraanvogelpoten\n korter geworden",
"output": "NO"
},
{
"input": " derevo na srub\na ptitsi bezzabotno\n gnezdishko tam vyut",
"output": "YES"
},
{
"input": "writing in the dark\nunaware that my pen\nhas run out of ink",
"output": "NO"
},
{
"input": "kusaaiu\nuieueua\nuo efaa",
"output": "YES"
},
{
"input": "v\nh\np",
"output": "NO"
},
{
"input": "i\ni\nu",
"output": "NO"
},
{
"input": "awmio eoj\nabdoolceegood\nwaadeuoy",
"output": "YES"
},
{
"input": "xzpnhhnqsjpxdboqojixmofawhdjcfbscq\nfoparnxnbzbveycoltwdrfbwwsuobyoz hfbrszy\nimtqryscsahrxpic agfjh wvpmczjjdrnwj mcggxcdo",
"output": "YES"
},
{
"input": "wxjcvccp cppwsjpzbd dhizbcnnllckybrnfyamhgkvkjtxxfzzzuyczmhedhztugpbgpvgh\nmdewztdoycbpxtp bsiw hknggnggykdkrlihvsaykzfiiw\ndewdztnngpsnn lfwfbvnwwmxoojknygqb hfe ibsrxsxr",
"output": "YES"
},
{
"input": "nbmtgyyfuxdvrhuhuhpcfywzrbclp znvxw synxmzymyxcntmhrjriqgdjh xkjckydbzjbvtjurnf\nhhnhxdknvamywhsrkprofnyzlcgtdyzzjdsfxyddvilnzjziz qmwfdvzckgcbrrxplxnxf mpxwxyrpesnewjrx ajxlfj\nvcczq hddzd cvefmhxwxxyqcwkr fdsndckmesqeq zyjbwbnbyhybd cta nsxzidl jpcvtzkldwd",
"output": "YES"
},
{
"input": "rvwdsgdsrutgjwscxz pkd qtpmfbqsmctuevxdj kjzknzghdvxzlaljcntg jxhvzn yciktbsbyscfypx x xhkxnfpdp\nwdfhvqgxbcts mnrwbr iqttsvigwdgvlxwhsmnyxnttedonxcfrtmdjjmacvqtkbmsnwwvvrlxwvtggeowtgsqld qj\nvsxcdhbzktrxbywpdvstr meykarwtkbm pkkbhvwvelclfmpngzxdmblhcvf qmabmweldplmczgbqgzbqnhvcdpnpjtch ",
"output": "YES"
},
{
"input": "brydyfsmtzzkpdsqvvztmprhqzbzqvgsblnz naait tdtiprjsttwusdykndwcccxfmzmrmfmzjywkpgbfnjpypgcbcfpsyfj k\nucwdfkfyxxxht lxvnovqnnsqutjsyagrplb jhvtwdptrwcqrovncdvqljjlrpxcfbxqgsfylbgmcjpvpl ccbcybmigpmjrxpu\nfgwtpcjeywgnxgbttgx htntpbk tkkpwbgxwtbxvcpkqbzetjdkcwad tftnjdxxjdvbpfibvxuglvx llyhgjvggtw jtjyphs",
"output": "YES"
},
{
"input": "nyc aqgqzjjlj mswgmjfcxlqdscheskchlzljlsbhyn iobxymwzykrsnljj\nnnebeaoiraga\nqpjximoqzswhyyszhzzrhfwhf iyxysdtcpmikkwpugwlxlhqfkn",
"output": "NO"
},
{
"input": "lzrkztgfe mlcnq ay ydmdzxh cdgcghxnkdgmgfzgahdjjmqkpdbskreswpnblnrc fmkwziiqrbskp\np oukeaz gvvy kghtrjlczyl qeqhgfgfej\nwfolhkmktvsjnrpzfxcxzqmfidtlzmuhxac wsncjgmkckrywvxmnjdpjpfydhk qlmdwphcvyngansqhl",
"output": "NO"
},
{
"input": "yxcboqmpwoevrdhvpxfzqmammak\njmhphkxppkqkszhqqtkvflarsxzla pbxlnnnafqbsnmznfj qmhoktgzix qpmrgzxqvmjxhskkksrtryehfnmrt dtzcvnvwp\nscwymuecjxhw rdgsffqywwhjpjbfcvcrnisfqllnbplpadfklayjguyvtrzhwblftclfmsr",
"output": "NO"
},
{
"input": "qfdwsr jsbrpfmn znplcx nhlselflytndzmgxqpgwhpi ghvbbxrkjdirfghcybhkkqdzmyacvrrcgsneyjlgzfvdmxyjmph\nylxlyrzs drbktzsniwcbahjkgohcghoaczsmtzhuwdryjwdijmxkmbmxv yyfrokdnsx\nyw xtwyzqlfxwxghugoyscqlx pljtz aldfskvxlsxqgbihzndhxkswkxqpwnfcxzfyvncstfpqf",
"output": "NO"
},
{
"input": "g rguhqhcrzmuqthtmwzhfyhpmqzzosa\nmhjimzvchkhejh irvzejhtjgaujkqfxhpdqjnxr dvqallgssktqvsxi\npcwbliftjcvuzrsqiswohi",
"output": "NO"
},
{
"input": " ngxtlq iehiise vgffqcpnmsoqzyseuqqtggokymol zn\nvjdjljazeujwoubkcvtsbepooxqzrueaauokhepiquuopfild\ngoabauauaeotoieufueeknudiilupouaiaexcoapapu",
"output": "NO"
},
{
"input": "ycnvnnqk mhrmhctpkfbc qbyvtjznmndqjzgbcxmvrpkfcll zwspfptmbxgrdv dsgkk nfytsqjrnfbhh pzdldzymvkdxxwh\nvnhjfwgdnyjptsmblyxmpzylsbjlmtkkwjcbqwjctqvrlqqkdsrktxlnslspvnn mdgsmzblhbnvpczmqkcffwhwljqkzmk hxcm\nrghnjvzcpprrgmtgytpkzyc mrdnnhpkwypwqbtzjyfwvrdwyjltbzxtbstzs xdjzdmx yjsqtzlrnvyssvglsdjrmsrfrcdpqt",
"output": "NO"
},
{
"input": "ioeeaioeiuoeaeieuuieooaouiuouiioaueeaiaiuoaoiioeeaauooiuuieeuaeeoauieeaiuoieiaieuoauaaoioooieueueuai\nuooaoeeaoiuuoeioaoouaououoeioiaeueoioaiouaeaoioiuuaueeuaiuoiueoiuaoeeieeouaeeaeeieioeoiiieuuueuuieuo\naeeouieeieoueaioeoioooiouaeeeiaaioueauaueiouuuaieuuioiaeiueauueaoieauauoeueuaiueuuaueeoueauaeaoieeoo",
"output": "NO"
},
{
"input": "mkgycgmrqtmlwj vddlbsgdptyfrcj nmrddjchydvbsyhgzytfzvwbhqmmhkpsyrcnknpsqrr wcnf wfvgskrnsmnrcqgcnc q\npn cclcyjjdp rqkpgyrvyyd hhynrochfa pthyffqvskazfshvzcmvhmsrxvquzkdvivlbfypwbgltfbjmhzljvgzmxjlrjdjq\nauuouieuaaauoooeoiuoaueuueauiuuuiaeuwuoiaeoeaieauieouoaeiiieiioiuuuiuauioiiuoeauueieaueiaioeuaoaaau",
"output": "NO"
},
{
"input": "ab\na\na",
"output": "NO"
},
{
"input": "on codeforcs\nbeta round is runnineg\na rustling of keys",
"output": "NO"
},
{
"input": "coooooooooooooooood\nd\nd",
"output": "NO"
},
{
"input": "on codeforces \nbeta round is runninog\n a rustling of keys",
"output": "NO"
},
{
"input": "aaaaaaaaaaaaaaa\na\na",
"output": "NO"
},
{
"input": "aaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\naaaaaccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc",
"output": "YES"
},
{
"input": "cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\ncccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc\ncccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc",
"output": "NO"
},
{
"input": "aaaaaa\naaaaaa\naaaaa",
"output": "NO"
}
] | 1,644,845,769 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 92 | 0 | n1 = input()
n2 = input()
n3 = input()
l1 = list()
l2 = list()
l3 = list()
v = "auioe"
n1.strip().split()
n2.strip().split()
n3.strip().split()
n1 = "".join(n1)
n2 = "".join(n2)
n3 = "".join(n3)
n1 = list(n1)
n2 = list(n2)
n3 = list(n3)
for i in n1:
if i in v:
l1.append(i)
for j in n2:
if j in v:
l2.append(j)
for h in n3:
if h in v:
l3.append(h)
if len(l1) == 5 and len(l2) == 7 and len(l3) == 5:
print("YES")
else:
print("NO") | Title: Haiku
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input Specification:
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output Specification:
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Demo Input:
['on codeforces \nbeta round is running\n a rustling of keys \n', 'how many gallons\nof edo s rain did you drink\n cuckoo\n']
Demo Output:
['YES', 'NO']
Note:
none | ```python
n1 = input()
n2 = input()
n3 = input()
l1 = list()
l2 = list()
l3 = list()
v = "auioe"
n1.strip().split()
n2.strip().split()
n3.strip().split()
n1 = "".join(n1)
n2 = "".join(n2)
n3 = "".join(n3)
n1 = list(n1)
n2 = list(n2)
n3 = list(n3)
for i in n1:
if i in v:
l1.append(i)
for j in n2:
if j in v:
l2.append(j)
for h in n3:
if h in v:
l3.append(h)
if len(l1) == 5 and len(l2) == 7 and len(l3) == 5:
print("YES")
else:
print("NO")
``` | 3.977 |
975 | B | Mancala | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes.
Initially, each hole has $a_i$ stones. When a player makes a move, he chooses a hole which contains a positive number of stones. He takes all the stones inside it and then redistributes these stones one by one in the next holes in a counter-clockwise direction.
Note that the counter-clockwise order means if the player takes the stones from hole $i$, he will put one stone in the $(i+1)$-th hole, then in the $(i+2)$-th, etc. If he puts a stone in the $14$-th hole, the next one will be put in the first hole.
After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli.
Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move. | The only line contains 14 integers $a_1, a_2, \ldots, a_{14}$ ($0 \leq a_i \leq 10^9$) — the number of stones in each hole.
It is guaranteed that for any $i$ ($1\leq i \leq 14$) $a_i$ is either zero or odd, and there is at least one stone in the board. | Output one integer, the maximum possible score after one move. | [
"0 1 1 0 0 0 0 0 0 7 0 0 0 0\n",
"5 1 1 1 1 0 0 0 0 0 0 0 0 0\n"
] | [
"4\n",
"8\n"
] | In the first test case the board after the move from the hole with $7$ stones will look like 1 2 2 0 0 0 0 0 0 0 1 1 1 1. Then the player collects the even numbers and ends up with a score equal to $4$. | 1,000 | [
{
"input": "0 1 1 0 0 0 0 0 0 7 0 0 0 0",
"output": "4"
},
{
"input": "5 1 1 1 1 0 0 0 0 0 0 0 0 0",
"output": "8"
},
{
"input": "10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 10001 1",
"output": "54294"
},
{
"input": "0 0 0 0 0 0 0 0 0 0 0 0 0 15",
"output": "2"
},
{
"input": "1 0 0 0 0 1 0 0 0 0 1 0 0 0",
"output": "0"
},
{
"input": "5 5 1 1 1 3 3 3 5 7 5 3 7 5",
"output": "38"
},
{
"input": "787 393 649 463 803 365 81 961 989 531 303 407 579 915",
"output": "7588"
},
{
"input": "8789651 4466447 1218733 6728667 1796977 6198853 8263135 6309291 8242907 7136751 3071237 5397369 6780785 9420869",
"output": "81063456"
},
{
"input": "0 0 0 0 0 0 0 0 0 0 0 0 0 29",
"output": "26"
},
{
"input": "282019717 109496191 150951267 609856495 953855615 569750143 6317733 255875779 645191029 572053369 290936613 338480779 879775193 177172893",
"output": "5841732816"
},
{
"input": "105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505 105413505",
"output": "120472578"
},
{
"input": "404418821 993626161 346204297 122439813 461187221 628048227 625919459 628611733 938993057 701270099 398043779 684205961 630975553 575964835",
"output": "8139909016"
},
{
"input": "170651077 730658441 824213789 583764177 129437345 717005779 675398017 314979709 380861369 265878463 746564659 797260041 506575735 335169317",
"output": "6770880638"
},
{
"input": "622585025 48249287 678950449 891575125 637411965 457739735 829353393 235216425 284006447 875591469 492839209 296444305 513776057 810057753",
"output": "7673796644"
},
{
"input": "475989857 930834747 786217439 927967137 489188151 869354161 276693267 56154399 131055697 509249443 143116853 426254423 44465165 105798821",
"output": "6172339560"
},
{
"input": "360122921 409370351 226220005 604004145 85173909 600403773 624052991 138163383 729239967 189036661 619842883 270087537 749500483 243727913",
"output": "5848946922"
},
{
"input": "997102881 755715147 273805839 436713689 547411799 72470207 522269145 647688957 137422311 422612659 197751751 679663349 821420227 387967237",
"output": "6900015198"
},
{
"input": "690518849 754551537 652949719 760695679 491633619 477564457 11669279 700467439 470069297 782338983 718169393 884421719 24619427 215745577",
"output": "7635414974"
},
{
"input": "248332749 486342237 662201929 917696895 555278549 252122023 850296207 463343655 832574345 954281071 168282553 825538865 996753493 461254663",
"output": "6400166934"
},
{
"input": "590789361 636464947 404477303 337309187 476703809 426863069 120608741 703406277 645444697 761482231 996635839 33459441 677458865 483861751",
"output": "7294965518"
},
{
"input": "297857621 238127103 749085829 139033277 597985489 202617713 982184715 183932743 278551059 297781685 330124279 338959601 682874531 187519685",
"output": "5201808164"
},
{
"input": "1 1 1 1 1 0 0 0 0 0 0 0 0 0",
"output": "2"
},
{
"input": "1 1 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "2"
},
{
"input": "1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "1 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "2"
},
{
"input": "0 0 0 0 0 0 0 0 0 0 0 0 1 1",
"output": "2"
}
] | 1,645,251,626 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 46 | 409,600 | from re import I
a = list(map(int,input().split()))
max = 0
for i in range(14):
if a[i]>0:
j = i+1
add_to_every = a[i]//14
rem = a[i]%14
temp =[x for x in a]
temp[i]=0
temp = [x+add_to_every for x in temp]
while rem>0:
temp[j%14]+=1
rem-=1
j+=1
sum = 0
for i in range(14):
if temp[i]%2==0:
sum+=temp[i]
#print(temp)
if sum>max:
max=sum
print(max) | Title: Mancala
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes.
Initially, each hole has $a_i$ stones. When a player makes a move, he chooses a hole which contains a positive number of stones. He takes all the stones inside it and then redistributes these stones one by one in the next holes in a counter-clockwise direction.
Note that the counter-clockwise order means if the player takes the stones from hole $i$, he will put one stone in the $(i+1)$-th hole, then in the $(i+2)$-th, etc. If he puts a stone in the $14$-th hole, the next one will be put in the first hole.
After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli.
Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move.
Input Specification:
The only line contains 14 integers $a_1, a_2, \ldots, a_{14}$ ($0 \leq a_i \leq 10^9$) — the number of stones in each hole.
It is guaranteed that for any $i$ ($1\leq i \leq 14$) $a_i$ is either zero or odd, and there is at least one stone in the board.
Output Specification:
Output one integer, the maximum possible score after one move.
Demo Input:
['0 1 1 0 0 0 0 0 0 7 0 0 0 0\n', '5 1 1 1 1 0 0 0 0 0 0 0 0 0\n']
Demo Output:
['4\n', '8\n']
Note:
In the first test case the board after the move from the hole with $7$ stones will look like 1 2 2 0 0 0 0 0 0 0 1 1 1 1. Then the player collects the even numbers and ends up with a score equal to $4$. | ```python
from re import I
a = list(map(int,input().split()))
max = 0
for i in range(14):
if a[i]>0:
j = i+1
add_to_every = a[i]//14
rem = a[i]%14
temp =[x for x in a]
temp[i]=0
temp = [x+add_to_every for x in temp]
while rem>0:
temp[j%14]+=1
rem-=1
j+=1
sum = 0
for i in range(14):
if temp[i]%2==0:
sum+=temp[i]
#print(temp)
if sum>max:
max=sum
print(max)
``` | 3 |
|
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,675,472,617 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 4,300,800 | import math
input (n)
input (m)
input (a)
X = max(math.ceil(n/a) , (n / a))
Y = max(math.ceil(m/a) , (m / a))
print (X * Y) | 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
import math
input (n)
input (m)
input (a)
X = max(math.ceil(n/a) , (n / a))
Y = max(math.ceil(m/a) , (m / a))
print (X * Y)
``` | -1 |
609 | A | USB Flash Drives | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives. | The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*. | Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives. | [
"3\n5\n2\n1\n3\n",
"3\n6\n2\n3\n2\n",
"2\n5\n5\n10\n"
] | [
"2\n",
"3\n",
"1\n"
] | In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second. | 0 | [
{
"input": "3\n5\n2\n1\n3",
"output": "2"
},
{
"input": "3\n6\n2\n3\n2",
"output": "3"
},
{
"input": "2\n5\n5\n10",
"output": "1"
},
{
"input": "5\n16\n8\n1\n3\n4\n9",
"output": "2"
},
{
"input": "10\n121\n10\n37\n74\n56\n42\n39\n6\n68\n8\n100",
"output": "2"
},
{
"input": "12\n4773\n325\n377\n192\n780\n881\n816\n839\n223\n215\n125\n952\n8",
"output": "7"
},
{
"input": "15\n7758\n182\n272\n763\n910\n24\n359\n583\n890\n735\n819\n66\n992\n440\n496\n227",
"output": "15"
},
{
"input": "30\n70\n6\n2\n10\n4\n7\n10\n5\n1\n8\n10\n4\n3\n5\n9\n3\n6\n6\n4\n2\n6\n5\n10\n1\n9\n7\n2\n1\n10\n7\n5",
"output": "8"
},
{
"input": "40\n15705\n702\n722\n105\n873\n417\n477\n794\n300\n869\n496\n572\n232\n456\n298\n473\n584\n486\n713\n934\n121\n303\n956\n934\n840\n358\n201\n861\n497\n131\n312\n957\n96\n914\n509\n60\n300\n722\n658\n820\n103",
"output": "21"
},
{
"input": "50\n18239\n300\n151\n770\n9\n200\n52\n247\n753\n523\n263\n744\n463\n540\n244\n608\n569\n771\n32\n425\n777\n624\n761\n628\n124\n405\n396\n726\n626\n679\n237\n229\n49\n512\n18\n671\n290\n768\n632\n739\n18\n136\n413\n117\n83\n413\n452\n767\n664\n203\n404",
"output": "31"
},
{
"input": "70\n149\n5\n3\n3\n4\n6\n1\n2\n9\n8\n3\n1\n8\n4\n4\n3\n6\n10\n7\n1\n10\n8\n4\n9\n3\n8\n3\n2\n5\n1\n8\n6\n9\n10\n4\n8\n6\n9\n9\n9\n3\n4\n2\n2\n5\n8\n9\n1\n10\n3\n4\n3\n1\n9\n3\n5\n1\n3\n7\n6\n9\n8\n9\n1\n7\n4\n4\n2\n3\n5\n7",
"output": "17"
},
{
"input": "70\n2731\n26\n75\n86\n94\n37\n25\n32\n35\n92\n1\n51\n73\n53\n66\n16\n80\n15\n81\n100\n87\n55\n48\n30\n71\n39\n87\n77\n25\n70\n22\n75\n23\n97\n16\n75\n95\n61\n61\n28\n10\n78\n54\n80\n51\n25\n24\n90\n58\n4\n77\n40\n54\n53\n47\n62\n30\n38\n71\n97\n71\n60\n58\n1\n21\n15\n55\n99\n34\n88\n99",
"output": "35"
},
{
"input": "70\n28625\n34\n132\n181\n232\n593\n413\n862\n887\n808\n18\n35\n89\n356\n640\n339\n280\n975\n82\n345\n398\n948\n372\n91\n755\n75\n153\n948\n603\n35\n694\n722\n293\n363\n884\n264\n813\n175\n169\n646\n138\n449\n488\n828\n417\n134\n84\n763\n288\n845\n801\n556\n972\n332\n564\n934\n699\n842\n942\n644\n203\n406\n140\n37\n9\n423\n546\n675\n491\n113\n587",
"output": "45"
},
{
"input": "80\n248\n3\n9\n4\n5\n10\n7\n2\n6\n2\n2\n8\n2\n1\n3\n7\n9\n2\n8\n4\n4\n8\n5\n4\n4\n10\n2\n1\n4\n8\n4\n10\n1\n2\n10\n2\n3\n3\n1\n1\n8\n9\n5\n10\n2\n8\n10\n5\n3\n6\n1\n7\n8\n9\n10\n5\n10\n10\n2\n10\n1\n2\n4\n1\n9\n4\n7\n10\n8\n5\n8\n1\n4\n2\n2\n3\n9\n9\n9\n10\n6",
"output": "27"
},
{
"input": "80\n2993\n18\n14\n73\n38\n14\n73\n77\n18\n81\n6\n96\n65\n77\n86\n76\n8\n16\n81\n83\n83\n34\n69\n58\n15\n19\n1\n16\n57\n95\n35\n5\n49\n8\n15\n47\n84\n99\n94\n93\n55\n43\n47\n51\n61\n57\n13\n7\n92\n14\n4\n83\n100\n60\n75\n41\n95\n74\n40\n1\n4\n95\n68\n59\n65\n15\n15\n75\n85\n46\n77\n26\n30\n51\n64\n75\n40\n22\n88\n68\n24",
"output": "38"
},
{
"input": "80\n37947\n117\n569\n702\n272\n573\n629\n90\n337\n673\n589\n576\n205\n11\n284\n645\n719\n777\n271\n567\n466\n251\n402\n3\n97\n288\n699\n208\n173\n530\n782\n266\n395\n957\n159\n463\n43\n316\n603\n197\n386\n132\n799\n778\n905\n784\n71\n851\n963\n883\n705\n454\n275\n425\n727\n223\n4\n870\n833\n431\n463\n85\n505\n800\n41\n954\n981\n242\n578\n336\n48\n858\n702\n349\n929\n646\n528\n993\n506\n274\n227",
"output": "70"
},
{
"input": "90\n413\n5\n8\n10\n7\n5\n7\n5\n7\n1\n7\n8\n4\n3\n9\n4\n1\n10\n3\n1\n10\n9\n3\n1\n8\n4\n7\n5\n2\n9\n3\n10\n10\n3\n6\n3\n3\n10\n7\n5\n1\n1\n2\n4\n8\n2\n5\n5\n3\n9\n5\n5\n3\n10\n2\n3\n8\n5\n9\n1\n3\n6\n5\n9\n2\n3\n7\n10\n3\n4\n4\n1\n5\n9\n2\n6\n9\n1\n1\n9\n9\n7\n7\n7\n8\n4\n5\n3\n4\n6\n9",
"output": "59"
},
{
"input": "90\n4226\n33\n43\n83\n46\n75\n14\n88\n36\n8\n25\n47\n4\n96\n19\n33\n49\n65\n17\n59\n72\n1\n55\n94\n92\n27\n33\n39\n14\n62\n79\n12\n89\n22\n86\n13\n19\n77\n53\n96\n74\n24\n25\n17\n64\n71\n81\n87\n52\n72\n55\n49\n74\n36\n65\n86\n91\n33\n61\n97\n38\n87\n61\n14\n73\n95\n43\n67\n42\n67\n22\n12\n62\n32\n96\n24\n49\n82\n46\n89\n36\n75\n91\n11\n10\n9\n33\n86\n28\n75\n39",
"output": "64"
},
{
"input": "90\n40579\n448\n977\n607\n745\n268\n826\n479\n59\n330\n609\n43\n301\n970\n726\n172\n632\n600\n181\n712\n195\n491\n312\n849\n722\n679\n682\n780\n131\n404\n293\n387\n567\n660\n54\n339\n111\n833\n612\n911\n869\n356\n884\n635\n126\n639\n712\n473\n663\n773\n435\n32\n973\n484\n662\n464\n699\n274\n919\n95\n904\n253\n589\n543\n454\n250\n349\n237\n829\n511\n536\n36\n45\n152\n626\n384\n199\n877\n941\n84\n781\n115\n20\n52\n726\n751\n920\n291\n571\n6\n199",
"output": "64"
},
{
"input": "100\n66\n7\n9\n10\n5\n2\n8\n6\n5\n4\n10\n10\n6\n5\n2\n2\n1\n1\n5\n8\n7\n8\n10\n5\n6\n6\n5\n9\n9\n6\n3\n8\n7\n10\n5\n9\n6\n7\n3\n5\n8\n6\n8\n9\n1\n1\n1\n2\n4\n5\n5\n1\n1\n2\n6\n7\n1\n5\n8\n7\n2\n1\n7\n10\n9\n10\n2\n4\n10\n4\n10\n10\n5\n3\n9\n1\n2\n1\n10\n5\n1\n7\n4\n4\n5\n7\n6\n10\n4\n7\n3\n4\n3\n6\n2\n5\n2\n4\n9\n5\n3",
"output": "7"
},
{
"input": "100\n4862\n20\n47\n85\n47\n76\n38\n48\n93\n91\n81\n31\n51\n23\n60\n59\n3\n73\n72\n57\n67\n54\n9\n42\n5\n32\n46\n72\n79\n95\n61\n79\n88\n33\n52\n97\n10\n3\n20\n79\n82\n93\n90\n38\n80\n18\n21\n43\n60\n73\n34\n75\n65\n10\n84\n100\n29\n94\n56\n22\n59\n95\n46\n22\n57\n69\n67\n90\n11\n10\n61\n27\n2\n48\n69\n86\n91\n69\n76\n36\n71\n18\n54\n90\n74\n69\n50\n46\n8\n5\n41\n96\n5\n14\n55\n85\n39\n6\n79\n75\n87",
"output": "70"
},
{
"input": "100\n45570\n14\n881\n678\n687\n993\n413\n760\n451\n426\n787\n503\n343\n234\n530\n294\n725\n941\n524\n574\n441\n798\n399\n360\n609\n376\n525\n229\n995\n478\n347\n47\n23\n468\n525\n749\n601\n235\n89\n995\n489\n1\n239\n415\n122\n671\n128\n357\n886\n401\n964\n212\n968\n210\n130\n871\n360\n661\n844\n414\n187\n21\n824\n266\n713\n126\n496\n916\n37\n193\n755\n894\n641\n300\n170\n176\n383\n488\n627\n61\n897\n33\n242\n419\n881\n698\n107\n391\n418\n774\n905\n87\n5\n896\n835\n318\n373\n916\n393\n91\n460",
"output": "78"
},
{
"input": "100\n522\n1\n5\n2\n4\n2\n6\n3\n4\n2\n10\n10\n6\n7\n9\n7\n1\n7\n2\n5\n3\n1\n5\n2\n3\n5\n1\n7\n10\n10\n4\n4\n10\n9\n10\n6\n2\n8\n2\n6\n10\n9\n2\n7\n5\n9\n4\n6\n10\n7\n3\n1\n1\n9\n5\n10\n9\n2\n8\n3\n7\n5\n4\n7\n5\n9\n10\n6\n2\n9\n2\n5\n10\n1\n7\n7\n10\n5\n6\n2\n9\n4\n7\n10\n10\n8\n3\n4\n9\n3\n6\n9\n10\n2\n9\n9\n3\n4\n1\n10\n2",
"output": "74"
},
{
"input": "100\n32294\n414\n116\n131\n649\n130\n476\n630\n605\n213\n117\n757\n42\n109\n85\n127\n635\n629\n994\n410\n764\n204\n161\n231\n577\n116\n936\n537\n565\n571\n317\n722\n819\n229\n284\n487\n649\n304\n628\n727\n816\n854\n91\n111\n549\n87\n374\n417\n3\n868\n882\n168\n743\n77\n534\n781\n75\n956\n910\n734\n507\n568\n802\n946\n891\n659\n116\n678\n375\n380\n430\n627\n873\n350\n930\n285\n6\n183\n96\n517\n81\n794\n235\n360\n551\n6\n28\n799\n226\n996\n894\n981\n551\n60\n40\n460\n479\n161\n318\n952\n433",
"output": "42"
},
{
"input": "100\n178\n71\n23\n84\n98\n8\n14\n4\n42\n56\n83\n87\n28\n22\n32\n50\n5\n96\n90\n1\n59\n74\n56\n96\n77\n88\n71\n38\n62\n36\n85\n1\n97\n98\n98\n32\n99\n42\n6\n81\n20\n49\n57\n71\n66\n9\n45\n41\n29\n28\n32\n68\n38\n29\n35\n29\n19\n27\n76\n85\n68\n68\n41\n32\n78\n72\n38\n19\n55\n83\n83\n25\n46\n62\n48\n26\n53\n14\n39\n31\n94\n84\n22\n39\n34\n96\n63\n37\n42\n6\n78\n76\n64\n16\n26\n6\n79\n53\n24\n29\n63",
"output": "2"
},
{
"input": "100\n885\n226\n266\n321\n72\n719\n29\n121\n533\n85\n672\n225\n830\n783\n822\n30\n791\n618\n166\n487\n922\n434\n814\n473\n5\n741\n947\n910\n305\n998\n49\n945\n588\n868\n809\n803\n168\n280\n614\n434\n634\n538\n591\n437\n540\n445\n313\n177\n171\n799\n778\n55\n617\n554\n583\n611\n12\n94\n599\n182\n765\n556\n965\n542\n35\n460\n177\n313\n485\n744\n384\n21\n52\n879\n792\n411\n614\n811\n565\n695\n428\n587\n631\n794\n461\n258\n193\n696\n936\n646\n756\n267\n55\n690\n730\n742\n734\n988\n235\n762\n440",
"output": "1"
},
{
"input": "100\n29\n9\n2\n10\n8\n6\n7\n7\n3\n3\n10\n4\n5\n2\n5\n1\n6\n3\n2\n5\n10\n10\n9\n1\n4\n5\n2\n2\n3\n1\n2\n2\n9\n6\n9\n7\n8\n8\n1\n5\n5\n3\n1\n5\n6\n1\n9\n2\n3\n8\n10\n8\n3\n2\n7\n1\n2\n1\n2\n8\n10\n5\n2\n3\n1\n10\n7\n1\n7\n4\n9\n6\n6\n4\n7\n1\n2\n7\n7\n9\n9\n7\n10\n4\n10\n8\n2\n1\n5\n5\n10\n5\n8\n1\n5\n6\n5\n1\n5\n6\n8",
"output": "3"
},
{
"input": "100\n644\n94\n69\n43\n36\n54\n93\n30\n74\n56\n95\n70\n49\n11\n36\n57\n30\n59\n3\n52\n59\n90\n82\n39\n67\n32\n8\n80\n64\n8\n65\n51\n48\n89\n90\n35\n4\n54\n66\n96\n68\n90\n30\n4\n13\n97\n41\n90\n85\n17\n45\n94\n31\n58\n4\n39\n76\n95\n92\n59\n67\n46\n96\n55\n82\n64\n20\n20\n83\n46\n37\n15\n60\n37\n79\n45\n47\n63\n73\n76\n31\n52\n36\n32\n49\n26\n61\n91\n31\n25\n62\n90\n65\n65\n5\n94\n7\n15\n97\n88\n68",
"output": "7"
},
{
"input": "100\n1756\n98\n229\n158\n281\n16\n169\n149\n239\n235\n182\n147\n215\n49\n270\n194\n242\n295\n289\n249\n19\n12\n144\n157\n92\n270\n122\n212\n97\n152\n14\n42\n12\n198\n98\n295\n154\n229\n191\n294\n5\n156\n43\n185\n184\n20\n125\n23\n10\n257\n244\n264\n79\n46\n277\n13\n22\n97\n212\n77\n293\n20\n51\n17\n109\n37\n68\n117\n51\n248\n10\n149\n179\n192\n239\n161\n13\n173\n297\n73\n43\n109\n288\n198\n81\n70\n254\n187\n277\n1\n295\n113\n95\n291\n293\n119\n205\n191\n37\n34\n116",
"output": "6"
},
{
"input": "100\n20562\n721\n452\n11\n703\n376\n183\n197\n203\n406\n642\n346\n446\n256\n760\n201\n360\n702\n707\n388\n779\n653\n610\n497\n768\n670\n134\n780\n306\n661\n180\n259\n256\n362\n6\n121\n415\n747\n170\n67\n439\n728\n193\n622\n481\n38\n225\n343\n303\n253\n436\n305\n68\n794\n247\n291\n600\n750\n188\n199\n757\n28\n776\n749\n253\n351\n53\n629\n129\n578\n209\n89\n651\n262\n638\n353\n469\n31\n144\n460\n176\n535\n562\n366\n639\n234\n577\n364\n761\n617\n303\n450\n778\n311\n289\n221\n274\n239\n626\n194\n36",
"output": "30"
},
{
"input": "100\n32630\n548\n21\n756\n138\n56\n719\n97\n86\n599\n531\n464\n137\n737\n239\n626\n179\n170\n271\n335\n178\n298\n597\n616\n104\n190\n216\n333\n606\n621\n310\n70\n286\n593\n655\n472\n556\n77\n625\n744\n395\n487\n762\n316\n469\n258\n475\n596\n651\n180\n737\n101\n511\n708\n163\n713\n475\n557\n649\n705\n583\n97\n447\n119\n91\n372\n194\n394\n667\n384\n635\n611\n746\n758\n613\n54\n508\n405\n188\n570\n254\n28\n599\n767\n115\n737\n194\n148\n686\n705\n236\n762\n613\n770\n633\n401\n773\n448\n274\n141\n718",
"output": "52"
},
{
"input": "1\n1\n1",
"output": "1"
}
] | 1,501,536,828 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 62 | 4,608,000 | n = int(input())
m = int(input())
x = []
for i in range(n):
val = int(input())
x.append(val)
x = sorted(x)[::-1]
for i, val in enumerate(x):
m -= val
if m <= 0:
print(i+1)
exit(0) | Title: USB Flash Drives
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
Input Specification:
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.
Output Specification:
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
Demo Input:
['3\n5\n2\n1\n3\n', '3\n6\n2\n3\n2\n', '2\n5\n5\n10\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second. | ```python
n = int(input())
m = int(input())
x = []
for i in range(n):
val = int(input())
x.append(val)
x = sorted(x)[::-1]
for i, val in enumerate(x):
m -= val
if m <= 0:
print(i+1)
exit(0)
``` | 3 |
|
575 | D | Tablecity | PROGRAMMING | 1,700 | [
"constructive algorithms",
"implementation"
] | null | null | There was a big bank robbery in Tablecity. In order to catch the thief, the President called none other than Albert – Tablecity’s Chief of Police. Albert does not know where the thief is located, but he does know how he moves.
Tablecity can be represented as 1000<=×<=2 grid, where every cell represents one district. Each district has its own unique name “(*X*,<=*Y*)”, where *X* and *Y* are the coordinates of the district in the grid. The thief’s movement is as
Every hour the thief will leave the district (*X*,<=*Y*) he is currently hiding in, and move to one of the districts: (*X*<=-<=1,<=*Y*), (*X*<=+<=1,<=*Y*), (*X*<=-<=1,<=*Y*<=-<=1), (*X*<=-<=1,<=*Y*<=+<=1), (*X*<=+<=1,<=*Y*<=-<=1), (*X*<=+<=1,<=*Y*<=+<=1) as long as it exists in Tablecity.
Below is an example of thief’s possible movements if he is located in district (7,1):
Albert has enough people so that every hour he can pick any two districts in Tablecity and fully investigate them, making sure that if the thief is located in one of them, he will get caught. Albert promised the President that the thief will be caught in no more than 2015 hours and needs your help in order to achieve that. | There is no input for this problem. | The first line of output contains integer *N* – duration of police search in hours. Each of the following *N* lines contains exactly 4 integers *X**i*1, *Y**i*1, *X**i*2, *Y**i*2 separated by spaces, that represent 2 districts (*X**i*1, *Y**i*1), (*X**i*2, *Y**i*2) which got investigated during i-th hour. Output is given in chronological order (i-th line contains districts investigated during i-th hour) and should guarantee that the thief is caught in no more than 2015 hours, regardless of thief’s initial position and movement.
- *N*<=≤<=2015 - 1<=≤<=*X*<=≤<=1000 - 1<=≤<=*Y*<=≤<=2 | [
"В этой задаче нет примеров ввода-вывода.\nThis problem doesn't have sample input and output."
] | [
"Смотрите замечание ниже.\nSee the note below."
] | Let's consider the following output:
2
5 1 50 2
8 1 80 2
This output is not guaranteed to catch the thief and is not correct. It is given to you only to show the expected output format. There exists a combination of an initial position and a movement strategy such that the police will not catch the thief.
Consider the following initial position and thief’s movement:
In the first hour, the thief is located in district (1,1). Police officers will search districts (5,1) and (50,2) and will not find him.
At the start of the second hour, the thief moves to district (2,2). Police officers will search districts (8,1) and (80,2) and will not find him.
Since there is no further investigation by the police, the thief escaped! | 0 | [
{
"input": "dummy",
"output": "2000\n1 1 1 2\n2 1 2 2\n3 1 3 2\n4 1 4 2\n5 1 5 2\n6 1 6 2\n7 1 7 2\n8 1 8 2\n9 1 9 2\n10 1 10 2\n11 1 11 2\n12 1 12 2\n13 1 13 2\n14 1 14 2\n15 1 15 2\n16 1 16 2\n17 1 17 2\n18 1 18 2\n19 1 19 2\n20 1 20 2\n21 1 21 2\n22 1 22 2\n23 1 23 2\n24 1 24 2\n25 1 25 2\n26 1 26 2\n27 1 27 2\n28 1 28 2\n29 1 29 2\n30 1 30 2\n31 1 31 2\n32 1 32 2\n33 1 33 2\n34 1 34 2\n35 1 35 2\n36 1 36 2\n37 1 37 2\n38 1 38 2\n39 1 39 2\n40 1 40 2\n41 1 41 2\n42 1 42 2\n43 1 43 2\n44 1 44 2\n45 1 45 2\n46 1 46 2\n47 1 47 2\n48 1 4..."
}
] | 1,441,527,877 | 1,477 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print('''2
5 1 50 2
8 1 80 1''') | Title: Tablecity
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There was a big bank robbery in Tablecity. In order to catch the thief, the President called none other than Albert – Tablecity’s Chief of Police. Albert does not know where the thief is located, but he does know how he moves.
Tablecity can be represented as 1000<=×<=2 grid, where every cell represents one district. Each district has its own unique name “(*X*,<=*Y*)”, where *X* and *Y* are the coordinates of the district in the grid. The thief’s movement is as
Every hour the thief will leave the district (*X*,<=*Y*) he is currently hiding in, and move to one of the districts: (*X*<=-<=1,<=*Y*), (*X*<=+<=1,<=*Y*), (*X*<=-<=1,<=*Y*<=-<=1), (*X*<=-<=1,<=*Y*<=+<=1), (*X*<=+<=1,<=*Y*<=-<=1), (*X*<=+<=1,<=*Y*<=+<=1) as long as it exists in Tablecity.
Below is an example of thief’s possible movements if he is located in district (7,1):
Albert has enough people so that every hour he can pick any two districts in Tablecity and fully investigate them, making sure that if the thief is located in one of them, he will get caught. Albert promised the President that the thief will be caught in no more than 2015 hours and needs your help in order to achieve that.
Input Specification:
There is no input for this problem.
Output Specification:
The first line of output contains integer *N* – duration of police search in hours. Each of the following *N* lines contains exactly 4 integers *X**i*1, *Y**i*1, *X**i*2, *Y**i*2 separated by spaces, that represent 2 districts (*X**i*1, *Y**i*1), (*X**i*2, *Y**i*2) which got investigated during i-th hour. Output is given in chronological order (i-th line contains districts investigated during i-th hour) and should guarantee that the thief is caught in no more than 2015 hours, regardless of thief’s initial position and movement.
- *N*<=≤<=2015 - 1<=≤<=*X*<=≤<=1000 - 1<=≤<=*Y*<=≤<=2
Demo Input:
["В этой задаче нет примеров ввода-вывода.\nThis problem doesn't have sample input and output."]
Demo Output:
['Смотрите замечание ниже.\nSee the note below.']
Note:
Let's consider the following output:
2
5 1 50 2
8 1 80 2
This output is not guaranteed to catch the thief and is not correct. It is given to you only to show the expected output format. There exists a combination of an initial position and a movement strategy such that the police will not catch the thief.
Consider the following initial position and thief’s movement:
In the first hour, the thief is located in district (1,1). Police officers will search districts (5,1) and (50,2) and will not find him.
At the start of the second hour, the thief moves to district (2,2). Police officers will search districts (8,1) and (80,2) and will not find him.
Since there is no further investigation by the police, the thief escaped! | ```python
print('''2
5 1 50 2
8 1 80 1''')
``` | 0 |
|
386 | A | Second-Price Auction | PROGRAMMING | 800 | [
"implementation"
] | null | null | In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different. | The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder. | The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based. | [
"2\n5 7\n",
"3\n10 2 8\n",
"6\n3 8 2 9 4 14\n"
] | [
"2 5\n",
"1 8\n",
"6 9\n"
] | none | 500 | [
{
"input": "2\n5 7",
"output": "2 5"
},
{
"input": "3\n10 2 8",
"output": "1 8"
},
{
"input": "6\n3 8 2 9 4 14",
"output": "6 9"
},
{
"input": "4\n4707 7586 4221 5842",
"output": "2 5842"
},
{
"input": "5\n3304 4227 4869 6937 6002",
"output": "4 6002"
},
{
"input": "6\n5083 3289 7708 5362 9031 7458",
"output": "5 7708"
},
{
"input": "7\n9038 6222 3392 1706 3778 1807 2657",
"output": "1 6222"
},
{
"input": "8\n7062 2194 4481 3864 7470 1814 8091 733",
"output": "7 7470"
},
{
"input": "9\n2678 5659 9199 2628 7906 7496 4524 2663 3408",
"output": "3 7906"
},
{
"input": "2\n3458 1504",
"output": "1 1504"
},
{
"input": "50\n9237 3904 407 9052 6657 9229 9752 3888 7732 2512 4614 1055 2355 7108 6506 6849 2529 8862 159 8630 7906 7941 960 8470 333 8659 54 9475 3163 5625 6393 6814 2656 3388 169 7918 4881 8468 9983 6281 6340 280 5108 2996 101 7617 3313 8172 326 1991",
"output": "39 9752"
},
{
"input": "100\n2515 3324 7975 6171 4240 1217 4829 5203 8603 6900 3031 4699 4732 6070 4221 3228 6497 7359 9130 4346 4619 1109 3945 5442 3271 16 9711 2045 6410 2301 3406 8125 6003 1892 1260 9661 3940 6692 4708 7027 4930 6925 1979 5361 4263 3144 867 8639 6230 5562 9714 3676 4231 3347 4733 4920 4881 3431 1059 7313 8912 3038 9308 72 9583 7009 3034 7425 2398 6141 3245 2495 2933 6710 8289 9806 1226 8393 7349 6462 1618 9613 3546 6012 2964 9995 1578 210 2123 4874 1252 8625 348 8020 803 7244 9080 5088 706 2602",
"output": "86 9806"
},
{
"input": "2\n2 1",
"output": "1 1"
},
{
"input": "2\n1 2",
"output": "2 1"
},
{
"input": "3\n10 20 30",
"output": "3 20"
},
{
"input": "3\n10 30 20",
"output": "2 20"
},
{
"input": "3\n20 10 30",
"output": "3 20"
},
{
"input": "3\n20 30 10",
"output": "2 20"
},
{
"input": "3\n30 10 20",
"output": "1 20"
},
{
"input": "3\n30 20 10",
"output": "1 20"
},
{
"input": "2\n1 10000",
"output": "2 1"
},
{
"input": "2\n10000 999",
"output": "1 999"
},
{
"input": "3\n3 4 1",
"output": "2 3"
},
{
"input": "6\n1 2 24 6 5 7",
"output": "3 7"
}
] | 1,672,190,898 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 46 | 0 | n = int(input())
moo = list(map(int, input().split()))
print(moo.index(max(moo))+1, end = " ")
moo.remove(max(moo))
print(max(moo)) | Title: Second-Price Auction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
Output Specification:
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
Demo Input:
['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n']
Demo Output:
['2 5\n', '1 8\n', '6 9\n']
Note:
none | ```python
n = int(input())
moo = list(map(int, input().split()))
print(moo.index(max(moo))+1, end = " ")
moo.remove(max(moo))
print(max(moo))
``` | 3 |
|
281 | A | Word Capitalization | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged. | A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103. | Output the given word after capitalization. | [
"ApPLe\n",
"konjac\n"
] | [
"ApPLe\n",
"Konjac\n"
] | none | 500 | [
{
"input": "ApPLe",
"output": "ApPLe"
},
{
"input": "konjac",
"output": "Konjac"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "A"
},
{
"input": "z",
"output": "Z"
},
{
"input": "ABACABA",
"output": "ABACABA"
},
{
"input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX",
"output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX"
},
{
"input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO",
"output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO"
},
{
"input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD",
"output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD"
},
{
"input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW",
"output": "QUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW"
},
{
"input": "kHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL",
"output": "KHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL"
},
{
"input": "aHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC",
"output": "AHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC"
},
{
"input": "sLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN",
"output": "SLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN"
},
{
"input": "iRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE",
"output": "IRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE"
},
{
"input": "cRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP",
"output": "CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP"
},
{
"input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG",
"output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG"
},
{
"input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV",
"output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV"
},
{
"input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI",
"output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI"
},
{
"input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX",
"output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX"
},
{
"input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ",
"output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ"
},
{
"input": "P",
"output": "P"
},
{
"input": "Xyzzy",
"output": "Xyzzy"
},
{
"input": "Zzz",
"output": "Zzz"
},
{
"input": "Zp",
"output": "Zp"
}
] | 1,699,618,963 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 92 | 0 | w = input()
p = w[0].upper() + w[1:]
print(p) | Title: Word Capitalization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
Input Specification:
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output Specification:
Output the given word after capitalization.
Demo Input:
['ApPLe\n', 'konjac\n']
Demo Output:
['ApPLe\n', 'Konjac\n']
Note:
none | ```python
w = input()
p = w[0].upper() + w[1:]
print(p)
``` | 3 |
|
869 | A | The Artful Expedient | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Rock... Paper!
After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.
A positive integer *n* is decided first. Both Koyomi and Karen independently choose *n* distinct positive integers, denoted by *x*1,<=*x*2,<=...,<=*x**n* and *y*1,<=*y*2,<=...,<=*y**n* respectively. They reveal their sequences, and repeat until all of 2*n* integers become distinct, which is the only final state to be kept and considered.
Then they count the number of ordered pairs (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) such that the value *x**i* xor *y**j* equals to one of the 2*n* integers. Here xor means the [bitwise exclusive or](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.
Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game. | The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=2<=000) — the length of both sequences.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=2·106) — the integers finally chosen by Koyomi.
The third line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (1<=≤<=*y**i*<=≤<=2·106) — the integers finally chosen by Karen.
Input guarantees that the given 2*n* integers are pairwise distinct, that is, no pair (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) exists such that one of the following holds: *x**i*<==<=*y**j*; *i*<=≠<=*j* and *x**i*<==<=*x**j*; *i*<=≠<=*j* and *y**i*<==<=*y**j*. | Output one line — the name of the winner, that is, "Koyomi" or "Karen" (without quotes). Please be aware of the capitalization. | [
"3\n1 2 3\n4 5 6\n",
"5\n2 4 6 8 10\n9 7 5 3 1\n"
] | [
"Karen\n",
"Karen\n"
] | In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again. | 500 | [
{
"input": "3\n1 2 3\n4 5 6",
"output": "Karen"
},
{
"input": "5\n2 4 6 8 10\n9 7 5 3 1",
"output": "Karen"
},
{
"input": "1\n1\n2000000",
"output": "Karen"
},
{
"input": "2\n97153 2000000\n1999998 254",
"output": "Karen"
},
{
"input": "15\n31 30 29 28 27 26 25 24 23 22 21 20 19 18 17\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15",
"output": "Karen"
},
{
"input": "30\n79656 68607 871714 1858841 237684 1177337 532141 161161 1111201 527235 323345 1979059 665353 507265 1290761 610606 1238375 743262 106355 1167830 180315 1233029 816465 752968 782570 1499881 1328457 1867240 13948 1302782\n322597 1868510 1958236 1348157 765908 1023636 874300 537124 631783 414906 886318 1931572 1381013 992451 1305644 1525745 716087 83173 303248 1572710 43084 333341 992413 267806 70390 644521 1014900 497068 178940 1920268",
"output": "Karen"
},
{
"input": "30\n1143673 436496 1214486 1315862 148404 724601 1430740 1433008 1654610 1635673 614673 1713408 1270999 1697 1463796 50027 525482 1659078 688200 842647 518551 877506 1017082 1807856 3280 759698 1208220 470180 829800 1960886\n1312613 1965095 967255 1289012 1950383 582960 856825 49684 808824 319418 1968270 190821 344545 211332 1219388 1773751 1876402 132626 541448 1584672 24276 1053225 1823073 1858232 1209173 1035991 1956373 1237148 1973608 848873",
"output": "Karen"
},
{
"input": "1\n2\n3",
"output": "Karen"
},
{
"input": "1\n1048576\n1020000",
"output": "Karen"
},
{
"input": "3\n9 33 69\n71 74 100",
"output": "Karen"
},
{
"input": "3\n1 2 3\n9 5 6",
"output": "Karen"
},
{
"input": "3\n1 7 8\n9 10 20",
"output": "Karen"
},
{
"input": "3\n1 3 2\n4 5 8",
"output": "Karen"
},
{
"input": "3\n2 1 100\n3 4 9",
"output": "Karen"
},
{
"input": "3\n3 1 100\n2 1000 100000",
"output": "Karen"
},
{
"input": "3\n1 2 5\n3 4 6",
"output": "Karen"
},
{
"input": "3\n3 1 8\n2 4 17",
"output": "Karen"
},
{
"input": "3\n1 5 6\n7 8 3",
"output": "Karen"
},
{
"input": "1\n1\n3",
"output": "Karen"
},
{
"input": "3\n1 3 10\n2 4 20",
"output": "Karen"
},
{
"input": "3\n7 8 10\n15 9 11",
"output": "Karen"
},
{
"input": "3\n5 6 8\n3 100 9",
"output": "Karen"
},
{
"input": "3\n1 2 3\n4 5 8",
"output": "Karen"
},
{
"input": "3\n1 2 19\n3 7 30",
"output": "Karen"
},
{
"input": "3\n1 2 3\n6 7 8",
"output": "Karen"
},
{
"input": "3\n1 4 55\n2 3 9",
"output": "Karen"
},
{
"input": "3\n1 100 200\n5 4 500",
"output": "Karen"
},
{
"input": "1\n6\n7",
"output": "Karen"
},
{
"input": "3\n1 3 5\n2 4 13",
"output": "Karen"
},
{
"input": "3\n3 1 100\n2 1000 10000",
"output": "Karen"
},
{
"input": "3\n1 3 9\n2 4 40",
"output": "Karen"
},
{
"input": "2\n1 199999\n1935807 2000000",
"output": "Karen"
},
{
"input": "3\n1 3 8\n2 4 24",
"output": "Karen"
},
{
"input": "2\n4 1\n7 3",
"output": "Karen"
},
{
"input": "3\n1 2 4\n3 7 8",
"output": "Karen"
},
{
"input": "3\n1 6 10000\n2 3 100000",
"output": "Karen"
}
] | 1,507,494,648 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<bits/stdc++.h>
using namespace std;
int main()
{
int a,i;
cin>>a;
int x[a],y[a];
for (i=0;i<a;i++)
cin>>x[a];
for (i=0;i<a;i++)
cin>>y[a];
puts("Karen");
return 0;
}
| Title: The Artful Expedient
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Rock... Paper!
After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.
A positive integer *n* is decided first. Both Koyomi and Karen independently choose *n* distinct positive integers, denoted by *x*1,<=*x*2,<=...,<=*x**n* and *y*1,<=*y*2,<=...,<=*y**n* respectively. They reveal their sequences, and repeat until all of 2*n* integers become distinct, which is the only final state to be kept and considered.
Then they count the number of ordered pairs (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) such that the value *x**i* xor *y**j* equals to one of the 2*n* integers. Here xor means the [bitwise exclusive or](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.
Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game.
Input Specification:
The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=2<=000) — the length of both sequences.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=2·106) — the integers finally chosen by Koyomi.
The third line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (1<=≤<=*y**i*<=≤<=2·106) — the integers finally chosen by Karen.
Input guarantees that the given 2*n* integers are pairwise distinct, that is, no pair (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) exists such that one of the following holds: *x**i*<==<=*y**j*; *i*<=≠<=*j* and *x**i*<==<=*x**j*; *i*<=≠<=*j* and *y**i*<==<=*y**j*.
Output Specification:
Output one line — the name of the winner, that is, "Koyomi" or "Karen" (without quotes). Please be aware of the capitalization.
Demo Input:
['3\n1 2 3\n4 5 6\n', '5\n2 4 6 8 10\n9 7 5 3 1\n']
Demo Output:
['Karen\n', 'Karen\n']
Note:
In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again. | ```python
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a,i;
cin>>a;
int x[a],y[a];
for (i=0;i<a;i++)
cin>>x[a];
for (i=0;i<a;i++)
cin>>y[a];
puts("Karen");
return 0;
}
``` | -1 |
|
588 | A | Duff and Meat | PROGRAMMING | 900 | [
"greedy"
] | null | null | Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days. | The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day. | Print the minimum money needed to keep Duff happy for *n* days, in one line. | [
"3\n1 3\n2 2\n3 1\n",
"3\n1 3\n2 1\n3 2\n"
] | [
"10\n",
"8\n"
] | In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day. | 750 | [
{
"input": "3\n1 3\n2 2\n3 1",
"output": "10"
},
{
"input": "3\n1 3\n2 1\n3 2",
"output": "8"
},
{
"input": "1\n39 52",
"output": "2028"
},
{
"input": "2\n25 56\n94 17",
"output": "2998"
},
{
"input": "5\n39 21\n95 89\n73 90\n9 55\n85 32",
"output": "6321"
},
{
"input": "12\n70 11\n74 27\n32 11\n26 83\n57 18\n97 28\n75 43\n75 21\n84 29\n16 2\n89 63\n21 88",
"output": "6742"
},
{
"input": "2\n100 68\n38 80",
"output": "9384"
}
] | 1,607,955,796 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 358 | 1,126,400 | n = int(input())
a = [0]*(n+3)
p = [0]*(n+3)
for i in range(n):
t1, t2 = map(int,input().split())
a[i] = t1
p[i] = t2
pmin = p[0]
kq = p[0]*a[0]
for i in range(1,n):
pmin = min(pmin, p[i])
kq+= pmin*a[i]
print(kq)
| Title: Duff and Meat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days.
Input Specification:
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day.
Output Specification:
Print the minimum money needed to keep Duff happy for *n* days, in one line.
Demo Input:
['3\n1 3\n2 2\n3 1\n', '3\n1 3\n2 1\n3 2\n']
Demo Output:
['10\n', '8\n']
Note:
In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day. | ```python
n = int(input())
a = [0]*(n+3)
p = [0]*(n+3)
for i in range(n):
t1, t2 = map(int,input().split())
a[i] = t1
p[i] = t2
pmin = p[0]
kq = p[0]*a[0]
for i in range(1,n):
pmin = min(pmin, p[i])
kq+= pmin*a[i]
print(kq)
``` | 3 |
|
618 | A | Slime Combining | PROGRAMMING | 800 | [
"implementation"
] | null | null | Your friend recently gave you some slimes for your birthday. You have *n* slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other *n*<=-<=1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value *v*, you combine them together to create a slime with value *v*<=+<=1.
You would like to see what the final state of the row is after you've added all *n* slimes. Please print the values of the slimes in the row from left to right. | The first line of the input will contain a single integer, *n* (1<=≤<=*n*<=≤<=100<=000). | Output a single line with *k* integers, where *k* is the number of slimes in the row after you've finished the procedure described in the problem statement. The *i*-th of these numbers should be the value of the *i*-th slime from the left. | [
"1\n",
"2\n",
"3\n",
"8\n"
] | [
"1\n",
"2\n",
"2 1\n",
"4\n"
] | In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1 1. 2 1. 2 1 1. 3 1. 3 1 1. 3 2 1. 3 2 1 1. 4 | 500 | [
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "2 1"
},
{
"input": "8",
"output": "4"
},
{
"input": "100000",
"output": "17 16 11 10 8 6"
},
{
"input": "12345",
"output": "14 13 6 5 4 1"
},
{
"input": "32",
"output": "6"
},
{
"input": "70958",
"output": "17 13 11 9 6 4 3 2"
},
{
"input": "97593",
"output": "17 15 14 13 12 11 9 6 5 4 1"
},
{
"input": "91706",
"output": "17 15 14 11 10 6 5 4 2"
},
{
"input": "85371",
"output": "17 15 12 11 9 7 6 5 4 2 1"
},
{
"input": "97205",
"output": "17 15 14 13 12 10 9 8 6 5 3 1"
},
{
"input": "34768",
"output": "16 11 10 9 8 7 5"
},
{
"input": "12705",
"output": "14 13 9 8 6 1"
},
{
"input": "30151",
"output": "15 14 13 11 9 8 7 3 2 1"
},
{
"input": "4974",
"output": "13 10 9 7 6 4 3 2"
},
{
"input": "32728",
"output": "15 14 13 12 11 10 9 8 7 5 4"
},
{
"input": "8192",
"output": "14"
},
{
"input": "65536",
"output": "17"
},
{
"input": "32",
"output": "6"
},
{
"input": "256",
"output": "9"
},
{
"input": "4096",
"output": "13"
},
{
"input": "33301",
"output": "16 10 5 3 1"
},
{
"input": "16725",
"output": "15 9 7 5 3 1"
},
{
"input": "149",
"output": "8 5 3 1"
},
{
"input": "16277",
"output": "14 13 12 11 10 9 8 5 3 1"
},
{
"input": "99701",
"output": "17 16 11 9 7 6 5 3 1"
}
] | 1,454,088,450 | 1,050 | Python 3 | WRONG_ANSWER | PRETESTS | 0 | 46 | 0 | from cmath import log
n = int(input())
k = n
a = 1
ans = []
while a <= n:
a *= 2
if n % a != 0:
a //= 2
ans.append(a)
n -= a
a *= 2
ans = ans[::-1]
print(len(ans))
print(*ans) | Title: Slime Combining
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend recently gave you some slimes for your birthday. You have *n* slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other *n*<=-<=1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value *v*, you combine them together to create a slime with value *v*<=+<=1.
You would like to see what the final state of the row is after you've added all *n* slimes. Please print the values of the slimes in the row from left to right.
Input Specification:
The first line of the input will contain a single integer, *n* (1<=≤<=*n*<=≤<=100<=000).
Output Specification:
Output a single line with *k* integers, where *k* is the number of slimes in the row after you've finished the procedure described in the problem statement. The *i*-th of these numbers should be the value of the *i*-th slime from the left.
Demo Input:
['1\n', '2\n', '3\n', '8\n']
Demo Output:
['1\n', '2\n', '2 1\n', '4\n']
Note:
In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.
In the second sample, we perform the following steps:
Initially we place a single slime in a row by itself. Thus, row is initially 1.
Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.
In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.
In the last sample, the steps look as follows:
1. 1 1. 2 1. 2 1 1. 3 1. 3 1 1. 3 2 1. 3 2 1 1. 4 | ```python
from cmath import log
n = int(input())
k = n
a = 1
ans = []
while a <= n:
a *= 2
if n % a != 0:
a //= 2
ans.append(a)
n -= a
a *= 2
ans = ans[::-1]
print(len(ans))
print(*ans)
``` | 0 |
|
346 | A | Alice and Bob | PROGRAMMING | 1,600 | [
"games",
"math",
"number theory"
] | null | null | It is so boring in the summer holiday, isn't it? So Alice and Bob have invented a new game to play. The rules are as follows. First, they get a set of *n* distinct integers. And then they take turns to make the following moves. During each move, either Alice or Bob (the player whose turn is the current) can choose two distinct integers *x* and *y* from the set, such that the set doesn't contain their absolute difference |*x*<=-<=*y*|. Then this player adds integer |*x*<=-<=*y*| to the set (so, the size of the set increases by one).
If the current player has no valid move, he (or she) loses the game. The question is who will finally win the game if both players play optimally. Remember that Alice always moves first. | The first line contains an integer *n* (2<=≤<=*n*<=≤<=100) — the initial number of elements in the set. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the set. | Print a single line with the winner's name. If Alice wins print "Alice", otherwise print "Bob" (without quotes). | [
"2\n2 3\n",
"2\n5 3\n",
"3\n5 6 7\n"
] | [
"Alice\n",
"Alice\n",
"Bob\n"
] | Consider the first test sample. Alice moves first, and the only move she can do is to choose 2 and 3, then to add 1 to the set. Next Bob moves, there is no valid move anymore, so the winner is Alice. | 500 | [
{
"input": "2\n2 3",
"output": "Alice"
},
{
"input": "2\n5 3",
"output": "Alice"
},
{
"input": "3\n5 6 7",
"output": "Bob"
},
{
"input": "10\n72 96 24 66 6 18 12 30 60 48",
"output": "Bob"
},
{
"input": "10\n78 66 6 60 18 84 36 96 72 48",
"output": "Bob"
},
{
"input": "10\n98 63 42 56 14 77 70 35 84 21",
"output": "Bob"
},
{
"input": "2\n1 1000000000",
"output": "Bob"
},
{
"input": "2\n1000000000 999999999",
"output": "Bob"
},
{
"input": "3\n2 4 6",
"output": "Bob"
},
{
"input": "2\n4 6",
"output": "Alice"
},
{
"input": "2\n2 6",
"output": "Alice"
},
{
"input": "2\n6 2",
"output": "Alice"
},
{
"input": "10\n100000000 200000000 300000000 400000000 500000000 600000000 700000000 800000000 900000000 1000000000",
"output": "Bob"
},
{
"input": "2\n1 2",
"output": "Bob"
},
{
"input": "10\n1 999999999 999999998 999999997 999999996 999999995 999999994 999999993 999999992 999999991",
"output": "Alice"
},
{
"input": "3\n6 14 21",
"output": "Bob"
},
{
"input": "3\n4 12 18",
"output": "Bob"
},
{
"input": "4\n2 3 15 30",
"output": "Bob"
},
{
"input": "2\n10 4",
"output": "Alice"
}
] | 1,668,977,891 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | len1 = int(input())
seq = list(map(int, input().split()))
def NWD(a, b):
while b > 0:
reszta = a % b
a, b = b, reszta
return a
nwd = NWD(seq[0], seq[1])
for n in range(2, len1):
nwd = NWD(nwd, seq[n])
print(max(seq) // nwd)
| Title: Alice and Bob
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It is so boring in the summer holiday, isn't it? So Alice and Bob have invented a new game to play. The rules are as follows. First, they get a set of *n* distinct integers. And then they take turns to make the following moves. During each move, either Alice or Bob (the player whose turn is the current) can choose two distinct integers *x* and *y* from the set, such that the set doesn't contain their absolute difference |*x*<=-<=*y*|. Then this player adds integer |*x*<=-<=*y*| to the set (so, the size of the set increases by one).
If the current player has no valid move, he (or she) loses the game. The question is who will finally win the game if both players play optimally. Remember that Alice always moves first.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100) — the initial number of elements in the set. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the set.
Output Specification:
Print a single line with the winner's name. If Alice wins print "Alice", otherwise print "Bob" (without quotes).
Demo Input:
['2\n2 3\n', '2\n5 3\n', '3\n5 6 7\n']
Demo Output:
['Alice\n', 'Alice\n', 'Bob\n']
Note:
Consider the first test sample. Alice moves first, and the only move she can do is to choose 2 and 3, then to add 1 to the set. Next Bob moves, there is no valid move anymore, so the winner is Alice. | ```python
len1 = int(input())
seq = list(map(int, input().split()))
def NWD(a, b):
while b > 0:
reszta = a % b
a, b = b, reszta
return a
nwd = NWD(seq[0], seq[1])
for n in range(2, len1):
nwd = NWD(nwd, seq[n])
print(max(seq) // nwd)
``` | 0 |
|
459 | B | Pashmak and Flowers | PROGRAMMING | 1,300 | [
"combinatorics",
"implementation",
"sortings"
] | null | null | Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way. | The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109). | The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively. | [
"2\n1 2\n",
"3\n1 4 5\n",
"5\n3 1 2 3 1\n"
] | [
"1 1",
"4 1",
"2 4"
] | In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers. | 500 | [
{
"input": "2\n1 2",
"output": "1 1"
},
{
"input": "3\n1 4 5",
"output": "4 1"
},
{
"input": "5\n3 1 2 3 1",
"output": "2 4"
},
{
"input": "2\n1 1",
"output": "0 1"
},
{
"input": "3\n1 1 1",
"output": "0 3"
},
{
"input": "4\n1 1 1 1",
"output": "0 6"
},
{
"input": "5\n1 1 1 1 1",
"output": "0 10"
},
{
"input": "5\n2 2 2 2 2",
"output": "0 10"
},
{
"input": "10\n2 2 2 2 2 2 2 2 2 2",
"output": "0 45"
},
{
"input": "3\n2 2 2",
"output": "0 3"
},
{
"input": "3\n3 3 3",
"output": "0 3"
},
{
"input": "2\n10000000 100000000",
"output": "90000000 1"
},
{
"input": "5\n5 5 5 5 5",
"output": "0 10"
},
{
"input": "5\n3 3 3 3 3",
"output": "0 10"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "0 15"
},
{
"input": "2\n5 6",
"output": "1 1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "0 45"
},
{
"input": "10\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "0 45"
},
{
"input": "4\n4 4 4 4",
"output": "0 6"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "0 21"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "0 55"
},
{
"input": "8\n8 8 8 8 8 8 8 8",
"output": "0 28"
},
{
"input": "2\n3 2",
"output": "1 1"
}
] | 1,672,478,482 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 58 | 171 | 27,545,600 | n=int(input())
a=list(map(int,input().split()))
a.sort()
a1=a.count(a[0])
a2=a.count(a[-1])
if a[0]==a[-1]:
c=n*(n-1)//2
else:
c=a1*a2
print(a[-1]-a[0],c)
| Title: Pashmak and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109).
Output Specification:
The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.
Demo Input:
['2\n1 2\n', '3\n1 4 5\n', '5\n3 1 2 3 1\n']
Demo Output:
['1 1', '4 1', '2 4']
Note:
In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers. | ```python
n=int(input())
a=list(map(int,input().split()))
a.sort()
a1=a.count(a[0])
a2=a.count(a[-1])
if a[0]==a[-1]:
c=n*(n-1)//2
else:
c=a1*a2
print(a[-1]-a[0],c)
``` | 3 |
|
630 | A | Again Twenty Five! | PROGRAMMING | 800 | [
"number theory"
] | null | null | The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City? | The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5. | Output the last two digits of 5*n* without spaces between them. | [
"2\n"
] | [
"25"
] | none | 0 | [
{
"input": "2",
"output": "25"
},
{
"input": "7",
"output": "25"
},
{
"input": "1000000000000000000",
"output": "25"
},
{
"input": "2000000000000000000",
"output": "25"
},
{
"input": "987654321012345678",
"output": "25"
}
] | 1,696,878,357 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 30 | 0 | a=int(input(""))
b=5**a
print(b)
| Title: Again Twenty Five!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.
Output Specification:
Output the last two digits of 5*n* without spaces between them.
Demo Input:
['2\n']
Demo Output:
['25']
Note:
none | ```python
a=int(input(""))
b=5**a
print(b)
``` | 0 |
|
294 | A | Shaass and Oskols | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots. | The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment. | On the *i*-th line of the output print the number of birds on the *i*-th wire. | [
"5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n",
"3\n2 4 1\n1\n2 2\n"
] | [
"0\n12\n5\n0\n16\n",
"3\n0\n3\n"
] | none | 500 | [
{
"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6",
"output": "0\n12\n5\n0\n16"
},
{
"input": "3\n2 4 1\n1\n2 2",
"output": "3\n0\n3"
},
{
"input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43",
"output": "0\n66\n57\n7\n0"
},
{
"input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51",
"output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52"
},
{
"input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1",
"output": "0\n0"
},
{
"input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5",
"output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21"
},
{
"input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0",
"output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95"
},
{
"input": "1\n100\n1\n1 100",
"output": "0"
},
{
"input": "1\n100\n1\n1 1",
"output": "0"
},
{
"input": "1\n50\n1\n1 25",
"output": "0"
},
{
"input": "2\n50 0\n1\n1 1",
"output": "0\n49"
},
{
"input": "1\n10\n0",
"output": "10"
}
] | 1,684,149,037 | 2,147,483,647 | PyPy 3 | OK | TESTS | 31 | 154 | 0 | # Test case 1
'''
1: 1 2 3 4 5 6 7 8 9 10
2: 1 2 3 4 5 6 7 8 9 10
3: 1 2 3 4 5 6 7 8 9 10
4: 1 2 3 4 5 6 7 8 9 10
5: 1 2 3 4 5 6 7 8 9 10
# 2 5
1: 1 2 3 4 1 2 3 4 5 6 7 8 9 10
2: 0
3: 1 2 3 4 5 6 7 8 9 10 6 7 8 9 10
4: 1 2 3 4 5 6 7 8 9 10
5: 1 2 3 4 5 6 7 8 9 10
# 3 13
1: 1 2 3 4 1 2 3 4 5 6 7 8 9 10
2: 1 2 3 4 5 6 7 8 9 10 6 7
3: 0
4: 1 2 3 4 5 6 7 8 9 10 9 10
5: 1 2 3 4 5 6 7 8 9 10
# 2 12
1: 1 2 3 4 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 6
2: 7
3: 0
4: 1 2 3 4 5 6 7 8 9 10 9 10
5: 1 2 3 4 5 6 7 8 9 10
# 1 13
1: 9
2: 7 10 1 2 3 4 5 6 7 8 9 10 6
3: 0
4: 1 2 3 4 5 6 7 8 9 10 9 10
5: 1 2 3 4 5 6 7 8 9 10
# 4 6
1: 0 (0)
2: 7 10 1 2 3 4 5 6 7 8 9 10 6
3: 1 2 3 4 5
4: 0
5: 1 2 3 4 5 6 7 8 9 10 7 8 9 10 9 10
'''
wire = int(input())
arr = list(map(int, input().split()))
m = int(input())
for i in range(m):
x, y = map(int, input().split())
index = x - 1
amount = arr[index]
jump_up = y - 1
jump_down = amount - y
arr[index] = 0
if index - 1 >= 0:
arr[index-1] += jump_up
if index + 1 < len(arr):
arr[index+1] += jump_down
for v in arr:
print(v) | Title: Shaass and Oskols
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
Input Specification:
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
Output Specification:
On the *i*-th line of the output print the number of birds on the *i*-th wire.
Demo Input:
['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n']
Demo Output:
['0\n12\n5\n0\n16\n', '3\n0\n3\n']
Note:
none | ```python
# Test case 1
'''
1: 1 2 3 4 5 6 7 8 9 10
2: 1 2 3 4 5 6 7 8 9 10
3: 1 2 3 4 5 6 7 8 9 10
4: 1 2 3 4 5 6 7 8 9 10
5: 1 2 3 4 5 6 7 8 9 10
# 2 5
1: 1 2 3 4 1 2 3 4 5 6 7 8 9 10
2: 0
3: 1 2 3 4 5 6 7 8 9 10 6 7 8 9 10
4: 1 2 3 4 5 6 7 8 9 10
5: 1 2 3 4 5 6 7 8 9 10
# 3 13
1: 1 2 3 4 1 2 3 4 5 6 7 8 9 10
2: 1 2 3 4 5 6 7 8 9 10 6 7
3: 0
4: 1 2 3 4 5 6 7 8 9 10 9 10
5: 1 2 3 4 5 6 7 8 9 10
# 2 12
1: 1 2 3 4 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 6
2: 7
3: 0
4: 1 2 3 4 5 6 7 8 9 10 9 10
5: 1 2 3 4 5 6 7 8 9 10
# 1 13
1: 9
2: 7 10 1 2 3 4 5 6 7 8 9 10 6
3: 0
4: 1 2 3 4 5 6 7 8 9 10 9 10
5: 1 2 3 4 5 6 7 8 9 10
# 4 6
1: 0 (0)
2: 7 10 1 2 3 4 5 6 7 8 9 10 6
3: 1 2 3 4 5
4: 0
5: 1 2 3 4 5 6 7 8 9 10 7 8 9 10 9 10
'''
wire = int(input())
arr = list(map(int, input().split()))
m = int(input())
for i in range(m):
x, y = map(int, input().split())
index = x - 1
amount = arr[index]
jump_up = y - 1
jump_down = amount - y
arr[index] = 0
if index - 1 >= 0:
arr[index-1] += jump_up
if index + 1 < len(arr):
arr[index+1] += jump_down
for v in arr:
print(v)
``` | 3 |
|
490 | A | Team Olympiad | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that? | The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child. | In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0. | [
"7\n1 3 1 3 2 1 2\n",
"4\n2 1 1 2\n"
] | [
"2\n3 5 2\n6 7 4\n",
"0\n"
] | none | 500 | [
{
"input": "7\n1 3 1 3 2 1 2",
"output": "2\n3 5 2\n6 7 4"
},
{
"input": "4\n2 1 1 2",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "1\n1 2 3"
},
{
"input": "12\n3 3 3 3 3 3 3 3 1 3 3 2",
"output": "1\n9 12 2"
},
{
"input": "60\n3 3 1 2 2 1 3 1 1 1 3 2 2 2 3 3 1 3 2 3 2 2 1 3 3 2 3 1 2 2 2 1 3 2 1 1 3 3 1 1 1 3 1 2 1 1 3 3 3 2 3 2 3 2 2 2 1 1 1 2",
"output": "20\n6 60 1\n17 44 20\n3 5 33\n36 21 42\n59 14 2\n58 26 49\n9 29 48\n23 19 24\n10 30 37\n41 54 15\n45 31 27\n57 55 38\n39 12 25\n35 34 11\n32 52 7\n8 50 18\n43 4 53\n46 56 51\n40 22 16\n28 13 47"
},
{
"input": "12\n3 1 1 1 1 1 1 2 1 1 1 1",
"output": "1\n3 8 1"
},
{
"input": "22\n2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 1 2 2 2 2",
"output": "1\n18 2 11"
},
{
"input": "138\n2 3 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 3 2 2 2 1 2 3 2 2 2 3 1 3 2 3 2 3 2 2 2 2 3 2 2 2 2 2 1 2 2 3 2 2 3 2 1 2 2 2 2 2 3 1 2 2 2 2 2 3 2 2 3 2 2 2 2 2 1 1 2 3 2 2 2 2 3 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 3 2 3 2 2 2 1 2 2 2 1 2 2 2 2 1 2 2 2 2 1 3",
"output": "18\n13 91 84\n34 90 48\n11 39 77\n78 129 50\n137 68 119\n132 122 138\n19 12 96\n40 7 2\n22 88 69\n107 73 46\n115 15 52\n127 106 87\n93 92 66\n71 112 117\n63 124 42\n17 70 101\n109 121 57\n123 25 36"
},
{
"input": "203\n2 2 1 2 1 2 2 2 1 2 2 1 1 3 1 2 1 2 1 1 2 3 1 1 2 3 3 2 2 2 1 2 1 1 1 1 1 3 1 1 2 1 1 2 2 2 1 2 2 2 1 2 3 2 1 1 2 2 1 2 1 2 2 1 1 2 2 2 1 1 2 2 1 2 1 2 2 3 2 1 2 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 2 2 2 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 2 2 2 1 2 2 1 3 2 1 1 1 2 1 1 2 1 1 2 2 2 1 1 2 2 2 1 2 1 3 2 1 2 2 2 1 1 1 2 2 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 2 1 1 1 1 1 1 2 2 3 1 1 2 3 1 1 1 1 1 1 2 2 1 1 1 2 2 3 2 1 3 1 1 1",
"output": "13\n188 72 14\n137 4 197\n158 76 122\n152 142 26\n104 119 179\n40 63 38\n12 1 78\n17 30 27\n189 60 53\n166 190 144\n129 7 183\n83 41 22\n121 81 200"
},
{
"input": "220\n1 1 3 1 3 1 1 3 1 3 3 3 3 1 3 3 1 3 3 3 3 3 1 1 1 3 1 1 1 3 2 3 3 3 1 1 3 3 1 1 3 3 3 3 1 3 3 1 1 1 2 3 1 1 1 2 3 3 3 2 3 1 1 3 1 1 1 3 2 1 3 2 3 1 1 3 3 3 1 3 1 1 1 3 3 2 1 3 2 1 1 3 3 1 1 1 2 1 1 3 2 1 2 1 1 1 3 1 3 3 1 2 3 3 3 3 1 3 1 1 1 1 2 3 1 1 1 1 1 1 3 2 3 1 3 1 3 1 1 3 1 3 1 3 1 3 1 3 3 2 3 1 3 3 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 2 1 1 1 3 3 1 3 3 3 1 1 1 3 1 3 3 1 1 1 2 3 1 1 3 1 1 1 1 2 3 1 1 2 3 3 1 3 1 3 3 3 3 1 3 2 3 1 1 3",
"output": "20\n198 89 20\n141 56 131\n166 204 19\n160 132 142\n111 112 195\n45 216 92\n6 31 109\n14 150 170\n199 60 18\n173 123 140\n134 69 156\n82 191 85\n126 200 80\n24 97 46\n62 86 149\n214 101 26\n79 171 78\n125 72 118\n172 103 162\n219 51 64"
},
{
"input": "61\n2 3 1 3 2 2 2 3 1 3 2 3 1 2 1 1 2 2 2 2 3 2 3 1 2 1 3 1 3 2 1 1 3 2 1 3 3 3 1 3 3 1 1 3 1 3 2 2 1 2 2 2 1 3 2 3 1 3 3 1 1",
"output": "20\n9 55 2\n24 34 27\n3 5 37\n35 17 41\n61 11 4\n60 19 54\n15 20 59\n26 14 29\n16 22 38\n43 50 12\n49 25 36\n57 51 40\n39 6 33\n32 30 10\n31 48 8\n13 47 23\n45 1 58\n53 52 56\n42 18 21\n28 7 44"
},
{
"input": "5\n1 2 2 3 3",
"output": "1\n1 3 4"
}
] | 1,675,299,922 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | b = input()
a = list(map(int, input().split()))
res = [[],[],[]]
for i in range(len(a)):
if(a[i] == 1):
res[0].append(i)
elif(a[i] == 2):
res[1].append(i)
else:
res[2].append(i)
j = min(len(i) for i in res)
print(j)
for i in range(j):
f = ""
for t in res:
f += str(t.pop()) + " "
print(f) | Title: Team Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
Output Specification:
In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
Demo Input:
['7\n1 3 1 3 2 1 2\n', '4\n2 1 1 2\n']
Demo Output:
['2\n3 5 2\n6 7 4\n', '0\n']
Note:
none | ```python
b = input()
a = list(map(int, input().split()))
res = [[],[],[]]
for i in range(len(a)):
if(a[i] == 1):
res[0].append(i)
elif(a[i] == 2):
res[1].append(i)
else:
res[2].append(i)
j = min(len(i) for i in res)
print(j)
for i in range(j):
f = ""
for t in res:
f += str(t.pop()) + " "
print(f)
``` | 0 |
|
446 | A | DZY Loves Sequences | PROGRAMMING | 1,600 | [
"dp",
"implementation",
"two pointers"
] | null | null | DZY has a sequence *a*, consisting of *n* integers.
We'll call a sequence *a**i*,<=*a**i*<=+<=1,<=...,<=*a**j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) a subsegment of the sequence *a*. The value (*j*<=-<=*i*<=+<=1) denotes the length of the subsegment.
Your task is to find the longest subsegment of *a*, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.
You only need to output the length of the subsegment you find. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). | In a single line print the answer to the problem — the maximum length of the required subsegment. | [
"6\n7 2 3 1 5 6\n"
] | [
"5\n"
] | You can choose subsegment *a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>, *a*<sub class="lower-index">6</sub> and change its 3rd element (that is *a*<sub class="lower-index">4</sub>) to 4. | 500 | [
{
"input": "6\n7 2 3 1 5 6",
"output": "5"
},
{
"input": "10\n424238336 649760493 681692778 714636916 719885387 804289384 846930887 957747794 596516650 189641422",
"output": "9"
},
{
"input": "50\n804289384 846930887 681692778 714636916 957747794 424238336 719885387 649760493 596516650 189641422 25202363 350490028 783368691 102520060 44897764 967513927 365180541 540383427 304089173 303455737 35005212 521595369 294702568 726956430 336465783 861021531 59961394 89018457 101513930 125898168 131176230 145174068 233665124 278722863 315634023 369133070 468703136 628175012 635723059 653377374 656478043 801979803 859484422 914544920 608413785 756898538 734575199 973594325 149798316 38664371",
"output": "19"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "2"
},
{
"input": "5\n1 2 3 4 1",
"output": "5"
},
{
"input": "10\n1 2 3 4 5 5 6 7 8 9",
"output": "6"
},
{
"input": "5\n1 1 1 1 1",
"output": "2"
},
{
"input": "5\n1 1 2 3 4",
"output": "5"
},
{
"input": "5\n1 2 3 1 6",
"output": "5"
},
{
"input": "1\n42",
"output": "1"
},
{
"input": "5\n1 2 42 3 4",
"output": "4"
},
{
"input": "5\n1 5 9 6 10",
"output": "4"
},
{
"input": "5\n5 2 3 4 5",
"output": "5"
},
{
"input": "3\n2 1 3",
"output": "3"
},
{
"input": "5\n1 2 3 3 4",
"output": "4"
},
{
"input": "8\n1 2 3 4 1 5 6 7",
"output": "5"
},
{
"input": "1\n3",
"output": "1"
},
{
"input": "3\n5 1 2",
"output": "3"
},
{
"input": "4\n1 4 3 4",
"output": "4"
},
{
"input": "6\n7 2 12 4 5 6",
"output": "5"
},
{
"input": "6\n7 2 3 1 4 5",
"output": "4"
},
{
"input": "6\n2 3 5 5 6 7",
"output": "6"
},
{
"input": "5\n2 4 7 6 8",
"output": "5"
},
{
"input": "3\n3 1 2",
"output": "3"
},
{
"input": "3\n1 1 2",
"output": "3"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "5\n4 1 2 3 4",
"output": "5"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6",
"output": "7"
},
{
"input": "4\n1 2 1 3",
"output": "3"
},
{
"input": "4\n4 3 1 2",
"output": "3"
},
{
"input": "6\n1 2 2 3 4 5",
"output": "5"
},
{
"input": "4\n1 1 1 2",
"output": "3"
},
{
"input": "4\n5 1 2 3",
"output": "4"
},
{
"input": "5\n9 1 2 3 4",
"output": "5"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "5\n1 3 2 4 5",
"output": "4"
},
{
"input": "6\n1 2 1 2 4 5",
"output": "5"
},
{
"input": "10\n1 1 5 3 2 9 9 7 7 6",
"output": "3"
},
{
"input": "6\n1 2 3 100000 100 101",
"output": "6"
},
{
"input": "4\n3 3 3 4",
"output": "3"
},
{
"input": "3\n4 3 5",
"output": "3"
},
{
"input": "5\n1 3 2 3 4",
"output": "4"
},
{
"input": "10\n1 2 3 4 5 10 10 11 12 13",
"output": "10"
},
{
"input": "7\n11 2 1 2 13 4 14",
"output": "5"
},
{
"input": "3\n5 1 3",
"output": "3"
},
{
"input": "4\n1 5 3 4",
"output": "4"
},
{
"input": "10\n1 2 3 4 100 6 7 8 9 10",
"output": "10"
},
{
"input": "3\n5 3 5",
"output": "3"
},
{
"input": "5\n100 100 7 8 9",
"output": "4"
},
{
"input": "5\n1 2 3 4 5",
"output": "5"
},
{
"input": "5\n1 2 4 4 5",
"output": "5"
},
{
"input": "6\n7 4 5 6 7 8",
"output": "6"
},
{
"input": "9\n3 4 1 6 3 4 5 6 7",
"output": "7"
},
{
"input": "3\n1000 1 2",
"output": "3"
},
{
"input": "3\n20 1 9",
"output": "3"
},
{
"input": "6\n7 2 3 1 4 6",
"output": "4"
},
{
"input": "3\n100 5 10",
"output": "3"
},
{
"input": "4\n2 2 2 3",
"output": "3"
},
{
"input": "6\n4 2 8 1 2 5",
"output": "4"
},
{
"input": "3\n25 1 6",
"output": "3"
},
{
"input": "10\n17 99 23 72 78 36 5 43 95 9",
"output": "5"
},
{
"input": "7\n21 16 22 21 11 13 19",
"output": "4"
},
{
"input": "5\n1 2 5 3 4",
"output": "4"
},
{
"input": "6\n2 2 2 3 4 5",
"output": "5"
},
{
"input": "5\n1 3 1 2 3",
"output": "4"
},
{
"input": "3\n81 33 64",
"output": "3"
},
{
"input": "7\n14 3 3 19 13 19 15",
"output": "4"
},
{
"input": "9\n1 2 3 4 5 42 7 8 9",
"output": "9"
},
{
"input": "5\n2 3 7 5 6",
"output": "5"
},
{
"input": "5\n1 3 3 4 5",
"output": "5"
},
{
"input": "6\n1 5 4 3 4 5",
"output": "4"
}
] | 1,655,546,554 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 62 | 3,379,200 | from collections import defaultdict as dd, deque as dq, Counter as ctr
import sys
from re import search
input = lambda: sys.stdin.readline().rstrip('\r\n')
from bisect import bisect_left as bl
from bisect import bisect_right as br
inp = lambda: int(sys.stdin.readline().rstrip("\r\n"))
mi = lambda x=int : map(x, sys.stdin.readline().rstrip("\r\n").split())
li = lambda x=int: list(mi(x))
lb = lambda: list(map(int, sys.stdin.readline().rstrip("\r\n")))
ls = lambda: list(sys.stdin.readline().rstrip("\r\n"))
bi = lambda n: bin(n).replace("0b", "")
def sbsq(a,s):
i=0
for x in s:
if i>=len(a):
break
if x==a[i]:
i+=1
return i==len(a)
yn = ['No', 'Yes']
YN = ['NO', 'YES']
YY = "YES"
NN = "NO"
yy = "Yes"
nn = "No"
inf= sys.maxsize
mod=1e9+7
from math import log2
import heapq as hq
def main(kase):
n=inp()
a=li()
if n==1:
print(1)
return
elif n==2:
print(2)
return
l=[0]*n
r=[0]*n
j=0
for i in range(n):
while j<n-1 and a[j+1]>a[j] or j<i:
j+=1
r[i]=j-i+1
j=n-1
for i in range(n-1,-1,-1):
while j>0 and a[j-1]<a[j] or j>i:
j-=1
l[i]=i-j+1
ans=0
for i in range(1,n-1):
if a[i-1]+1<a[i+1]:
ans=max(ans,l[i-1]+1+r[i+1] )
print(ans)
pass
if __name__ == "__main__":
test_Cases=1
# test_Cases=inp()
for i in range(test_Cases):
main(i) | Title: DZY Loves Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY has a sequence *a*, consisting of *n* integers.
We'll call a sequence *a**i*,<=*a**i*<=+<=1,<=...,<=*a**j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) a subsegment of the sequence *a*. The value (*j*<=-<=*i*<=+<=1) denotes the length of the subsegment.
Your task is to find the longest subsegment of *a*, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.
You only need to output the length of the subsegment you find.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
In a single line print the answer to the problem — the maximum length of the required subsegment.
Demo Input:
['6\n7 2 3 1 5 6\n']
Demo Output:
['5\n']
Note:
You can choose subsegment *a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>, *a*<sub class="lower-index">6</sub> and change its 3rd element (that is *a*<sub class="lower-index">4</sub>) to 4. | ```python
from collections import defaultdict as dd, deque as dq, Counter as ctr
import sys
from re import search
input = lambda: sys.stdin.readline().rstrip('\r\n')
from bisect import bisect_left as bl
from bisect import bisect_right as br
inp = lambda: int(sys.stdin.readline().rstrip("\r\n"))
mi = lambda x=int : map(x, sys.stdin.readline().rstrip("\r\n").split())
li = lambda x=int: list(mi(x))
lb = lambda: list(map(int, sys.stdin.readline().rstrip("\r\n")))
ls = lambda: list(sys.stdin.readline().rstrip("\r\n"))
bi = lambda n: bin(n).replace("0b", "")
def sbsq(a,s):
i=0
for x in s:
if i>=len(a):
break
if x==a[i]:
i+=1
return i==len(a)
yn = ['No', 'Yes']
YN = ['NO', 'YES']
YY = "YES"
NN = "NO"
yy = "Yes"
nn = "No"
inf= sys.maxsize
mod=1e9+7
from math import log2
import heapq as hq
def main(kase):
n=inp()
a=li()
if n==1:
print(1)
return
elif n==2:
print(2)
return
l=[0]*n
r=[0]*n
j=0
for i in range(n):
while j<n-1 and a[j+1]>a[j] or j<i:
j+=1
r[i]=j-i+1
j=n-1
for i in range(n-1,-1,-1):
while j>0 and a[j-1]<a[j] or j>i:
j-=1
l[i]=i-j+1
ans=0
for i in range(1,n-1):
if a[i-1]+1<a[i+1]:
ans=max(ans,l[i-1]+1+r[i+1] )
print(ans)
pass
if __name__ == "__main__":
test_Cases=1
# test_Cases=inp()
for i in range(test_Cases):
main(i)
``` | 0 |
|
895 | A | Pizza Separation | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360. | Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya. | [
"4\n90 90 90 90\n",
"3\n100 100 160\n",
"1\n360\n",
"4\n170 30 150 10\n"
] | [
"0\n",
"40\n",
"360\n",
"0\n"
] | In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | 500 | [
{
"input": "4\n90 90 90 90",
"output": "0"
},
{
"input": "3\n100 100 160",
"output": "40"
},
{
"input": "1\n360",
"output": "360"
},
{
"input": "4\n170 30 150 10",
"output": "0"
},
{
"input": "5\n10 10 10 10 320",
"output": "280"
},
{
"input": "8\n45 45 45 45 45 45 45 45",
"output": "0"
},
{
"input": "3\n120 120 120",
"output": "120"
},
{
"input": "5\n110 90 70 50 40",
"output": "40"
},
{
"input": "2\n170 190",
"output": "20"
},
{
"input": "15\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 10",
"output": "10"
},
{
"input": "5\n30 60 180 60 30",
"output": "0"
},
{
"input": "2\n359 1",
"output": "358"
},
{
"input": "5\n100 100 30 100 30",
"output": "40"
},
{
"input": "5\n36 34 35 11 244",
"output": "128"
},
{
"input": "5\n96 94 95 71 4",
"output": "18"
},
{
"input": "2\n85 275",
"output": "190"
},
{
"input": "3\n281 67 12",
"output": "202"
},
{
"input": "5\n211 113 25 9 2",
"output": "62"
},
{
"input": "13\n286 58 6 1 1 1 1 1 1 1 1 1 1",
"output": "212"
},
{
"input": "15\n172 69 41 67 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "20\n226 96 2 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "92"
},
{
"input": "50\n148 53 32 11 4 56 8 2 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "0"
},
{
"input": "3\n1 1 358",
"output": "356"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 341",
"output": "322"
},
{
"input": "33\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 1 328",
"output": "296"
},
{
"input": "70\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 1 291",
"output": "222"
},
{
"input": "130\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 231",
"output": "102"
},
{
"input": "200\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 161",
"output": "0"
},
{
"input": "222\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 139",
"output": "0"
},
{
"input": "10\n8 3 11 4 1 10 10 1 8 304",
"output": "248"
},
{
"input": "12\n8 7 7 3 11 2 10 1 10 8 10 283",
"output": "206"
},
{
"input": "13\n10 8 9 10 5 9 4 1 10 11 1 7 275",
"output": "190"
},
{
"input": "14\n1 6 3 11 9 5 9 8 5 6 7 3 7 280",
"output": "200"
},
{
"input": "15\n10 11 5 4 11 5 4 1 5 4 5 5 9 6 275",
"output": "190"
},
{
"input": "30\n8 7 5 8 3 7 2 4 3 8 11 3 9 11 2 4 1 4 5 6 11 5 8 3 6 3 11 2 11 189",
"output": "18"
},
{
"input": "70\n5 3 6 8 9 2 8 9 11 5 2 8 9 11 7 6 6 9 7 11 7 6 3 8 2 4 4 8 4 3 2 2 3 5 6 5 11 2 7 7 5 8 10 5 2 1 10 9 4 10 7 1 8 10 9 1 5 1 1 1 2 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "29\n2 10 1 5 7 2 9 11 9 9 10 8 4 11 2 5 4 1 4 9 6 10 8 3 1 3 8 9 189",
"output": "18"
},
{
"input": "35\n3 4 11 4 4 2 3 4 3 9 7 10 2 7 8 3 11 3 6 4 6 7 11 10 8 7 6 7 2 8 5 3 2 2 168",
"output": "0"
},
{
"input": "60\n4 10 3 10 6 3 11 8 11 9 3 5 9 2 6 5 6 9 4 10 1 1 3 7 2 10 5 5 3 10 5 2 1 2 9 11 11 9 11 4 11 7 5 6 10 9 3 4 7 8 7 3 6 7 8 5 1 1 1 5",
"output": "0"
},
{
"input": "71\n3 11 8 1 10 1 7 9 6 4 11 10 11 2 4 1 11 7 9 10 11 4 8 7 11 3 8 4 1 8 4 2 9 9 7 10 10 9 5 7 9 7 2 1 7 6 5 11 5 9 4 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "63\n2 11 5 8 7 9 9 8 10 5 9 10 11 8 10 2 3 5 3 7 5 10 2 9 4 8 1 8 5 9 7 7 1 8 7 7 9 10 10 10 8 7 7 2 2 8 9 7 10 8 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "81\n5 8 7 11 2 7 1 1 5 8 7 2 3 11 4 9 7 6 4 4 2 1 1 7 9 4 1 8 3 1 4 10 7 9 9 8 11 3 4 3 10 8 6 4 7 2 4 3 6 11 11 10 7 10 2 10 8 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": "0"
},
{
"input": "47\n5 3 7 4 2 7 8 1 9 10 5 11 10 7 7 5 1 3 2 11 3 8 6 1 6 10 8 3 2 10 5 6 8 6 9 7 10 9 7 4 8 11 10 1 5 11 68",
"output": "0"
},
{
"input": "100\n5 8 9 3 2 3 9 8 11 10 4 8 1 1 1 1 6 5 10 9 5 3 7 7 2 11 10 2 3 2 2 8 7 3 5 5 10 9 2 5 10 6 7 7 4 7 7 8 2 8 9 9 2 4 1 1 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "0"
},
{
"input": "120\n9 11 3 7 3 7 9 1 10 7 11 4 1 5 3 5 6 3 1 11 8 8 11 7 3 5 1 9 1 7 10 10 10 10 9 5 4 8 2 8 2 1 4 5 3 11 3 5 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "0"
},
{
"input": "200\n7 7 9 8 2 8 5 8 3 9 7 10 2 9 11 8 11 7 5 2 6 3 11 9 5 1 10 2 1 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "0"
},
{
"input": "220\n3 2 8 1 3 5 5 11 1 5 2 6 9 2 2 6 8 10 7 1 3 2 10 9 10 10 4 10 9 5 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "0"
},
{
"input": "6\n27 15 28 34 41 215",
"output": "70"
},
{
"input": "7\n41 38 41 31 22 41 146",
"output": "14"
},
{
"input": "8\n24 27 34 23 29 23 30 170",
"output": "20"
},
{
"input": "9\n11 11 20 20 33 32 35 26 172",
"output": "6"
},
{
"input": "10\n36 13 28 13 33 34 23 25 34 121",
"output": "0"
},
{
"input": "11\n19 37 13 41 37 15 32 12 19 35 100",
"output": "10"
},
{
"input": "12\n37 25 34 38 21 24 34 38 11 29 28 41",
"output": "2"
},
{
"input": "13\n24 40 20 26 25 29 39 29 35 28 19 18 28",
"output": "2"
},
{
"input": "14\n11 21 40 19 28 34 13 16 23 30 34 22 25 44",
"output": "4"
},
{
"input": "3\n95 91 174",
"output": "12"
},
{
"input": "4\n82 75 78 125",
"output": "46"
},
{
"input": "6\n87 75 88 94 15 1",
"output": "4"
},
{
"input": "10\n27 52 58 64 45 64 1 19 2 28",
"output": "12"
},
{
"input": "50\n14 12 11 8 1 6 11 6 7 8 4 11 4 5 7 3 5 4 7 24 10 2 3 4 6 13 2 1 8 7 5 13 10 8 5 20 1 2 23 7 14 3 4 4 2 8 8 2 6 1",
"output": "0"
},
{
"input": "100\n3 3 4 3 3 6 3 2 8 2 13 3 1 1 2 1 3 4 1 7 1 2 2 6 3 2 10 3 1 2 5 6 2 3 3 2 3 11 8 3 2 6 1 3 3 4 7 7 2 2 1 2 6 3 3 2 3 1 3 8 2 6 4 2 1 12 2 2 2 1 4 1 4 1 3 1 3 1 5 2 6 6 7 1 2 3 2 4 4 2 5 9 8 2 4 6 5 1 1 3",
"output": "0"
},
{
"input": "150\n1 5 1 2 2 2 1 4 2 2 2 3 1 2 1 2 2 2 2 1 2 2 2 1 5 3 4 1 3 4 5 2 4 2 1 2 2 1 1 2 3 2 4 2 2 3 3 1 1 5 2 3 2 1 9 2 1 1 2 1 4 1 1 3 2 2 2 1 2 2 2 1 3 3 4 2 2 1 3 3 3 1 4 3 4 1 2 2 1 1 1 2 2 5 4 1 1 1 2 1 2 3 2 2 6 3 3 3 1 2 1 1 2 8 2 2 4 3 4 5 3 1 4 2 2 2 2 1 4 4 1 1 2 2 4 9 6 3 1 1 2 1 3 4 1 3 2 2 2 1",
"output": "0"
},
{
"input": "200\n1 2 1 3 1 3 1 2 1 4 6 1 2 2 2 2 1 1 1 1 3 2 1 2 2 2 1 2 2 2 2 1 1 1 3 2 3 1 1 2 1 1 2 1 1 1 1 1 1 2 1 2 2 4 1 3 1 2 1 2 2 1 2 1 3 1 1 2 2 1 1 1 1 2 4 1 2 1 1 1 2 1 3 1 1 3 1 2 2 4 1 1 2 1 2 1 2 2 2 2 1 1 2 1 2 1 3 3 1 1 1 2 1 3 3 1 2 1 3 1 3 3 1 2 2 1 4 1 2 2 1 2 2 4 2 5 1 2 2 1 2 1 2 1 5 2 1 2 2 1 2 4 1 2 2 4 2 3 2 3 1 2 1 1 2 2 2 1 1 2 1 4 1 2 1 1 2 1 2 3 1 1 1 2 2 3 1 3 2 2 3 1 2 1 2 1 1 2 1 2",
"output": "0"
},
{
"input": "5\n35 80 45 100 100",
"output": "40"
},
{
"input": "4\n90 179 90 1",
"output": "2"
},
{
"input": "5\n50 50 20 160 80",
"output": "0"
},
{
"input": "5\n30 175 30 5 120",
"output": "10"
},
{
"input": "4\n170 30 10 150",
"output": "20"
},
{
"input": "6\n90 30 90 30 90 30",
"output": "60"
},
{
"input": "4\n70 80 110 100",
"output": "20"
},
{
"input": "7\n35 45 70 100 10 10 90",
"output": "0"
},
{
"input": "6\n50 90 10 90 20 100",
"output": "20"
},
{
"input": "6\n10 155 162 1 26 6",
"output": "18"
},
{
"input": "7\n80 90 80 45 10 10 45",
"output": "20"
},
{
"input": "4\n18 36 162 144",
"output": "36"
},
{
"input": "5\n20 50 50 160 80",
"output": "40"
},
{
"input": "5\n10 30 140 20 160",
"output": "0"
},
{
"input": "6\n90 80 60 50 40 40",
"output": "20"
},
{
"input": "9\n40 20 20 20 20 20 20 40 160",
"output": "40"
},
{
"input": "4\n90 54 90 126",
"output": "72"
},
{
"input": "4\n150 170 30 10",
"output": "20"
},
{
"input": "8\n130 12 13 85 41 67 5 7",
"output": "26"
},
{
"input": "7\n70 170 20 10 30 30 30",
"output": "20"
},
{
"input": "8\n100 100 50 50 15 15 15 15",
"output": "40"
},
{
"input": "4\n100 70 80 110",
"output": "20"
},
{
"input": "5\n160 130 40 20 10",
"output": "20"
},
{
"input": "4\n20 149 151 40",
"output": "22"
},
{
"input": "4\n100 10 100 150",
"output": "60"
},
{
"input": "6\n19 64 105 168 1 3",
"output": "16"
},
{
"input": "8\n10 10 70 70 90 90 10 10",
"output": "0"
}
] | 1,516,901,119 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 61 | 5,632,000 | N=int(input())
L=list(map(int,(input().split(" "))))
sum1=0
i=0
rr=360
b=10**9
while(sum1!=180 and i<N):
sum1=sum1+L[i]
rr=rr-L[i]
hh=abs(rr-sum1)
if(hh<b):
if(sum1<rr):
b=hh
else:
b=hh
break
else:
rr=rr+L[i]
sum1=sum1-L[i]
i=i+1
sum2=360-sum1
diff=abs(sum1-sum2)
if(diff in L or ((diff//2) in L)):
diff=0
print(diff) | Title: Pizza Separation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
Output Specification:
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.
Demo Input:
['4\n90 90 90 90\n', '3\n100 100 160\n', '1\n360\n', '4\n170 30 150 10\n']
Demo Output:
['0\n', '40\n', '360\n', '0\n']
Note:
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | ```python
N=int(input())
L=list(map(int,(input().split(" "))))
sum1=0
i=0
rr=360
b=10**9
while(sum1!=180 and i<N):
sum1=sum1+L[i]
rr=rr-L[i]
hh=abs(rr-sum1)
if(hh<b):
if(sum1<rr):
b=hh
else:
b=hh
break
else:
rr=rr+L[i]
sum1=sum1-L[i]
i=i+1
sum2=360-sum1
diff=abs(sum1-sum2)
if(diff in L or ((diff//2) in L)):
diff=0
print(diff)
``` | 0 |
|
992 | A | Nastya and an Array | PROGRAMMING | 800 | [
"implementation",
"sortings"
] | null | null | Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array. | Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero. | [
"5\n1 1 1 1 1\n",
"3\n2 0 -1\n",
"4\n5 -6 -5 1\n"
] | [
"1\n",
"2\n",
"4\n"
] | In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element. | 500 | [
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "3\n2 0 -1",
"output": "2"
},
{
"input": "4\n5 -6 -5 1",
"output": "4"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n21794 -79194",
"output": "2"
},
{
"input": "3\n-63526 95085 -5239",
"output": "3"
},
{
"input": "3\n0 53372 -20572",
"output": "2"
},
{
"input": "13\n-2075 -32242 27034 -37618 -96962 82203 64846 48249 -71761 28908 -21222 -61370 46899",
"output": "13"
},
{
"input": "5\n806 0 1308 1954 683",
"output": "4"
},
{
"input": "8\n-26 0 -249 -289 -126 -206 288 -11",
"output": "7"
},
{
"input": "10\n2 2 2 1 2 -1 0 2 -1 1",
"output": "3"
},
{
"input": "1\n8",
"output": "1"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "10"
},
{
"input": "5\n2 0 -1 0 0",
"output": "2"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n-1 0",
"output": "1"
},
{
"input": "4\n0 0 0 0",
"output": "0"
},
{
"input": "8\n10 9 -1 0 0 3 2 3",
"output": "5"
},
{
"input": "5\n5 0 1 2 3",
"output": "4"
},
{
"input": "3\n1 1 0",
"output": "1"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "5\n1 2 0 0 0",
"output": "2"
},
{
"input": "5\n1 0 0 0 0",
"output": "1"
},
{
"input": "5\n4 5 6 0 0",
"output": "3"
},
{
"input": "4\n-1 0 0 1",
"output": "2"
},
{
"input": "5\n3 0 0 4 5",
"output": "3"
},
{
"input": "3\n0 0 2",
"output": "1"
},
{
"input": "3\n1 0 0",
"output": "1"
},
{
"input": "4\n0 0 0 4",
"output": "1"
},
{
"input": "5\n-1 0 0 0 0",
"output": "1"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "10\n0 0 0 0 0 1 2 3 0 0",
"output": "3"
},
{
"input": "4\n0 1 2 3",
"output": "3"
},
{
"input": "3\n0 1 2",
"output": "2"
},
{
"input": "4\n2 0 0 -1",
"output": "2"
}
] | 1,694,666,091 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 1,000 | 13,619,200 | def magic_array(array):
n = len(array)
time = 0
while True:
non_zero_count = 0
minimum = float('inf')
for i in range(n):
if array[i] != 0:
non_zero_count += 1
minimum = min(minimum, array[i])
if non_zero_count == 0:
return time
for i in range(n):
if array[i] != 0:
array[i] += -minimum
time += 1
if __name__ == '__main__':
a=int(input())
array = list(map(int,input().split()))
time = magic_array(array)
print(time) | Title: Nastya and an Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
Output Specification:
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
Demo Input:
['5\n1 1 1 1 1\n', '3\n2 0 -1\n', '4\n5 -6 -5 1\n']
Demo Output:
['1\n', '2\n', '4\n']
Note:
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element. | ```python
def magic_array(array):
n = len(array)
time = 0
while True:
non_zero_count = 0
minimum = float('inf')
for i in range(n):
if array[i] != 0:
non_zero_count += 1
minimum = min(minimum, array[i])
if non_zero_count == 0:
return time
for i in range(n):
if array[i] != 0:
array[i] += -minimum
time += 1
if __name__ == '__main__':
a=int(input())
array = list(map(int,input().split()))
time = magic_array(array)
print(time)
``` | 0 |
|
416 | A | Guess a number! | PROGRAMMING | 1,400 | [
"greedy",
"implementation",
"two pointers"
] | null | null | A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show.
The rules are simple. The host thinks of an integer *y* and the participants guess it by asking questions to the host. There are four types of acceptable questions:
- Is it true that *y* is strictly larger than number *x*? - Is it true that *y* is strictly smaller than number *x*? - Is it true that *y* is larger than or equal to number *x*? - Is it true that *y* is smaller than or equal to number *x*?
On each question the host answers truthfully, "yes" or "no".
Given the sequence of questions and answers, find any integer value of *y* that meets the criteria of all answers. If there isn't such value, print "Impossible". | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10000) — the number of questions (and answers). Next *n* lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is:
- ">" (for the first type queries), - "<" (for the second type queries), - ">=" (for the third type queries), - "<=" (for the fourth type queries).
All values of *x* are integer and meet the inequation <=-<=109<=≤<=*x*<=≤<=109. The answer is an English letter "Y" (for "yes") or "N" (for "no").
Consequtive elements in lines are separated by a single space. | Print any of such integers *y*, that the answers to all the queries are correct. The printed number *y* must meet the inequation <=-<=2·109<=≤<=*y*<=≤<=2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). | [
"4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N\n",
"2\n> 100 Y\n< -100 Y\n"
] | [
"17\n",
"Impossible\n"
] | none | 500 | [
{
"input": "4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N",
"output": "17"
},
{
"input": "2\n> 100 Y\n< -100 Y",
"output": "Impossible"
},
{
"input": "4\n< 1 N\n> 1 N\n> 1 N\n> 1 N",
"output": "1"
},
{
"input": "4\n<= 1 Y\n>= 1 Y\n>= 1 Y\n<= 1 Y",
"output": "1"
},
{
"input": "4\n< 10 Y\n> -6 Y\n< 10 Y\n< -10 N",
"output": "-5"
},
{
"input": "1\n< 1 N",
"output": "1361956"
},
{
"input": "1\n<= 1 Y",
"output": "-1998638045"
},
{
"input": "1\n> 1 N",
"output": "-1998638045"
},
{
"input": "1\n>= 1 Y",
"output": "1361956"
},
{
"input": "4\n< 1 N\n< 1 N\n< 1 N\n<= 1 Y",
"output": "1"
},
{
"input": "4\n< 1 N\n>= 1 Y\n< 1 N\n< 1 N",
"output": "1361956"
},
{
"input": "4\n> 1 N\n<= 1 Y\n<= 1 Y\n> 1 N",
"output": "-1998638045"
},
{
"input": "4\n>= 1 Y\n> 1 N\n>= 1 Y\n>= 1 Y",
"output": "1"
},
{
"input": "4\n<= 9 Y\n< 3 Y\n< 2 Y\n< 2 Y",
"output": "-1998638045"
},
{
"input": "4\n< 0 N\n< -7 N\n>= 8 N\n>= -5 Y",
"output": "3"
},
{
"input": "4\n<= -6 N\n<= -8 N\n<= 3 Y\n<= 7 Y",
"output": "-2"
},
{
"input": "4\n>= 7 N\n<= -1 N\n>= 5 N\n<= -10 N",
"output": "0"
},
{
"input": "4\n> 5 N\n>= -5 Y\n> -9 Y\n> -9 Y",
"output": "-4"
},
{
"input": "10\n<= -60 N\n>= -59 Y\n> 22 Y\n> 95 N\n<= 91 Y\n> 77 Y\n>= -59 Y\n> -25 Y\n> -22 Y\n>= 52 Y",
"output": "85"
},
{
"input": "10\n>= -18 Y\n>= -35 Y\n> -94 Y\n< -23 N\n< -69 N\n< -68 N\n< 82 Y\n> 92 N\n< 29 Y\n>= -25 Y",
"output": "18"
},
{
"input": "10\n>= 18 Y\n<= -32 N\n>= 85 N\n<= 98 Y\n<= -43 N\n<= -79 N\n>= 97 N\n< -38 N\n< -55 N\n<= -93 N",
"output": "64"
},
{
"input": "10\n<= 2 Y\n< -33 Y\n> 6 N\n> -6 N\n< -28 Y\n> -62 Y\n< 57 Y\n<= 24 Y\n> 23 N\n> -25 N",
"output": "-54"
},
{
"input": "10\n<= -31 N\n>= 66 N\n<= 0 Y\n> -95 Y\n< 27 Y\n< -42 N\n> 3 N\n< 6 Y\n>= -42 Y\n> -70 Y",
"output": "-29"
},
{
"input": "10\n>= 54 N\n<= -52 N\n>= 64 N\n> 65 N\n< 37 Y\n> -84 Y\n>= -94 Y\n>= -95 Y\n> -72 Y\n<= 18 N",
"output": "22"
},
{
"input": "10\n> -24 N\n<= -5 Y\n<= -33 Y\n> 45 N\n> -59 Y\n> -21 N\n<= -48 N\n> 40 N\n< 12 Y\n>= 14 N",
"output": "-47"
},
{
"input": "10\n>= 91 Y\n>= -68 Y\n< 92 N\n>= -15 Y\n> 51 Y\n<= 14 N\n> 17 Y\n< 94 Y\n>= 49 Y\n> -36 Y",
"output": "93"
},
{
"input": "1\n< -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n< 1 Y",
"output": "-1998638045"
},
{
"input": "1\n>= -999999999 Y",
"output": "-998638044"
},
{
"input": "1\n> 100000 Y",
"output": "1461956"
},
{
"input": "1\n<= 999999999 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 1000000000 N",
"output": "1001361956"
},
{
"input": "4\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n>= 1000000000 Y",
"output": "1001361955"
},
{
"input": "1\n<= 999999999 N",
"output": "1001361955"
},
{
"input": "1\n<= 100 Y",
"output": "-1998638045"
},
{
"input": "1\n> 1000000000 Y",
"output": "1001361956"
},
{
"input": "1\n<= 1 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n<= -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n<= -999999999 Y",
"output": "-1998638045"
},
{
"input": "1\n> 100 Y",
"output": "1362056"
},
{
"input": "2\n< -1000000000 Y\n< 3 Y",
"output": "-1998638045"
},
{
"input": "1\n<= -1000000 Y",
"output": "-1998638045"
},
{
"input": "8\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 15 Y",
"output": "-1998638045"
},
{
"input": "1\n< 1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n> 10000000 Y",
"output": "11361956"
},
{
"input": "1\n< 0 Y",
"output": "-1998638045"
},
{
"input": "1\n< 100 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 5 Y",
"output": "-1998638045"
},
{
"input": "3\n> 5 Y\n> 0 Y\n< 4 Y",
"output": "Impossible"
},
{
"input": "1\n>= -1000000000 N",
"output": "-1998638045"
},
{
"input": "3\n>= 1 Y\n<= 1 Y\n> 10 Y",
"output": "Impossible"
},
{
"input": "2\n> 1 Y\n< 2 Y",
"output": "Impossible"
},
{
"input": "3\n>= 5 Y\n<= 5 Y\n< 5 Y",
"output": "Impossible"
},
{
"input": "2\n>= 5 N\n> 5 Y",
"output": "Impossible"
},
{
"input": "3\n>= 4 Y\n> 4 Y\n<= 4 Y",
"output": "Impossible"
},
{
"input": "2\n>= 4 Y\n> 4 Y",
"output": "1361960"
}
] | 1,589,260,812 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 202 | 2,048,000 | def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
n=int(input())
c=[0]*n
p=[0]*n
ind=[0]*n
for i in range(n):
c[i],p[i]=map(int,input().split())
ind[i]=i
c=sort_list(c,p)
ind=sort_list(ind,p)
ind.reverse()
p.sort(reverse=True)
c.reverse()
ans=0
cost=0
k=int(input())
l=list(map(int,input().split()))
d=dict()
for i in range(n):
ind1=-1
cur=999999999999999999
for j in range(k):
if c[i]<=l[j] and l[j]<=cur:
ind1=j
cur=l[j]
if cur!=999999999999999999:
ans+=1
cost+=p[i]
l[ind1]=999999999999999999999999
d.update({ind[i]:ind1+1})
print(ans,cost)
for i in sorted(d.keys()):
print(i+1,d[i]+1) | Title: Guess a number!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show.
The rules are simple. The host thinks of an integer *y* and the participants guess it by asking questions to the host. There are four types of acceptable questions:
- Is it true that *y* is strictly larger than number *x*? - Is it true that *y* is strictly smaller than number *x*? - Is it true that *y* is larger than or equal to number *x*? - Is it true that *y* is smaller than or equal to number *x*?
On each question the host answers truthfully, "yes" or "no".
Given the sequence of questions and answers, find any integer value of *y* that meets the criteria of all answers. If there isn't such value, print "Impossible".
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10000) — the number of questions (and answers). Next *n* lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is:
- ">" (for the first type queries), - "<" (for the second type queries), - ">=" (for the third type queries), - "<=" (for the fourth type queries).
All values of *x* are integer and meet the inequation <=-<=109<=≤<=*x*<=≤<=109. The answer is an English letter "Y" (for "yes") or "N" (for "no").
Consequtive elements in lines are separated by a single space.
Output Specification:
Print any of such integers *y*, that the answers to all the queries are correct. The printed number *y* must meet the inequation <=-<=2·109<=≤<=*y*<=≤<=2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes).
Demo Input:
['4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N\n', '2\n> 100 Y\n< -100 Y\n']
Demo Output:
['17\n', 'Impossible\n']
Note:
none | ```python
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
n=int(input())
c=[0]*n
p=[0]*n
ind=[0]*n
for i in range(n):
c[i],p[i]=map(int,input().split())
ind[i]=i
c=sort_list(c,p)
ind=sort_list(ind,p)
ind.reverse()
p.sort(reverse=True)
c.reverse()
ans=0
cost=0
k=int(input())
l=list(map(int,input().split()))
d=dict()
for i in range(n):
ind1=-1
cur=999999999999999999
for j in range(k):
if c[i]<=l[j] and l[j]<=cur:
ind1=j
cur=l[j]
if cur!=999999999999999999:
ans+=1
cost+=p[i]
l[ind1]=999999999999999999999999
d.update({ind[i]:ind1+1})
print(ans,cost)
for i in sorted(d.keys()):
print(i+1,d[i]+1)
``` | -1 |
|
767 | A | Snacktower | PROGRAMMING | 1,100 | [
"data structures",
"implementation"
] | null | null | According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*. | Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty. | [
"3\n3 1 2\n",
"5\n4 5 1 2 3\n"
] | [
"3\n \n2 1",
"5 4\n \n \n3 2 1\n"
] | In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | 500 | [
{
"input": "3\n3 1 2",
"output": "3 \n\n2 1 "
},
{
"input": "5\n4 5 1 2 3",
"output": "5 4 \n\n\n3 2 1 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "10 \n9 8 \n7 6 5 4 3 2 1 "
},
{
"input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7",
"output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 "
},
{
"input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39",
"output": "100 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n99 98 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 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 \n39 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 "
},
{
"input": "2\n2 1",
"output": "2 \n1 "
}
] | 1,695,662,827 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 2,000 | 13,516,800 | num = int(input())
our_arr = list(map(int,input().split()))
stacked = []
waiting = []
haha = num
x = ""
for i in range(num):
waiting.append(our_arr[i])
if haha not in waiting:
print("")
if haha in waiting:
while haha in waiting:
stacked.append(haha)
haha -=1
print(' '.join(map(str, stacked)))
stacked.clear() | Title: Snacktower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
Output Specification:
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
Demo Input:
['3\n3 1 2\n', '5\n4 5 1 2 3\n']
Demo Output:
['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n']
Note:
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | ```python
num = int(input())
our_arr = list(map(int,input().split()))
stacked = []
waiting = []
haha = num
x = ""
for i in range(num):
waiting.append(our_arr[i])
if haha not in waiting:
print("")
if haha in waiting:
while haha in waiting:
stacked.append(haha)
haha -=1
print(' '.join(map(str, stacked)))
stacked.clear()
``` | 0 |
|
264 | A | Escape from Stones | PROGRAMMING | 1,200 | [
"constructive algorithms",
"data structures",
"implementation",
"two pointers"
] | null | null | Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*].
You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls. | The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r". | Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left. | [
"llrlr\n",
"rrlll\n",
"lrlrr\n"
] | [
"3\n5\n4\n2\n1\n",
"1\n2\n5\n4\n3\n",
"2\n4\n5\n3\n1\n"
] | In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1. | 500 | [
{
"input": "llrlr",
"output": "3\n5\n4\n2\n1"
},
{
"input": "rrlll",
"output": "1\n2\n5\n4\n3"
},
{
"input": "lrlrr",
"output": "2\n4\n5\n3\n1"
},
{
"input": "lllrlrllrl",
"output": "4\n6\n9\n10\n8\n7\n5\n3\n2\n1"
},
{
"input": "llrlrrrlrr",
"output": "3\n5\n6\n7\n9\n10\n8\n4\n2\n1"
},
{
"input": "rlrrrllrrr",
"output": "1\n3\n4\n5\n8\n9\n10\n7\n6\n2"
},
{
"input": "lrrlrrllrrrrllllllrr",
"output": "2\n3\n5\n6\n9\n10\n11\n12\n19\n20\n18\n17\n16\n15\n14\n13\n8\n7\n4\n1"
},
{
"input": "rlrrrlrrrllrrllrlrll",
"output": "1\n3\n4\n5\n7\n8\n9\n12\n13\n16\n18\n20\n19\n17\n15\n14\n11\n10\n6\n2"
},
{
"input": "lllrrlrlrllrrrrrllrl",
"output": "4\n5\n7\n9\n12\n13\n14\n15\n16\n19\n20\n18\n17\n11\n10\n8\n6\n3\n2\n1"
},
{
"input": "rrrllrrrlllrlllrlrrr",
"output": "1\n2\n3\n6\n7\n8\n12\n16\n18\n19\n20\n17\n15\n14\n13\n11\n10\n9\n5\n4"
},
{
"input": "rrlllrrrlrrlrrrlllrlrlrrrlllrllrrllrllrrlrlrrllllrlrrrrlrlllrlrrrlrlrllrlrlrrlrrllrrrlrlrlllrrllllrl",
"output": "1\n2\n6\n7\n8\n10\n11\n13\n14\n15\n19\n21\n23\n24\n25\n29\n32\n33\n36\n39\n40\n42\n44\n45\n50\n52\n53\n54\n55\n57\n61\n63\n64\n65\n67\n69\n72\n74\n76\n77\n79\n80\n83\n84\n85\n87\n89\n93\n94\n99\n100\n98\n97\n96\n95\n92\n91\n90\n88\n86\n82\n81\n78\n75\n73\n71\n70\n68\n66\n62\n60\n59\n58\n56\n51\n49\n48\n47\n46\n43\n41\n38\n37\n35\n34\n31\n30\n28\n27\n26\n22\n20\n18\n17\n16\n12\n9\n5\n4\n3"
},
{
"input": "llrlrlllrrllrllllrlrrlrlrrllrlrlrrlrrrrrrlllrrlrrrrrlrrrlrlrlrrlllllrrrrllrrlrlrrrllllrlrrlrrlrlrrll",
"output": "3\n5\n9\n10\n13\n18\n20\n21\n23\n25\n26\n29\n31\n33\n34\n36\n37\n38\n39\n40\n41\n45\n46\n48\n49\n50\n51\n52\n54\n55\n56\n58\n60\n62\n63\n69\n70\n71\n72\n75\n76\n78\n80\n81\n82\n87\n89\n90\n92\n93\n95\n97\n98\n100\n99\n96\n94\n91\n88\n86\n85\n84\n83\n79\n77\n74\n73\n68\n67\n66\n65\n64\n61\n59\n57\n53\n47\n44\n43\n42\n35\n32\n30\n28\n27\n24\n22\n19\n17\n16\n15\n14\n12\n11\n8\n7\n6\n4\n2\n1"
},
{
"input": "llrrrrllrrlllrlrllrlrllllllrrrrrrrrllrrrrrrllrlrrrlllrrrrrrlllllllrrlrrllrrrllllrrlllrrrlrlrrlrlrllr",
"output": "3\n4\n5\n6\n9\n10\n14\n16\n19\n21\n28\n29\n30\n31\n32\n33\n34\n35\n38\n39\n40\n41\n42\n43\n46\n48\n49\n50\n54\n55\n56\n57\n58\n59\n67\n68\n70\n71\n74\n75\n76\n81\n82\n86\n87\n88\n90\n92\n93\n95\n97\n100\n99\n98\n96\n94\n91\n89\n85\n84\n83\n80\n79\n78\n77\n73\n72\n69\n66\n65\n64\n63\n62\n61\n60\n53\n52\n51\n47\n45\n44\n37\n36\n27\n26\n25\n24\n23\n22\n20\n18\n17\n15\n13\n12\n11\n8\n7\n2\n1"
},
{
"input": "lllllrllrrlllrrrllrrrrlrrlrllllrrrrrllrlrllllllrrlrllrlrllrlrrlrlrrlrrrlrrrrllrlrrrrrrrllrllrrlrllrl",
"output": "6\n9\n10\n14\n15\n16\n19\n20\n21\n22\n24\n25\n27\n32\n33\n34\n35\n36\n39\n41\n48\n49\n51\n54\n56\n59\n61\n62\n64\n66\n67\n69\n70\n71\n73\n74\n75\n76\n79\n81\n82\n83\n84\n85\n86\n87\n90\n93\n94\n96\n99\n100\n98\n97\n95\n92\n91\n89\n88\n80\n78\n77\n72\n68\n65\n63\n60\n58\n57\n55\n53\n52\n50\n47\n46\n45\n44\n43\n42\n40\n38\n37\n31\n30\n29\n28\n26\n23\n18\n17\n13\n12\n11\n8\n7\n5\n4\n3\n2\n1"
},
{
"input": "llrlrlrlrlrlrrlllllllrllllrllrrrlllrrllrllrrlllrrlllrlrrllllrrlllrrllrrllllrrlllrlllrrrllrrrrrrllrrl",
"output": "3\n5\n7\n9\n11\n13\n14\n22\n27\n30\n31\n32\n36\n37\n40\n43\n44\n48\n49\n53\n55\n56\n61\n62\n66\n67\n70\n71\n76\n77\n81\n85\n86\n87\n90\n91\n92\n93\n94\n95\n98\n99\n100\n97\n96\n89\n88\n84\n83\n82\n80\n79\n78\n75\n74\n73\n72\n69\n68\n65\n64\n63\n60\n59\n58\n57\n54\n52\n51\n50\n47\n46\n45\n42\n41\n39\n38\n35\n34\n33\n29\n28\n26\n25\n24\n23\n21\n20\n19\n18\n17\n16\n15\n12\n10\n8\n6\n4\n2\n1"
},
{
"input": "l",
"output": "1"
},
{
"input": "r",
"output": "1"
}
] | 1,643,039,639 | 2,147,483,647 | Python 3 | OK | TESTS | 57 | 1,715 | 30,003,200 | def main():
line = input()
l = [None] * len(line)
ptr1 = 0
ptr2 = len(line)-1
counter = 0
for c in line:
counter += 1
if c == 'l':
l[ptr2] = counter
ptr2 -= 1
else:
l[ptr1] = counter
ptr1 += 1
for n in l:
print(n)
if __name__ == "__main__":
main()
| Title: Escape from Stones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*].
You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls.
Input Specification:
The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r".
Output Specification:
Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left.
Demo Input:
['llrlr\n', 'rrlll\n', 'lrlrr\n']
Demo Output:
['3\n5\n4\n2\n1\n', '1\n2\n5\n4\n3\n', '2\n4\n5\n3\n1\n']
Note:
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1. | ```python
def main():
line = input()
l = [None] * len(line)
ptr1 = 0
ptr2 = len(line)-1
counter = 0
for c in line:
counter += 1
if c == 'l':
l[ptr2] = counter
ptr2 -= 1
else:
l[ptr1] = counter
ptr1 += 1
for n in l:
print(n)
if __name__ == "__main__":
main()
``` | 3 |
|
908 | A | New Year and Counting Cards | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true. | The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit. | Print a single integer, the minimum number of cards you must turn over to verify your claim. | [
"ee\n",
"z\n",
"0ay1\n"
] | [
"2\n",
"0\n",
"2\n"
] | In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | 500 | [
{
"input": "ee",
"output": "2"
},
{
"input": "z",
"output": "0"
},
{
"input": "0ay1",
"output": "2"
},
{
"input": "0abcdefghijklmnopqrstuvwxyz1234567896",
"output": "10"
},
{
"input": "0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b",
"output": "18"
},
{
"input": "01234567890123456789012345678901234567890123456789",
"output": "25"
},
{
"input": "qwertyuioplkjhgfdsazxcvbnmqwertyuioplkjhgfdsazxcvb",
"output": "10"
},
{
"input": "cjw2dwmr10pku4yxohe0wglktd",
"output": "4"
},
{
"input": "6z2tx805jie8cfybwtfqvmlveec3iak5z5u3lu62vbxyqht6",
"output": "13"
},
{
"input": "kaq7jyialrfp4ilkni90eq8v3amcbygon7py0hb8z26fbl8ss1",
"output": "13"
},
{
"input": "hpwn50zgbmct80k9rizjqg40nycgs0acwikjqt11nr6m61krfs",
"output": "8"
},
{
"input": "l3rw91a4m25l8iytxyeuixsegzcbm4h41ornf3pixkrmwznrzc",
"output": "14"
},
{
"input": "2222",
"output": "0"
},
{
"input": "13579",
"output": "5"
},
{
"input": "1",
"output": "1"
},
{
"input": "0",
"output": "0"
},
{
"input": "a",
"output": "1"
},
{
"input": "y",
"output": "0"
},
{
"input": "w",
"output": "0"
},
{
"input": "oo",
"output": "2"
},
{
"input": "oy",
"output": "1"
},
{
"input": "yo",
"output": "1"
},
{
"input": "yy",
"output": "0"
},
{
"input": "a0",
"output": "1"
},
{
"input": "a9",
"output": "2"
},
{
"input": "y0",
"output": "0"
},
{
"input": "y7",
"output": "1"
},
{
"input": "0a",
"output": "1"
},
{
"input": "3a",
"output": "2"
},
{
"input": "06",
"output": "0"
},
{
"input": "07",
"output": "1"
},
{
"input": "70",
"output": "1"
},
{
"input": "77",
"output": "2"
},
{
"input": "13570",
"output": "4"
},
{
"input": "0000000000011111",
"output": "5"
},
{
"input": "1357",
"output": "4"
},
{
"input": "uuuuuuuuuuuuuuuuuu",
"output": "18"
},
{
"input": "gabieurat",
"output": "5"
}
] | 1,598,679,623 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 124 | 0 | s=input()
c=0
for i in range(len(s)):
tmp=s[i]
if s[i]=="1":
c+=1
elif s[i]=='a' or s[i]=='e' or s[i]=='i' or s[i]=='o' or s[i]=='u':
c+=1
print(c) | Title: New Year and Counting Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input Specification:
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
Output Specification:
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Demo Input:
['ee\n', 'z\n', '0ay1\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | ```python
s=input()
c=0
for i in range(len(s)):
tmp=s[i]
if s[i]=="1":
c+=1
elif s[i]=='a' or s[i]=='e' or s[i]=='i' or s[i]=='o' or s[i]=='u':
c+=1
print(c)
``` | 0 |
|
47 | B | Coins | PROGRAMMING | 1,200 | [
"implementation"
] | B. Coins | 2 | 256 | One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal. | The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B. | It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights. | [
"A>B\nC<B\nA>C\n",
"A<B\nB>C\nC>A\n"
] | [
"CBA",
"ACB"
] | none | 1,000 | [
{
"input": "A>B\nC<B\nA>C",
"output": "CBA"
},
{
"input": "A<B\nB>C\nC>A",
"output": "ACB"
},
{
"input": "A<C\nB<A\nB>C",
"output": "Impossible"
},
{
"input": "A<B\nA<C\nB>C",
"output": "ACB"
},
{
"input": "B>A\nC<B\nC>A",
"output": "ACB"
},
{
"input": "A>B\nB>C\nC<A",
"output": "CBA"
},
{
"input": "A>C\nA>B\nB<C",
"output": "BCA"
},
{
"input": "C<B\nB>A\nA<C",
"output": "ACB"
},
{
"input": "C<B\nA>B\nC<A",
"output": "CBA"
},
{
"input": "C>B\nB>A\nA<C",
"output": "ABC"
},
{
"input": "C<B\nB<A\nC>A",
"output": "Impossible"
},
{
"input": "B<C\nC<A\nA>B",
"output": "BCA"
},
{
"input": "A>B\nC<B\nC<A",
"output": "CBA"
},
{
"input": "B>A\nC>B\nA>C",
"output": "Impossible"
},
{
"input": "B<A\nC>B\nC>A",
"output": "BAC"
},
{
"input": "A<B\nC>B\nA<C",
"output": "ABC"
},
{
"input": "A<B\nC<A\nB<C",
"output": "Impossible"
},
{
"input": "A>C\nC<B\nB>A",
"output": "CAB"
},
{
"input": "C>A\nA<B\nB>C",
"output": "ACB"
},
{
"input": "C>A\nC<B\nB>A",
"output": "ACB"
},
{
"input": "B>C\nB>A\nA<C",
"output": "ACB"
},
{
"input": "C<B\nC<A\nB<A",
"output": "CBA"
},
{
"input": "A<C\nA<B\nB>C",
"output": "ACB"
},
{
"input": "B>A\nA>C\nB>C",
"output": "CAB"
},
{
"input": "B<A\nA<C\nC<B",
"output": "Impossible"
},
{
"input": "A<C\nB>C\nA>B",
"output": "Impossible"
},
{
"input": "B>A\nC<A\nC>B",
"output": "Impossible"
},
{
"input": "A>C\nC>B\nB<A",
"output": "BCA"
},
{
"input": "B<C\nB<A\nA>C",
"output": "BCA"
},
{
"input": "A>B\nC>B\nA<C",
"output": "BAC"
},
{
"input": "C<B\nC<A\nB<A",
"output": "CBA"
},
{
"input": "A<C\nA>B\nB>C",
"output": "Impossible"
},
{
"input": "B>A\nB>C\nA<C",
"output": "ACB"
},
{
"input": "B>C\nC<A\nB<A",
"output": "CBA"
},
{
"input": "C>A\nB>A\nB>C",
"output": "ACB"
},
{
"input": "B<A\nB>C\nA<C",
"output": "Impossible"
},
{
"input": "B<C\nA<B\nC>A",
"output": "ABC"
},
{
"input": "C<B\nB>A\nA>C",
"output": "CAB"
},
{
"input": "A>B\nC>A\nB<C",
"output": "BAC"
},
{
"input": "A>B\nB>C\nC<A",
"output": "CBA"
},
{
"input": "B>C\nC>A\nA>B",
"output": "Impossible"
},
{
"input": "B<A\nB>C\nC<A",
"output": "CBA"
},
{
"input": "C>B\nB>A\nC>A",
"output": "ABC"
},
{
"input": "A>C\nA>B\nB>C",
"output": "CBA"
},
{
"input": "B<C\nA>B\nA<C",
"output": "BAC"
},
{
"input": "C>A\nC<B\nB>A",
"output": "ACB"
},
{
"input": "C>B\nA<B\nC<A",
"output": "Impossible"
},
{
"input": "A<C\nC<B\nA>B",
"output": "Impossible"
},
{
"input": "B>C\nA<B\nA<C",
"output": "ACB"
},
{
"input": "C>A\nA<B\nC>B",
"output": "ABC"
}
] | 1,690,729,915 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 92 | 0 | def find_coin_order(weighing_results):
relationships = {'A': set(), 'B': set(), 'C': set()}
for result in weighing_results:
coin1, sign, coin2 = result[0], result[1], result[2]
if sign == '>':
relationships[coin1].add(coin2)
else:
relationships[coin2].add(coin1)
heaviest = [coin for coin in relationships if not relationships[coin]]
lightest = [coin for coin in relationships if len(relationships[coin]) == 2]
if not heaviest or not lightest:
return "Impossible"
else:
middle = list(set(relationships.keys()) - set(heaviest + lightest))[0]
return heaviest[0] + middle + lightest[0]
# Read the input
weighing_results = []
for _ in range(3):
result = input().strip()
weighing_results.append(result)
output = find_coin_order(weighing_results)
print(output) | Title: Coins
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
Input Specification:
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
Output Specification:
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
Demo Input:
['A>B\nC<B\nA>C\n', 'A<B\nB>C\nC>A\n']
Demo Output:
['CBA', 'ACB']
Note:
none | ```python
def find_coin_order(weighing_results):
relationships = {'A': set(), 'B': set(), 'C': set()}
for result in weighing_results:
coin1, sign, coin2 = result[0], result[1], result[2]
if sign == '>':
relationships[coin1].add(coin2)
else:
relationships[coin2].add(coin1)
heaviest = [coin for coin in relationships if not relationships[coin]]
lightest = [coin for coin in relationships if len(relationships[coin]) == 2]
if not heaviest or not lightest:
return "Impossible"
else:
middle = list(set(relationships.keys()) - set(heaviest + lightest))[0]
return heaviest[0] + middle + lightest[0]
# Read the input
weighing_results = []
for _ in range(3):
result = input().strip()
weighing_results.append(result)
output = find_coin_order(weighing_results)
print(output)
``` | 3.977 |
946 | B | Weird Subtraction Process | PROGRAMMING | 1,100 | [
"math",
"number theory"
] | null | null | You have two variables *a* and *b*. Consider the following sequence of actions performed with these variables:
1. If *a*<==<=0 or *b*<==<=0, end the process. Otherwise, go to step 2;1. If *a*<=≥<=2·*b*, then set the value of *a* to *a*<=-<=2·*b*, and repeat step 1. Otherwise, go to step 3;1. If *b*<=≥<=2·*a*, then set the value of *b* to *b*<=-<=2·*a*, and repeat step 1. Otherwise, end the process.
Initially the values of *a* and *b* are positive integers, and so the process will be finite.
You have to determine the values of *a* and *b* after the process ends. | The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1018). *n* is the initial value of variable *a*, and *m* is the initial value of variable *b*. | Print two integers — the values of *a* and *b* after the end of the process. | [
"12 5\n",
"31 12\n"
] | [
"0 1\n",
"7 12\n"
] | Explanations to the samples:
1. *a* = 12, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 1 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 0, *b* = 1;1. *a* = 31, *b* = 12 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 7, *b* = 12. | 0 | [
{
"input": "12 5",
"output": "0 1"
},
{
"input": "31 12",
"output": "7 12"
},
{
"input": "1000000000000000000 7",
"output": "8 7"
},
{
"input": "31960284556200 8515664064180",
"output": "14928956427840 8515664064180"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1000000000000000000 1000000000000000000"
},
{
"input": "1 1000",
"output": "1 0"
},
{
"input": "1 1000000",
"output": "1 0"
},
{
"input": "1 1000000000000000",
"output": "1 0"
},
{
"input": "1 99999999999999999",
"output": "1 1"
},
{
"input": "1 4",
"output": "1 0"
},
{
"input": "1000000000000001 500000000000000",
"output": "1 0"
},
{
"input": "1 1000000000000000000",
"output": "1 0"
},
{
"input": "2 4",
"output": "2 0"
},
{
"input": "2 1",
"output": "0 1"
},
{
"input": "6 19",
"output": "6 7"
},
{
"input": "22 5",
"output": "0 1"
},
{
"input": "10000000000000000 100000000000000001",
"output": "0 1"
},
{
"input": "1 1000000000000",
"output": "1 0"
},
{
"input": "2 1000000000000000",
"output": "2 0"
},
{
"input": "2 10",
"output": "2 2"
},
{
"input": "51 100",
"output": "51 100"
},
{
"input": "3 1000000000000000000",
"output": "3 4"
},
{
"input": "1000000000000000000 3",
"output": "4 3"
},
{
"input": "1 10000000000000000",
"output": "1 0"
},
{
"input": "8796203 7556",
"output": "1019 1442"
},
{
"input": "5 22",
"output": "1 0"
},
{
"input": "1000000000000000000 1",
"output": "0 1"
},
{
"input": "1 100000000000",
"output": "1 0"
},
{
"input": "2 1000000000000",
"output": "2 0"
},
{
"input": "5 4567865432345678",
"output": "5 8"
},
{
"input": "576460752303423487 288230376151711743",
"output": "1 1"
},
{
"input": "499999999999999999 1000000000000000000",
"output": "3 2"
},
{
"input": "1 9999999999999",
"output": "1 1"
},
{
"input": "103 1000000000000000000",
"output": "103 196"
},
{
"input": "7 1",
"output": "1 1"
},
{
"input": "100000000000000001 10000000000000000",
"output": "1 0"
},
{
"input": "5 10",
"output": "5 0"
},
{
"input": "7 11",
"output": "7 11"
},
{
"input": "1 123456789123456",
"output": "1 0"
},
{
"input": "5000000000000 100000000000001",
"output": "0 1"
},
{
"input": "1000000000000000 1",
"output": "0 1"
},
{
"input": "1000000000000000000 499999999999999999",
"output": "2 3"
},
{
"input": "10 5",
"output": "0 5"
},
{
"input": "9 18917827189272",
"output": "9 0"
},
{
"input": "179 100000000000497000",
"output": "179 270"
},
{
"input": "5 100000000000001",
"output": "1 1"
},
{
"input": "5 20",
"output": "5 0"
},
{
"input": "100000001 50000000",
"output": "1 0"
},
{
"input": "345869461223138161 835002744095575440",
"output": "1 0"
},
{
"input": "8589934592 4294967296",
"output": "0 4294967296"
},
{
"input": "4 8",
"output": "4 0"
},
{
"input": "1 100000000000000000",
"output": "1 0"
},
{
"input": "1000000000000000000 333333333333333",
"output": "1000 1333"
},
{
"input": "25 12",
"output": "1 0"
},
{
"input": "24 54",
"output": "0 6"
},
{
"input": "6 12",
"output": "6 0"
},
{
"input": "129200000000305 547300000001292",
"output": "1 0"
},
{
"input": "1000000000000000000 49999999999999999",
"output": "20 39"
},
{
"input": "1 2",
"output": "1 0"
},
{
"input": "1 123456789876",
"output": "1 0"
},
{
"input": "2 3",
"output": "2 3"
},
{
"input": "1 3",
"output": "1 1"
},
{
"input": "1 1",
"output": "1 1"
},
{
"input": "19 46",
"output": "3 2"
},
{
"input": "3 6",
"output": "3 0"
},
{
"input": "129 1000000000000000000",
"output": "1 0"
},
{
"input": "12 29",
"output": "0 1"
},
{
"input": "8589934592 2147483648",
"output": "0 2147483648"
},
{
"input": "2147483648 8589934592",
"output": "2147483648 0"
},
{
"input": "5 6",
"output": "5 6"
},
{
"input": "1000000000000000000 2",
"output": "0 2"
},
{
"input": "2 7",
"output": "2 3"
},
{
"input": "17174219820754872 61797504734333370",
"output": "17174219820754872 27449065092823626"
},
{
"input": "49 100",
"output": "1 0"
},
{
"input": "7 17",
"output": "1 1"
},
{
"input": "1000000000000000000 10000001",
"output": "0 1"
},
{
"input": "49999999999999999 2",
"output": "3 2"
},
{
"input": "49999999999999999 1",
"output": "1 1"
},
{
"input": "576460752303423487 2",
"output": "3 2"
},
{
"input": "19395 19395",
"output": "19395 19395"
},
{
"input": "19394 19394",
"output": "19394 19394"
}
] | 1,563,868,566 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 124 | 0 | a, b=map (int, input ().split ())
while True:
if (a==0 or b==0) or (max(a, b)<2*min(a,b)):
print (a, b)
break
elif a>=2*b:
a=a-(a//(2*b))*2*b
else:
b=b-(b//(2*a))*2*a | Title: Weird Subtraction Process
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have two variables *a* and *b*. Consider the following sequence of actions performed with these variables:
1. If *a*<==<=0 or *b*<==<=0, end the process. Otherwise, go to step 2;1. If *a*<=≥<=2·*b*, then set the value of *a* to *a*<=-<=2·*b*, and repeat step 1. Otherwise, go to step 3;1. If *b*<=≥<=2·*a*, then set the value of *b* to *b*<=-<=2·*a*, and repeat step 1. Otherwise, end the process.
Initially the values of *a* and *b* are positive integers, and so the process will be finite.
You have to determine the values of *a* and *b* after the process ends.
Input Specification:
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1018). *n* is the initial value of variable *a*, and *m* is the initial value of variable *b*.
Output Specification:
Print two integers — the values of *a* and *b* after the end of the process.
Demo Input:
['12 5\n', '31 12\n']
Demo Output:
['0 1\n', '7 12\n']
Note:
Explanations to the samples:
1. *a* = 12, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 5 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 2, *b* = 1 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 0, *b* = 1;1. *a* = 31, *b* = 12 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> *a* = 7, *b* = 12. | ```python
a, b=map (int, input ().split ())
while True:
if (a==0 or b==0) or (max(a, b)<2*min(a,b)):
print (a, b)
break
elif a>=2*b:
a=a-(a//(2*b))*2*b
else:
b=b-(b//(2*a))*2*a
``` | 3 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below. | Input will consist of a single integer *A* (1<=≤<=*A*<=≤<=105), the desired number of ways. | In the first line print integers *N* and *M* (1<=≤<=*N*<=≤<=106,<=1<=≤<=*M*<=≤<=10), the amount of change to be made, and the number of denominations, respectively.
Then print *M* integers *D*1,<=*D*2,<=...,<=*D**M* (1<=≤<=*D**i*<=≤<=106), the denominations of the coins. All denominations must be distinct: for any *i*<=≠<=*j* we must have *D**i*<=≠<=*D**j*.
If there are multiple tests, print any of them. You can print denominations in atbitrary order. | [
"18\n",
"3\n",
"314\n"
] | [
"30 4\n1 5 10 25\n",
"20 2\n5 2\n",
"183 4\n6 5 2 139\n"
] | none | 0 | [
{
"input": "18",
"output": "30 4\n1 5 10 25"
},
{
"input": "3",
"output": "20 2\n5 2"
},
{
"input": "314",
"output": "183 4\n6 5 2 139"
},
{
"input": "1023",
"output": "2045 2\n1 2"
},
{
"input": "100000",
"output": "199999 2\n1 2"
},
{
"input": "1",
"output": "1 2\n1 2"
},
{
"input": "2",
"output": "3 2\n1 2"
},
{
"input": "3",
"output": "20 2\n5 2"
},
{
"input": "4",
"output": "7 2\n1 2"
},
{
"input": "5",
"output": "9 2\n1 2"
},
{
"input": "6",
"output": "11 2\n1 2"
},
{
"input": "7",
"output": "13 2\n1 2"
},
{
"input": "8",
"output": "15 2\n1 2"
},
{
"input": "9",
"output": "17 2\n1 2"
},
{
"input": "10",
"output": "19 2\n1 2"
},
{
"input": "11",
"output": "21 2\n1 2"
},
{
"input": "12",
"output": "23 2\n1 2"
},
{
"input": "13",
"output": "25 2\n1 2"
},
{
"input": "14",
"output": "27 2\n1 2"
},
{
"input": "15",
"output": "29 2\n1 2"
},
{
"input": "16",
"output": "31 2\n1 2"
},
{
"input": "17",
"output": "33 2\n1 2"
},
{
"input": "18",
"output": "30 4\n1 5 10 25"
},
{
"input": "19",
"output": "37 2\n1 2"
},
{
"input": "20",
"output": "39 2\n1 2"
},
{
"input": "239",
"output": "477 2\n1 2"
},
{
"input": "4063",
"output": "8125 2\n1 2"
},
{
"input": "65536",
"output": "131071 2\n1 2"
},
{
"input": "65535",
"output": "131069 2\n1 2"
},
{
"input": "65537",
"output": "131073 2\n1 2"
},
{
"input": "99991",
"output": "199981 2\n1 2"
},
{
"input": "99990",
"output": "199979 2\n1 2"
},
{
"input": "99992",
"output": "199983 2\n1 2"
},
{
"input": "99971",
"output": "199941 2\n1 2"
},
{
"input": "99990",
"output": "199979 2\n1 2"
},
{
"input": "2",
"output": "3 2\n1 2"
},
{
"input": "99991",
"output": "199981 2\n1 2"
}
] | 1,553,820,854 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 124 | 0 | n = int(input())
print (max(1,n*2-2),2,end=" \n")
print (1,2,end=" ") | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.
People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?
As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.
Input Specification:
Input will consist of a single integer *A* (1<=≤<=*A*<=≤<=105), the desired number of ways.
Output Specification:
In the first line print integers *N* and *M* (1<=≤<=*N*<=≤<=106,<=1<=≤<=*M*<=≤<=10), the amount of change to be made, and the number of denominations, respectively.
Then print *M* integers *D*1,<=*D*2,<=...,<=*D**M* (1<=≤<=*D**i*<=≤<=106), the denominations of the coins. All denominations must be distinct: for any *i*<=≠<=*j* we must have *D**i*<=≠<=*D**j*.
If there are multiple tests, print any of them. You can print denominations in atbitrary order.
Demo Input:
['18\n', '3\n', '314\n']
Demo Output:
['30 4\n1 5 10 25\n', '20 2\n5 2\n', '183 4\n6 5 2 139\n']
Note:
none | ```python
n = int(input())
print (max(1,n*2-2),2,end=" \n")
print (1,2,end=" ")
``` | 3 |
|
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,671,830,567 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 31 | 0 | input = list(input())
n = int(input[0])
m = int(input[2])
a = int(input[4])
f = 0
if n // a != n / a:
f += (n//a) + 1
else:
f += n/a
if m // a != m/a:
f = f*((m//a) + 1)
else:
f += f(m/a)
print(f)
| 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
input = list(input())
n = int(input[0])
m = int(input[2])
a = int(input[4])
f = 0
if n // a != n / a:
f += (n//a) + 1
else:
f += n/a
if m // a != m/a:
f = f*((m//a) + 1)
else:
f += f(m/a)
print(f)
``` | -1 |
88 | A | Chord | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | A. Chord | 2 | 256 | Vasya studies music.
He has learned lots of interesting stuff. For example, he knows that there are 12 notes: C, C#, D, D#, E, F, F#, G, G#, A, B, H. He also knows that the notes are repeated cyclically: after H goes C again, and before C stands H. We will consider the C note in the row's beginning and the C note after the H similar and we will identify them with each other. The distance between the notes along the musical scale is measured in tones: between two consecutive notes there's exactly one semitone, that is, 0.5 tone. The distance is taken from the lowest tone to the uppest one, that is, the distance between C and E is 4 semitones and between E and C is 8 semitones
Vasya also knows what a chord is. A chord is an unordered set of no less than three notes. However, for now Vasya only works with triads, that is with the chords that consist of exactly three notes. He can already distinguish between two types of triads — major and minor.
Let's define a major triad. Let the triad consist of notes *X*, *Y* and *Z*. If we can order the notes so as the distance along the musical scale between *X* and *Y* equals 4 semitones and the distance between *Y* and *Z* is 3 semitones, then the triad is major. The distance between *X* and *Z*, accordingly, equals 7 semitones.
A minor triad is different in that the distance between *X* and *Y* should be 3 semitones and between *Y* and *Z* — 4 semitones.
For example, the triad "C E G" is major: between C and E are 4 semitones, and between E and G are 3 semitones. And the triplet "C# B F" is minor, because if we order the notes as "B C# F", than between B and C# will be 3 semitones, and between C# and F — 4 semitones.
Help Vasya classify the triad the teacher has given to him. | The only line contains 3 space-separated notes in the above-given notation. | Print "major" if the chord is major, "minor" if it is minor, and "strange" if the teacher gave Vasya some weird chord which is neither major nor minor. Vasya promises you that the answer will always be unambiguous. That is, there are no chords that are both major and minor simultaneously. | [
"C E G\n",
"C# B F\n",
"A B H\n"
] | [
"major\n",
"minor\n",
"strange\n"
] | none | 500 | [
{
"input": "C E G",
"output": "major"
},
{
"input": "C# B F",
"output": "minor"
},
{
"input": "A B H",
"output": "strange"
},
{
"input": "G H E",
"output": "minor"
},
{
"input": "D# B G",
"output": "major"
},
{
"input": "D# B F#",
"output": "minor"
},
{
"input": "F H E",
"output": "strange"
},
{
"input": "B F# G",
"output": "strange"
},
{
"input": "F# H C",
"output": "strange"
},
{
"input": "C# F C",
"output": "strange"
},
{
"input": "G# C# E",
"output": "minor"
},
{
"input": "D# H G#",
"output": "minor"
},
{
"input": "C F A",
"output": "major"
},
{
"input": "H E G#",
"output": "major"
},
{
"input": "G D# B",
"output": "major"
},
{
"input": "E C G",
"output": "major"
},
{
"input": "G# C# F",
"output": "major"
},
{
"input": "D# C G#",
"output": "major"
},
{
"input": "C# F B",
"output": "minor"
},
{
"input": "D# C G",
"output": "minor"
},
{
"input": "A D F",
"output": "minor"
},
{
"input": "F# H D",
"output": "minor"
},
{
"input": "D A F",
"output": "minor"
},
{
"input": "D A F#",
"output": "major"
},
{
"input": "C# B F",
"output": "minor"
},
{
"input": "A C F",
"output": "major"
},
{
"input": "D F# H",
"output": "minor"
},
{
"input": "H G# D#",
"output": "minor"
},
{
"input": "A D F#",
"output": "major"
},
{
"input": "H E G#",
"output": "major"
},
{
"input": "D# B F#",
"output": "minor"
},
{
"input": "D# H F#",
"output": "major"
},
{
"input": "A D F#",
"output": "major"
},
{
"input": "B G D#",
"output": "major"
},
{
"input": "E A C#",
"output": "major"
},
{
"input": "D H G",
"output": "major"
},
{
"input": "H D F#",
"output": "minor"
},
{
"input": "G D# C",
"output": "minor"
},
{
"input": "H D G",
"output": "major"
},
{
"input": "E C G",
"output": "major"
},
{
"input": "D# A E",
"output": "strange"
},
{
"input": "A F E",
"output": "strange"
},
{
"input": "C E F",
"output": "strange"
},
{
"input": "A B C",
"output": "strange"
},
{
"input": "E F D#",
"output": "strange"
},
{
"input": "C G# G#",
"output": "strange"
},
{
"input": "F D# G#",
"output": "strange"
},
{
"input": "B G D#",
"output": "major"
},
{
"input": "E E G#",
"output": "strange"
},
{
"input": "A G H",
"output": "strange"
},
{
"input": "F E A",
"output": "strange"
},
{
"input": "D B E",
"output": "strange"
},
{
"input": "G# C# D",
"output": "strange"
},
{
"input": "D# D# F#",
"output": "strange"
},
{
"input": "H B G",
"output": "strange"
},
{
"input": "D C B",
"output": "strange"
},
{
"input": "D B B",
"output": "strange"
},
{
"input": "C A H",
"output": "strange"
},
{
"input": "F# H F#",
"output": "strange"
},
{
"input": "A F F#",
"output": "strange"
},
{
"input": "C D C",
"output": "strange"
},
{
"input": "G F# A",
"output": "strange"
},
{
"input": "C C# D",
"output": "strange"
},
{
"input": "C A E",
"output": "minor"
},
{
"input": "A H B",
"output": "strange"
},
{
"input": "B E F#",
"output": "strange"
},
{
"input": "G# G# A",
"output": "strange"
},
{
"input": "B C# C#",
"output": "strange"
},
{
"input": "C G D#",
"output": "minor"
},
{
"input": "C B D#",
"output": "strange"
},
{
"input": "F H F",
"output": "strange"
},
{
"input": "E G# C",
"output": "strange"
},
{
"input": "F# F# F#",
"output": "strange"
},
{
"input": "F C F",
"output": "strange"
},
{
"input": "A F D",
"output": "minor"
}
] | 1,587,970,600 | 2,147,483,647 | PyPy 3 | OK | TESTS | 75 | 310 | 0 | lt = ['C', 'C#', 'D', 'D#', 'E', 'F', 'F#', 'G', 'G#', 'A' ,'B', 'H']
major = [[4,3], [3,5], [5,4]]
minor = [[3,4], [4,5], [5,3]]
x,y,z = input().split()
ind = [lt.index(p) for p in (x,y,z)]
ind.sort()
diff = [ind[1]-ind[0], ind[2]-ind[1]]
if diff in major:
print('major')
elif diff in minor:
print('minor')
else:
print('strange') | Title: Chord
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya studies music.
He has learned lots of interesting stuff. For example, he knows that there are 12 notes: C, C#, D, D#, E, F, F#, G, G#, A, B, H. He also knows that the notes are repeated cyclically: after H goes C again, and before C stands H. We will consider the C note in the row's beginning and the C note after the H similar and we will identify them with each other. The distance between the notes along the musical scale is measured in tones: between two consecutive notes there's exactly one semitone, that is, 0.5 tone. The distance is taken from the lowest tone to the uppest one, that is, the distance between C and E is 4 semitones and between E and C is 8 semitones
Vasya also knows what a chord is. A chord is an unordered set of no less than three notes. However, for now Vasya only works with triads, that is with the chords that consist of exactly three notes. He can already distinguish between two types of triads — major and minor.
Let's define a major triad. Let the triad consist of notes *X*, *Y* and *Z*. If we can order the notes so as the distance along the musical scale between *X* and *Y* equals 4 semitones and the distance between *Y* and *Z* is 3 semitones, then the triad is major. The distance between *X* and *Z*, accordingly, equals 7 semitones.
A minor triad is different in that the distance between *X* and *Y* should be 3 semitones and between *Y* and *Z* — 4 semitones.
For example, the triad "C E G" is major: between C and E are 4 semitones, and between E and G are 3 semitones. And the triplet "C# B F" is minor, because if we order the notes as "B C# F", than between B and C# will be 3 semitones, and between C# and F — 4 semitones.
Help Vasya classify the triad the teacher has given to him.
Input Specification:
The only line contains 3 space-separated notes in the above-given notation.
Output Specification:
Print "major" if the chord is major, "minor" if it is minor, and "strange" if the teacher gave Vasya some weird chord which is neither major nor minor. Vasya promises you that the answer will always be unambiguous. That is, there are no chords that are both major and minor simultaneously.
Demo Input:
['C E G\n', 'C# B F\n', 'A B H\n']
Demo Output:
['major\n', 'minor\n', 'strange\n']
Note:
none | ```python
lt = ['C', 'C#', 'D', 'D#', 'E', 'F', 'F#', 'G', 'G#', 'A' ,'B', 'H']
major = [[4,3], [3,5], [5,4]]
minor = [[3,4], [4,5], [5,3]]
x,y,z = input().split()
ind = [lt.index(p) for p in (x,y,z)]
ind.sort()
diff = [ind[1]-ind[0], ind[2]-ind[1]]
if diff in major:
print('major')
elif diff in minor:
print('minor')
else:
print('strange')
``` | 3.9225 |
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"
}
] | 1,617,990,723 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 154 | 0 | a, b = map(int, input().split())
mul = a * b
print(mul//2)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
a, b = map(int, input().split())
mul = a * b
print(mul//2)
``` | 3.9615 |
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"
}
] | 1,605,198,156 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 109 | 0 | from math import log2
n = int(input())
m = int(input())
if log2(m) >= n:
print(m % 2**n)
else:
print(m) | 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
from math import log2
n = int(input())
m = int(input())
if log2(m) >= n:
print(m % 2**n)
else:
print(m)
``` | 3 |
|
302 | A | Eugeny and Array | PROGRAMMING | 800 | [
"implementation"
] | null | null | Eugeny has array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* integers. Each integer *a**i* equals to -1, or to 1. Also, he has *m* queries:
- Query number *i* is given as a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). - The response to the query will be integer 1, if the elements of array *a* can be rearranged so as the sum *a**l**i*<=+<=*a**l**i*<=+<=1<=+<=...<=+<=*a**r**i*<==<=0, otherwise the response to the query will be integer 0.
Help Eugeny, answer all his queries. | The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*a**i*<==<=-1,<=1). Next *m* lines contain Eugene's queries. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). | Print *m* integers — the responses to Eugene's queries in the order they occur in the input. | [
"2 3\n1 -1\n1 1\n1 2\n2 2\n",
"5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5\n"
] | [
"0\n1\n0\n",
"0\n1\n0\n1\n0\n"
] | none | 500 | [
{
"input": "2 3\n1 -1\n1 1\n1 2\n2 2",
"output": "0\n1\n0"
},
{
"input": "5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5",
"output": "0\n1\n0\n1\n0"
},
{
"input": "3 3\n1 1 1\n2 2\n1 1\n1 1",
"output": "0\n0\n0"
},
{
"input": "4 4\n-1 -1 -1 -1\n1 3\n1 2\n1 2\n1 1",
"output": "0\n0\n0\n0"
},
{
"input": "5 5\n-1 -1 -1 -1 -1\n1 1\n1 1\n3 4\n1 1\n1 4",
"output": "0\n0\n0\n0\n0"
},
{
"input": "6 6\n-1 -1 1 -1 -1 1\n1 1\n3 4\n1 1\n1 1\n1 3\n1 4",
"output": "0\n1\n0\n0\n0\n1"
},
{
"input": "7 7\n-1 -1 -1 1 -1 -1 -1\n1 1\n2 7\n1 3\n1 5\n4 7\n1 7\n6 7",
"output": "0\n0\n0\n0\n0\n0\n1"
},
{
"input": "8 8\n1 1 1 1 1 1 1 1\n5 8\n2 6\n2 3\n1 7\n7 7\n1 6\n1 8\n1 3",
"output": "0\n0\n0\n0\n0\n0\n0\n0"
},
{
"input": "9 9\n-1 1 1 1 1 1 1 1 1\n1 7\n5 6\n1 4\n1 1\n1 1\n6 8\n1 1\n6 7\n3 5",
"output": "0\n1\n0\n0\n0\n0\n0\n1\n0"
},
{
"input": "10 10\n-1 1 -1 1 -1 -1 -1 -1 -1 -1\n6 7\n2 5\n3 6\n1 3\n3 5\n4 5\n3 4\n1 6\n1 1\n1 1",
"output": "1\n1\n1\n0\n0\n1\n1\n0\n0\n0"
},
{
"input": "1 1\n-1\n1 1",
"output": "0"
},
{
"input": "1 1\n1\n1 1",
"output": "0"
}
] | 1,606,921,543 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 108 | 307,200 | def rearr(a,m,l,r):
req = abs(r-l)+1
if(req%2 != 0):
return 0
pos = req//2
neg = req//2
for i in a:
if(i==1):
pos = pos-1
else:
neg = neg-1
if(pos <= 0 and neg <= 0):
return 1
return 0
n,m = input().split()
n = int(n)
m = int(m)
for i in range(m):
l,r = input().split()
l = int(l)
r = int(r)
print(rearr(a,m,l,r)) | Title: Eugeny and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eugeny has array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* integers. Each integer *a**i* equals to -1, or to 1. Also, he has *m* queries:
- Query number *i* is given as a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). - The response to the query will be integer 1, if the elements of array *a* can be rearranged so as the sum *a**l**i*<=+<=*a**l**i*<=+<=1<=+<=...<=+<=*a**r**i*<==<=0, otherwise the response to the query will be integer 0.
Help Eugeny, answer all his queries.
Input Specification:
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*a**i*<==<=-1,<=1). Next *m* lines contain Eugene's queries. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
Output Specification:
Print *m* integers — the responses to Eugene's queries in the order they occur in the input.
Demo Input:
['2 3\n1 -1\n1 1\n1 2\n2 2\n', '5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5\n']
Demo Output:
['0\n1\n0\n', '0\n1\n0\n1\n0\n']
Note:
none | ```python
def rearr(a,m,l,r):
req = abs(r-l)+1
if(req%2 != 0):
return 0
pos = req//2
neg = req//2
for i in a:
if(i==1):
pos = pos-1
else:
neg = neg-1
if(pos <= 0 and neg <= 0):
return 1
return 0
n,m = input().split()
n = int(n)
m = int(m)
for i in range(m):
l,r = input().split()
l = int(l)
r = int(r)
print(rearr(a,m,l,r))
``` | -1 |
|
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"
}
] | 1,607,539,016 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 42 | 109 | 0 | x,y = [int(x) for x in input().split()]
c=y-1
if c>x or c<0 : print("no")
elif (x-c)%2==0: print("yes")
else: print("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 = [int(x) for x in input().split()]
c=y-1
if c>x or c<0 : print("no")
elif (x-c)%2==0: print("yes")
else: print("no")
``` | 0 |
|
716 | A | Crazy Computer | PROGRAMMING | 800 | [
"implementation"
] | null | null | ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything. | The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word. | Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*. | [
"6 5\n1 3 8 14 19 20\n",
"6 1\n1 3 5 7 9 10\n"
] | [
"3",
"2"
] | The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1. | 500 | [
{
"input": "6 5\n1 3 8 14 19 20",
"output": "3"
},
{
"input": "6 1\n1 3 5 7 9 10",
"output": "2"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "5 5\n1 7 12 13 14",
"output": "4"
},
{
"input": "2 1000000000\n1 1000000000",
"output": "2"
},
{
"input": "3 5\n1 10 20",
"output": "1"
},
{
"input": "3 10\n1 2 3",
"output": "3"
},
{
"input": "2 1\n1 100",
"output": "1"
},
{
"input": "3 1\n1 2 10",
"output": "1"
},
{
"input": "2 1\n1 2",
"output": "2"
}
] | 1,578,833,066 | 2,147,483,647 | PyPy 3 | OK | TESTS | 81 | 233 | 11,059,200 | n,m = map(int,input().split())
A = list(map(int, input().split(' ')[:n]))
temp = 1
for i in range(n-1):
if((A[i+1] - A[i])<=m):
temp += 1
else:
temp = 1
print(temp) | Title: Crazy Computer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.
Input Specification:
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word.
Output Specification:
Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*.
Demo Input:
['6 5\n1 3 8 14 19 20\n', '6 1\n1 3 5 7 9 10\n']
Demo Output:
['3', '2']
Note:
The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1. | ```python
n,m = map(int,input().split())
A = list(map(int, input().split(' ')[:n]))
temp = 1
for i in range(n-1):
if((A[i+1] - A[i])<=m):
temp += 1
else:
temp = 1
print(temp)
``` | 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"
}
] | 1,586,351,150 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 93 | 0 | n,c=list(map(int,input().split()))
p=list(map(int,input().split()))
m=0
for i in range(n-1):
if p[i]>p[i+1]:
m=max(m,p[i]-p[i+1]-c)
print(m) | 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
n,c=list(map(int,input().split()))
p=list(map(int,input().split()))
m=0
for i in range(n-1):
if p[i]>p[i+1]:
m=max(m,p[i]-p[i+1]-c)
print(m)
``` | 3 |
|
554 | A | Kyoya and Photobooks | PROGRAMMING | 900 | [
"brute force",
"math",
"strings"
] | null | null | Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem. | The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters. | Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make. | [
"a\n",
"hi\n"
] | [
"51\n",
"76\n"
] | In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets. | 250 | [
{
"input": "a",
"output": "51"
},
{
"input": "hi",
"output": "76"
},
{
"input": "y",
"output": "51"
},
{
"input": "kgan",
"output": "126"
},
{
"input": "zoabkyuvus",
"output": "276"
},
{
"input": "spyemhyznjieyhhbk",
"output": "451"
},
{
"input": "xulsyfkuizjauadjjopu",
"output": "526"
},
{
"input": "e",
"output": "51"
},
{
"input": "zv",
"output": "76"
},
{
"input": "jgv",
"output": "101"
},
{
"input": "zsfo",
"output": "126"
},
{
"input": "jselr",
"output": "151"
},
{
"input": "dwemig",
"output": "176"
},
{
"input": "mddoxsf",
"output": "201"
},
{
"input": "jgirkrmi",
"output": "226"
},
{
"input": "spkxurcum",
"output": "251"
},
{
"input": "fykkiubdkt",
"output": "276"
},
{
"input": "fznbcxsxygs",
"output": "301"
},
{
"input": "qcrvrdqcbtou",
"output": "326"
},
{
"input": "qktrbjzrqgmlr",
"output": "351"
},
{
"input": "foamodbvptlxxg",
"output": "376"
},
{
"input": "ydzpjhsidipricw",
"output": "401"
},
{
"input": "lpfpndmjfvqejdgf",
"output": "426"
},
{
"input": "ofkvparuvjtggnmab",
"output": "451"
},
{
"input": "xxncfutrtxcwdzwbgs",
"output": "476"
},
{
"input": "zovhffccflkgqncsdte",
"output": "501"
},
{
"input": "cskgsxywlvfeicoueglr",
"output": "526"
},
{
"input": "gggggggggggggggggggg",
"output": "526"
},
{
"input": "qdqdddqddqqddqddqdqd",
"output": "526"
},
{
"input": "takttttaakaaktakttkt",
"output": "526"
},
{
"input": "coccoooogogcgocccmcg",
"output": "526"
},
{
"input": "kskkskkkssksssk",
"output": "401"
},
{
"input": "lllllllllllllll",
"output": "401"
}
] | 1,435,333,011 | 2,147,483,647 | PyPy 3 | OK | TESTS | 33 | 108 | 0 | s = input()
n = len(s)
ans = n * 25 + 26
print(ans) | Title: Kyoya and Photobooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input Specification:
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
Output Specification:
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Demo Input:
['a\n', 'hi\n']
Demo Output:
['51\n', '76\n']
Note:
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets. | ```python
s = input()
n = len(s)
ans = n * 25 + 26
print(ans)
``` | 3 |
|
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,626,897,934 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 154 | 6,758,400 | n = int(input())
a = list(map(int, input().split()))
c = []
nc = []
for i in range(len(a)):
if a[i] % 2 == 0:
c.append(i+1)
else:
nc.append(i+1)
if len(c) == 1:
print(c[0])
else:
print(nc[0]) | Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
n = int(input())
a = list(map(int, input().split()))
c = []
nc = []
for i in range(len(a)):
if a[i] % 2 == 0:
c.append(i+1)
else:
nc.append(i+1)
if len(c) == 1:
print(c[0])
else:
print(nc[0])
``` | 3.948911 |
802 | G | Fake News (easy) | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it... | The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z). | Output YES if the string *s* contains heidi as a subsequence and NO otherwise. | [
"abcheaibcdi\n",
"hiedi\n"
] | [
"YES",
"NO"
] | A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*. | 0 | [
{
"input": "abcheaibcdi",
"output": "YES"
},
{
"input": "hiedi",
"output": "NO"
},
{
"input": "ihied",
"output": "NO"
},
{
"input": "diehi",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "iheid",
"output": "NO"
},
{
"input": "eihdi",
"output": "NO"
},
{
"input": "ehdii",
"output": "NO"
},
{
"input": "edhii",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "ehdii",
"output": "NO"
},
{
"input": "eufyajkssayhjhqcwxmctecaeepjwmfoscqprpcxsqfwnlgzsmmuwuoruantipholrauvxydfvftwfzhnckxswussvlidcojiciflpvkcxkkcmmvtfvxrkwcpeelwsuzqgamamdtdgzscmikvojfvqehblmjczkvtdeymgertgkwfwfukafqlfdhtedcctixhyetdypswgagrpyto",
"output": "YES"
},
{
"input": "arfbvxgdvqzuloojjrwoyqqbxamxybaqltfimofulusfebodjkwwrgwcppkwiodtpjaraglyplgerrpqjkpoggjmfxhwtqrijpijrcyxnoodvwpyjfpvqaoazllbrpzananbrvvybboedidtuvqquklkpeflfaltukjhzjgiofombhbmqbihgtapswykfvlgdoapjqntvqsaohmbvnphvyyhvhavslamczuqifxnwknkaenqmlvetrqogqxmlptgrmqvxzdxdmwobjesmgxckpmawtioavwdngyiwkzypfnxcovwzdohshwlavwsthdssiadhiwmhpvgkrbezm",
"output": "YES"
},
{
"input": "zcectngbqnejjjtsfrluummmqabzqbyccshjqbrjthzhlbmzjfxugvjouwhumsgrnopiyakfadjnbsesamhynsbfbfunupwbxvohfmpwlcpxhovwpfpciclatgmiufwdvtsqrsdcymvkldpnhfeisrzhyhhlkwdzthgprvkpyldeysvbmcibqkpudyrraqdlxpjecvwcvuiklcrsbgvqasmxmtxqzmawcjtozioqlfflinnxpeexbzloaeqjvglbdeufultpjqexvjjjkzemtzuzmxvawilcqdrcjzpqyhtwfphuonzwkotthsaxrmwtnlmcdylxqcfffyndqeouztluqwlhnkkvzwcfiscikv",
"output": "YES"
},
{
"input": "plqaykgovxkvsiahdbglktdlhcqwelxxmtlyymrsyubxdskvyjkrowvcbpdofpjqspsrgpakdczletxujzlsegepzleipiyycpinzxgwjsgslnxsotouddgfcybozfpjhhocpybfjbaywsehbcfrayvancbrumdfngqytnhihyxnlvilrqyhnxeckprqafofelospffhtwguzjbbjlzbqrtiielbvzutzgpqxosiaqznndgobcluuqlhmffiowkjdlkokehtjdyjvmxsiyxureflmdomerfekxdvtitvwzmdsdzplkpbtafxqfpudnhfqpoiwvjnylanunmagoweobdvfjgepbsymfutrjarlxclhgavpytiiqwvojrptofuvlohzeguxdsrihsbucelhhuedltnnjgzxwyblbqvnoliiydfinzlogbvucwykryzcyibnniggbkdkdcdgcsbvvnavtyhtkanrblpvomvjs",
"output": "YES"
},
{
"input": "fbldqzggeunkpwcfirxanmntbfrudijltoertsdvcvcmbwodbibsrxendzebvxwydpasaqnisrijctsuatihxxygbeovhxjdptdcppkvfytdpjspvrannxavmkmisqtygntxkdlousdypyfkrpzapysfpdbyprufwzhunlsfugojddkmxzinatiwfxdqmgyrnjnxvrclhxyuwxtshoqdjptmeecvgmrlvuwqtmnfnfeeiwcavwnqmyustawbjodzwsqmnjxhpqmgpysierlwbbdzcwprpsexyvreewcmlbvaiytjlxdqdaqftefdlmtmmjcwvfejshymhnouoshdzqcwzxpzupkbcievodzqkqvyjuuxxwepxjalvkzufnveji",
"output": "YES"
},
{
"input": "htsyljgoelbbuipivuzrhmfpkgderqpoprlxdpasxhpmxvaztccldtmujjzjmcpdvsdghzpretlsyyiljhjznseaacruriufswuvizwwuvdioazophhyytvbiogttnnouauxllbdn",
"output": "YES"
},
{
"input": "ikmxzqdzxqlvgeojsnhqzciujslwjyzzexnregabdqztpplosdakimjxmuqccbnwvzbajoiqgdobccwnrwmixohrbdarhoeeelzbpigiybtesybwefpcfx",
"output": "YES"
},
{
"input": "bpvbpjvbdfiodsmahxpcubjxdykesubnypalhypantshkjffmxjmelblqnjdmtaltneuyudyevkgedkqrdmrfeemgpghwrifcwincfixokfgurhqbcfzeajrgkgpwqwsepudxulywowwxzdxkumsicsvnzfxspmjpaixgejeaoyoibegosqoyoydmphfpbutrrewyjecowjckvpcceoamtfbitdneuwqfvnagswlskmsmkhmxyfsrpqwhxzocyffiumcy",
"output": "YES"
},
{
"input": "vllsexwrazvlfvhvrtqeohvzzresjdiuhomfpgqcxpqdevplecuaepixhlijatxzegciizpvyvxuembiplwklahlqibykfideysjygagjbgqkbhdhkatddcwlxboinfuomnpc",
"output": "YES"
},
{
"input": "pnjdwpxmvfoqkjtbhquqcuredrkwqzzfjmdvpnbqtypzdovemhhclkvigjvtprrpzbrbcbatkucaqteuciuozytsptvsskkeplaxdaqmjkmef",
"output": "NO"
},
{
"input": "jpwfhvlxvsdhtuozvlmnfiotrgapgjxtcsgcjnodcztupysvvvmjpzqkpommadppdrykuqkcpzojcwvlogvkddedwbggkrhuvtsvdiokehlkdlnukcufjvqxnikcdawvexxwffxtriqbdmkahxdtygodzohwtdmmuvmatdkvweqvaehaxiefpevkvqpyxsrhtmgjsdfcwzqobibeduooldrmglbinrepmunizheqzvgqvpdskhxfidxfnbisyizhepwyrcykcmjxnkyfjgrqlkixcvysa",
"output": "YES"
},
{
"input": "aftcrvuumeqbfvaqlltscnuhkpcifrrhnutjinxdhhdbzvizlrapzjdatuaynoplgjketupgaejciosofuhcgcjdcucarfvtsofgubtphijciswsvidnvpztlaarydkeqxzwdhfbmullkimerukusbrdnnujviydldrwhdfllsjtziwfeaiqotbiprespmxjulnyunkdtcghrzvhtcychkwatqqmladxpvmvlkzscthylbzkpgwlzfjqwarqvdeyngekqvrhrftpxnkfcibbowvnqdkulcdydspcubwlgoyinpnzgidbgunparnueddzwtzdiavbprbbg",
"output": "YES"
},
{
"input": "oagjghsidigeh",
"output": "NO"
},
{
"input": "chdhzpfzabupskiusjoefrwmjmqkbmdgboicnszkhdrlegeqjsldurmbshijadlwsycselhlnudndpdhcnhruhhvsgbthpruiqfirxkhpqhzhqdfpyozolbionodypfcqfeqbkcgmqkizgeyyelzeoothexcoaahedgrvoemqcwccbvoeqawqeuusyjxmgjkpfwcdttfmwunzuwvsihliexlzygqcgpbdiawfvqukikhbjerjkyhpcknlndaystrgsinghlmekbvhntcpypmchcwoglsmwwdulqneuabuuuvtyrnjxfcgoothalwkzzfxakneusezgnnepkpipzromqubraiggqndliz",
"output": "YES"
},
{
"input": "lgirxqkrkgjcutpqitmffvbujcljkqardlalyigxorscczuzikoylcxenryhskoavymexysvmhbsvhtycjlmzhijpuvcjshyfeycvvcfyzytzoyvxajpqdjtfiatnvxnyeqtfcagfftafllhhjhplbdsrfpctkqpinpdfrtlzyjllfbeffputywcckupyslkbbzpgcnxgbmhtqeqqehpdaokkjtatrhyiuusjhwgiiiikxpzdueasemosmmccoakafgvxduwiuflovhhfhffgnnjhoperhhjtvocpqytjxkmrknnknqeglffhfuplopmktykxuvcmbwpoeisrlyyhdpxfvzseucofyhziuiikihpqheqdyzwigeaqzhxzvporgisxgvhyicqyejovqloibhbunsvsunpvmdckkbuokitdzleilfwutcvuuytpupizinfjrzhxudsmjcjyfcpfgthujjowdwtgbvi",
"output": "YES"
},
{
"input": "uuehrvufgerqbzyzksmqnewacotuimawhlbycdbsmhshrsbqwybbkwjwsrkwptvlbbwjiivqugzrxxwgidrcrhrwsmwgeoleptfamzefgaeyxouxocrpvomjrazmxrnffdwrrmblgdiabdncvfougtmjgvvazasnygdrigbsrieoonirlivfyodvulouslxosswgpdexuldmkdbpdlgutiotvxjyecbrsvbmqxrlcpcipjjncduyqtohlzybvlemmfdeubihwlwqglkgjvnwrbgydcpwklmjeewqklmqdbajqgrpnynaxfvxjzgibqerxyhnxenrmcdqaaeksbzyrcaepozqpetaurlhjuxxhwppuhgoihxdxbmxeiahyaqkbknktlzkheaarjoqqrsyeducvoygwalgarldcdlqogfvsncejssmx",
"output": "YES"
},
{
"input": "iiopulfjxoitgiusqrhgbkiyzinphjtclodbkkydetylvuimkhdkklmyoacmekdvjpuxcrvqnjhqhhbfenlpzpwijtykqziocilvtpqhxuyrphdlamawjuzgjwiebkqyrzyqgtkcrhntjxqmcgkrqcslogjegfrivzidfedeegwbbsopvxvdoididlpypjogxaodtueebbwuwilacunqpozczcgrpaxxrtogpxgfkudtxchoravrrdtimhenwmnafxaruuojduxxglefejycfcyxllfimkszmbrhcwpnwjqgwvzysmlaaifdxfjjpgjmcksiigpanappjdloiolojmcqbnpnjjzaufdpjaknylmyvolhwypygwepmqwpiglpcnpypnudhdzpdvgzosyjthzcwtytxq",
"output": "YES"
}
] | 1,616,270,254 | 2,147,483,647 | Python 3 | OK | TESTS | 58 | 62 | 0 | s = input()
h = 'heidi'
j = 0
for i in s:
if i == h[j]:
j += 1
if j == 5:
break
print('YES' if j == 5 else 'NO')
| Title: Fake News (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
Input Specification:
The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).
Output Specification:
Output YES if the string *s* contains heidi as a subsequence and NO otherwise.
Demo Input:
['abcheaibcdi\n', 'hiedi\n']
Demo Output:
['YES', 'NO']
Note:
A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*. | ```python
s = input()
h = 'heidi'
j = 0
for i in s:
if i == h[j]:
j += 1
if j == 5:
break
print('YES' if j == 5 else 'NO')
``` | 3 |
|
535 | B | Tavas and SaDDas | PROGRAMMING | 1,100 | [
"bitmasks",
"brute force",
"combinatorics",
"implementation"
] | null | null | Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back. | The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109). | Print the index of *n* among all lucky numbers. | [
"4\n",
"7\n",
"77\n"
] | [
"1\n",
"2\n",
"6\n"
] | none | 1,000 | [
{
"input": "4",
"output": "1"
},
{
"input": "7",
"output": "2"
},
{
"input": "77",
"output": "6"
},
{
"input": "4",
"output": "1"
},
{
"input": "474744",
"output": "83"
},
{
"input": "777774",
"output": "125"
},
{
"input": "447",
"output": "8"
},
{
"input": "774",
"output": "13"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447747",
"output": "140"
},
{
"input": "7747474",
"output": "233"
},
{
"input": "4444",
"output": "15"
},
{
"input": "4447",
"output": "16"
},
{
"input": "7",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447744",
"output": "139"
},
{
"input": "77474",
"output": "57"
},
{
"input": "7747",
"output": "28"
},
{
"input": "444",
"output": "7"
},
{
"input": "7",
"output": "2"
},
{
"input": "7774477",
"output": "242"
},
{
"input": "4477774",
"output": "157"
},
{
"input": "7444",
"output": "23"
},
{
"input": "7474747",
"output": "212"
},
{
"input": "77",
"output": "6"
},
{
"input": "774477",
"output": "114"
},
{
"input": "7",
"output": "2"
},
{
"input": "47",
"output": "4"
},
{
"input": "747777",
"output": "110"
},
{
"input": "444444444",
"output": "511"
},
{
"input": "777777777",
"output": "1022"
},
{
"input": "477477447",
"output": "728"
},
{
"input": "777744747",
"output": "996"
}
] | 1,605,693,279 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 78 | 0 |
a=(input())
l=list(map(int,a))
c=len(l)-1
value=2**c
#print(value)
for i in range(len(l)):
if l[i]!=4:
y=len(l)-(i+1)
value=value+2**y
print(value+1)
| Title: Tavas and SaDDas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
Input Specification:
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the index of *n* among all lucky numbers.
Demo Input:
['4\n', '7\n', '77\n']
Demo Output:
['1\n', '2\n', '6\n']
Note:
none | ```python
a=(input())
l=list(map(int,a))
c=len(l)-1
value=2**c
#print(value)
for i in range(len(l)):
if l[i]!=4:
y=len(l)-(i+1)
value=value+2**y
print(value+1)
``` | 0 |
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