prob_desc_description stringlengths 63 3.8k | prob_desc_output_spec stringlengths 17 1.47k ⌀ | lang_cluster stringclasses 2
values | src_uid stringlengths 32 32 | code_uid stringlengths 32 32 | lang stringclasses 7
values | prob_desc_output_to stringclasses 3
values | prob_desc_memory_limit stringclasses 19
values | file_name stringclasses 111
values | tags listlengths 0 11 | prob_desc_created_at stringlengths 10 10 | prob_desc_sample_inputs stringlengths 2 802 | prob_desc_notes stringlengths 4 3k ⌀ | exec_outcome stringclasses 1
value | difficulty int64 -1 3.5k ⌀ | prob_desc_input_from stringclasses 3
values | prob_desc_time_limit stringclasses 27
values | prob_desc_input_spec stringlengths 28 2.42k ⌀ | prob_desc_sample_outputs stringlengths 2 796 | source_code stringlengths 42 65.5k | hidden_unit_tests stringclasses 1
value |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
You are given an array $$$a$$$ consisting of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You want to split it into exactly $$$k$$$ non-empty non-intersecting subsegments such that each subsegment has odd sum (i. e. for each subsegment, the sum of all elements that belong to this subsegment is odd). It is impossible to... | For each query, print the answer to it. If it is impossible to divide the initial array into exactly $$$k$$$ subsegments in such a way that each of them will have odd sum of elements, print "NO" in the first line. Otherwise, print "YES" in the first line and any possible division of the array in the second line. The di... | C | 7f5269f3357827b9d8682d70befd3de1 | 8d75c67aba11115071aef59cd2ba1ba3 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"constructive algorithms",
"math"
] | 1563978900 | ["3\n5 3\n7 18 3 14 1\n5 4\n1 2 3 4 5\n6 2\n1 2 8 4 10 2"] | null | PASSED | 1,200 | standard input | 3 seconds | The first line contains one integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Then $$$q$$$ queries follow. The first line of the query contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of subsegments, respectivel... | ["YES\n1 3 5\nNO\nNO"] | #include<stdio.h>
int main()
{
long int T,n,i,j,k,l,Sum=0,c;
scanf("%ld",&T);
for(i=0;i<T;i++)
{
c=0;
scanf("%ld",&n);
scanf("%ld",&k);
long int ar[n];
long int br[n];
for(j=0;j<n;j++)
{
scanf("%ld",&ar[j]);
if(c==k-1)
{
Sum=Sum+ar[j];
continue;
}
Sum=Sum+ar[j];
if(Sum%2!=0)
{
b... | |
You are given an array $$$a$$$ consisting of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You want to split it into exactly $$$k$$$ non-empty non-intersecting subsegments such that each subsegment has odd sum (i. e. for each subsegment, the sum of all elements that belong to this subsegment is odd). It is impossible to... | For each query, print the answer to it. If it is impossible to divide the initial array into exactly $$$k$$$ subsegments in such a way that each of them will have odd sum of elements, print "NO" in the first line. Otherwise, print "YES" in the first line and any possible division of the array in the second line. The di... | C | 7f5269f3357827b9d8682d70befd3de1 | 63bdff0eb5b118a8adcdadd6a9c2126f | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"constructive algorithms",
"math"
] | 1563978900 | ["3\n5 3\n7 18 3 14 1\n5 4\n1 2 3 4 5\n6 2\n1 2 8 4 10 2"] | null | PASSED | 1,200 | standard input | 3 seconds | The first line contains one integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Then $$$q$$$ queries follow. The first line of the query contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of subsegments, respectivel... | ["YES\n1 3 5\nNO\nNO"] | #include <stdio.h>
typedef long long ll;
#define max(l,r) ((l)>(r)?l:r)
#define min(l,r) ((l)<(r)?l:r)
#define swap(l,r) {ll tp=l;l=r;r=tp;}
#define rep(i,n) for(int i=0;i<(int)(n);i++)
int n,t,k;
int a[202020];
int solve(){
int ans = 0;
int odd = 0;
rep(i,n)
if(a[i]%2)
odd++;
... | |
You are given an array $$$a$$$ consisting of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You want to split it into exactly $$$k$$$ non-empty non-intersecting subsegments such that each subsegment has odd sum (i. e. for each subsegment, the sum of all elements that belong to this subsegment is odd). It is impossible to... | For each query, print the answer to it. If it is impossible to divide the initial array into exactly $$$k$$$ subsegments in such a way that each of them will have odd sum of elements, print "NO" in the first line. Otherwise, print "YES" in the first line and any possible division of the array in the second line. The di... | C | 7f5269f3357827b9d8682d70befd3de1 | 843abaa52c7a5aa3e5b14b5b4b8182d9 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"constructive algorithms",
"math"
] | 1563978900 | ["3\n5 3\n7 18 3 14 1\n5 4\n1 2 3 4 5\n6 2\n1 2 8 4 10 2"] | null | PASSED | 1,200 | standard input | 3 seconds | The first line contains one integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Then $$$q$$$ queries follow. The first line of the query contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of subsegments, respectivel... | ["YES\n1 3 5\nNO\nNO"] | #include <stdio.h>
#define READ(n) int (n); scanf("%d",&(n))
#define getInt(a) scanf("%d", &a)
#define FOR(i,L,R) for (int i = L; i < R; i++)
int main(){
READ(q);
while(q--){
READ(n); READ(k);
int arr[n], oddCnt = 0;
FOR(i, 0, n){
READ(a);
arr[i] = a % 2;
... | |
You are given an array $$$a$$$ consisting of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You want to split it into exactly $$$k$$$ non-empty non-intersecting subsegments such that each subsegment has odd sum (i. e. for each subsegment, the sum of all elements that belong to this subsegment is odd). It is impossible to... | For each query, print the answer to it. If it is impossible to divide the initial array into exactly $$$k$$$ subsegments in such a way that each of them will have odd sum of elements, print "NO" in the first line. Otherwise, print "YES" in the first line and any possible division of the array in the second line. The di... | C | 7f5269f3357827b9d8682d70befd3de1 | 58c1c4c7a500e5f234cbe22847d56c51 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"constructive algorithms",
"math"
] | 1563978900 | ["3\n5 3\n7 18 3 14 1\n5 4\n1 2 3 4 5\n6 2\n1 2 8 4 10 2"] | null | PASSED | 1,200 | standard input | 3 seconds | The first line contains one integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Then $$$q$$$ queries follow. The first line of the query contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of subsegments, respectivel... | ["YES\n1 3 5\nNO\nNO"] | #include<stdio.h>
int main(){
int t;
scanf("%d", &t);
while(t--){
int n, k;
scanf("%d %d", &n, &k);
int a[n];
int b[n];
int count = 0;
for(int i=0; i<n; i++){
scanf("%d", &a[i]);
if(a[i]%2 == 1){
b[count] = i;
count++;
}
}
if(count >=k && k%2 == c... | |
You are given an array $$$a$$$ consisting of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You want to split it into exactly $$$k$$$ non-empty non-intersecting subsegments such that each subsegment has odd sum (i. e. for each subsegment, the sum of all elements that belong to this subsegment is odd). It is impossible to... | For each query, print the answer to it. If it is impossible to divide the initial array into exactly $$$k$$$ subsegments in such a way that each of them will have odd sum of elements, print "NO" in the first line. Otherwise, print "YES" in the first line and any possible division of the array in the second line. The di... | C | 7f5269f3357827b9d8682d70befd3de1 | dc8e414860b0c72ba52fdb04dfe20008 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"constructive algorithms",
"math"
] | 1563978900 | ["3\n5 3\n7 18 3 14 1\n5 4\n1 2 3 4 5\n6 2\n1 2 8 4 10 2"] | null | PASSED | 1,200 | standard input | 3 seconds | The first line contains one integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Then $$$q$$$ queries follow. The first line of the query contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of subsegments, respectivel... | ["YES\n1 3 5\nNO\nNO"] | #include<stdio.h>
int main()
{
int q,n,k,a,b,i,o=0,t[200001],j=0;
scanf("%d",&q);
while(q--)
{
scanf("%d%d",&n,&k);
for(i=0; i<n; i++)
{
scanf("%d",&a);
if(a&1)
o++,t[j++]=i+1;
}
if(o<k||((k+o)&1))
printf("NO\n")... | |
You are given an array $$$a$$$ consisting of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You want to split it into exactly $$$k$$$ non-empty non-intersecting subsegments such that each subsegment has odd sum (i. e. for each subsegment, the sum of all elements that belong to this subsegment is odd). It is impossible to... | For each query, print the answer to it. If it is impossible to divide the initial array into exactly $$$k$$$ subsegments in such a way that each of them will have odd sum of elements, print "NO" in the first line. Otherwise, print "YES" in the first line and any possible division of the array in the second line. The di... | C | 7f5269f3357827b9d8682d70befd3de1 | ee73fb60e42e178afac280a47661fd2f | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"constructive algorithms",
"math"
] | 1563978900 | ["3\n5 3\n7 18 3 14 1\n5 4\n1 2 3 4 5\n6 2\n1 2 8 4 10 2"] | null | PASSED | 1,200 | standard input | 3 seconds | The first line contains one integer $$$q$$$ ($$$1 \le q \le 2 \cdot 10^5$$$) — the number of queries. Then $$$q$$$ queries follow. The first line of the query contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le k \le n \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of subsegments, respectivel... | ["YES\n1 3 5\nNO\nNO"] | #include <stdio.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,k,K[200000],o=0;
scanf("%d %d",&n,&k);
int a[200000];
long long sum=0;
int i=0;
while(i<n)
{
scanf("%d",&a[i]);
sum=(long long)sum+a[i];
... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | 3daf5000b5ebb47e754421ae8c987988 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include <stdio.h>
int main(void)
{
int n;
int hh,mm;
scanf("%d",&n);
scanf("%d:%d", &hh, &mm);
if( n == 12)
{
while(hh<1)
hh+=10;
while(hh>12)
hh-=10;
while(mm>=60)
mm-=60;
}
else if( n == 24)
{
while(hh<... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | 6491916d048e8aa381d1a1faa1f7b616 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | /* Problem: 722A - Broken Clock */
/* Solver: Gusztav Szmolik */
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main ()
{
unsigned short tf;
unsigned char tm[7];
unsigned short h;
unsigned short m;
if (scanf("%hu",&tf) != 1)
return -1;
if (tf != 12 && tf != 24)
... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | 180a29068d293b57af3ed0144797026c | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include<stdio.h>
int main() {
char h1, h2, m1, m2;
int form;
scanf("%d\n", &form);
scanf("%c%c:%c%c", &h1,&h2,&m1,&m2);
if (form == 12) {
if (h1 > '1') {
if (h2 != '0') {
h1 = '0';
}
else {
h1 = '1';
}
}
if (h1 == '0'&&h2 == '0') {
h2 = '1';
}
if(h1 == '1'&&h2 > '2'){
h2 = '0... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | cb08db9b7a25fc2f6bf5655c903f1add | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include <stdio.h>
#include <stdlib.h>
int main(void)
{
int format;
char hh[3], mm[3];
scanf("%d\n", &format);
hh[0]=getchar();
hh[1]=getchar();
getchar();
mm[0]=getchar();
mm[1]=getchar();
hh[2]='\0';
mm[2]='\0';
int m=atoi(mm), h=atoi(hh);
if(m>59)
mm[0]='0';
... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | 2b338822baa846554fc9c1be6f51d941 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include <stdio.h>
#include <string.h>
int main()
{
//freopen ("lel.in","r",stdin);
int format;
int h,m;
scanf("%d",&format);
scanf("%2d:%2d",&h,&m);
if(format==24)
{
if(h>23)
h=h%10;
if(m>59)
m=m%10;
}
else if(format==12)
{ if(h==0)
h=1;
if(h>12 && h%10!=0)
h=h%10;
else if(h>12) h=10;
if(m... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | a6f3ada5415f077f7ffd35801b6f049a | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include<stdio.h>
int main()
{
int a,h,m;
scanf ("%d\n", &a);
a = a / 12;
scanf ("%d:%d", &h, &m);
if ( a == 1)
{
if (h > 12) h = h % 10;
if (h == 0) h = 10;
if (m > 59) m = m % 10;
}
else
{
if (h > 24) h = h % 10;
if (h == 24) h = 14;
if (m > 59) m = m % 10;
}
printf ("%02d:%02d", h, m );
ret... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | e6368f3076e5ac06a287a48281756e1c | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include <stdio.h>
int main (){
int format,h1,h2,m1,m2,hh,mm;
scanf ("%d",&format);
scanf ("%1d%1d:%1d%1d",&h1,&h2,&m1,&m2);
hh=h1*10+h2;
mm=m1*10+m2;
if (format==24){
if (hh>=24)
h1=0;
if (mm>=60)
m1=0;
}else {
if (hh> 12)
h1=0;
if (mm>=60)
m1=0;
if (h1==0 && h2==0)
... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | 9b3fbfa2c4be558959c5d364aaa9700f | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include <stdio.h>
int main(){
int n;
char h1, h2, m1, m2;
scanf("%d\n", &n);
scanf("%c%c:%c%c", &h1, &h2, &m1, &m2);
if(n == 24){
if(h1 == '2' && h2 > '3'){
h2 = '0';
}
else if(h1 > '2'){
h1 = '0';
}
}
else{
if(h1 == '0' && h2 == '0'){
h1 = '1';
}
else if(h1... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | ecd5079a190e9ebe11d31abe9e9f79eb | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include<stdio.h>
int main()
{
int n;
int a,b;
scanf("%d",&n);
scanf("%d:%d",&a,&b);
if(n==24)
{
if(0<=a&&a<=23)
{
if(a<=9)
printf("0");
printf("%d",a);
}
else
{
if(a/10==2)
printf("20");
else
{
a%=10;
printf("%d",10+a);
}
}
}
else if(n==12)
{
if(1<=a&&a<=12)
... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | 4a4917ba77eeaf7307328b5ae5ea656c | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include <stdio.h>
#include <stdlib.h>
int main()
{
int stat;
scanf("%d", &stat);
char s[10];
scanf("%s", &s);
if (s[3] > '5') s[3] = '5';
if (stat == 12){
if ((s[0] > '1') && (s[1] != '0')) s[0] = '0';
else if (s[1] == '0') s[0] = '1';
else if ((s[0] == '1') && (s[1] > ... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | 6194cad9500aac975311ac44b4f0ec9a | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #define _CRT_SECURE_NO_WARNINGS
#include <stdlib.h>
#include <stdio.h>
int main()
{
int m;
char s[100];
int hour=0, min = 0;
scanf("%d", &m);
scanf("%s", s);
hour = 10 * (s[0] - '0') + s[1] - '0';
min = 10 * (s[3] - '0') + s[4] - '0';
if (m == 24) {
if (hour > 23)
hour %=10;
if (min > 59)
min %= 10;
... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | f01f02805dd75f194d9a60814463a755 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include<stdio.h>
int main()
{
int f;
int h, m;
scanf("%d", &f);
scanf("%d:%d", &h,&m);
if(m>59)
m=m%10;
if(f==12){
if(h==0)
h=1;
if(h>12){
if(h%10==0)
h=10;
else
h=h%10;
}
}
else if(f... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | 3376071714d819562e5b51580bf2e504 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include<stdio.h>
int main()
{
int h,a,b,c,d;
scanf("%d",&h);
scanf("%1d%1d:%1d%1d",&a,&b,&c,&d);
if(h==12)
{
if(a==1 && b>2) a=0;
else if(b==0 && a>1) a=1;
else if(a>1 && b>0) a=0;
else if(a==0 && b==0) b=1;
}
if(h==24)
{
if(... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | 87ead4856a67dde8cd389d230d098045 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include <stdio.h>
#include <string.h>
char ss[7];
int main()
{
int t;
scanf("%d", &t);
scanf("%s", ss);
if (ss[3] - '0' > 5) ss[3] = '5';
if (t == 12)
{
if (ss[0] == '0'&&ss[1] == '0') ss[1] = '1';
else if (ss[0] == '1'&&ss[1] - '0' > 2) ss[1] = '2';
else if (ss[0] - '0' > 1 && ss[1] - '0' <= 2) ss[0] = '1... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | fd404e93fcedc8fdeedb7ef90e9e3cba | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include <stdio.h>
//#define DEBUG
int main(void) {
#ifdef DEBUG
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
int f;
char t[10];
scanf("%d\n%s", &f, t);
if (t[3] > '5')
t[3] = '0';
if (f == 12) {
if (t[0] == '0' && t[1] == '0')
t[1] = '1';
else if (t[0] == '1... | |
You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.You are given a time in format HH:MM that is currently displayed on the broken cl... | The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them. | C | 88d56c1e3a7ffa94354ce0c70d8e958f | c0ca0a1d6fcc78541bc24e6388d55187 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1475330700 | ["24\n17:30", "12\n17:30", "24\n99:99"] | null | PASSED | 1,300 | standard input | 1 second | The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively. The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes. | ["17:30", "07:30", "09:09"] | #include <stdio.h>
int main()
{
int a, h1, h2;
scanf("%d", &a);
scanf("%d:%d", &h1, &h2);
if (h2 > 59)
h2 = h2 % 10;
if (a == 12)
if (h1 > 12||h1 == 0)
if (h1%10 == 0)
h1 = 10;
else
h1 = h1 % 10;
if (a == 24)
if (!(h1 <= 23))
h1 =h1 % 10;
printf("%02d:%02d\n", h1, h2);
return 0;
}
| |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 4ed0d6fd1effbd3c5b64fd6a5c4fc066 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
#include<math.h>
int cmp(const void*a,const void*b)
{
return *(int*)b-*(int*)a;
}
int main()
{
int i,j,k1=0,k2=0,k3=0,n,n1=0,n2=0,n3=0,judge=0;
int c[5001],a[5001],b[5001],d[5001];
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&c[i]);
if(c[i]==1)
{
a[n1]=i+1;
n1++;
}
else if(c[i]... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 506b46418a71b032acada012c3a6dd72 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
#include<math.h>
int cmp(const void*a,const void*b)
{
return *(int*)b-*(int*)a;
}
int main()
{
int i,j,k,a,num,n,n1=0,n2=0,n3=0,judge=0;
int c[5001],b[4][5001];
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&c[i]);
if(c[i]==1)
{
b[1][n1]=i+1;
n1++;
}
else if(c[i]==2)
{
b[2][... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | b8eeb1bf65a68a336c6f5170184ce20b | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
#include<math.h>
int cmp(const void*a,const void*b)
{
return *(int*)b-*(int*)a;
}
int main()
{
int i,j,k,a,n,n1=0,n2=0,n3=0,judge=0;
int c[5001],b[3][2],d[5001];
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&c[i]);
if(c[i]==1)
{
n1++;
}
else if(c[i]==2)
{
n2++;
}
else
{
... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 7083d4553353124078d639a35790ca2a | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
int main()
{
int n,i,t,one[5002],two[5005],three[5005],on=0,tw=0,thr=0;
scanf("%d",&n);
for(i=1; i<=n; i++)
{
scanf("%d",&t);
if(t==1)
{
on++;
one[on]=i;
}
if(t==2)
{
tw++;
two[tw]=i;
... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 45237a8acfaae043554c384fe6b3b175 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
int main(){
int ones=0,twoes=0,threes=0;
int n,checker,j=1,n1=0,n2=1,n3=2,w;
scanf("%d",&n);
int a[n*3];
for(int i=1;i<=n;i++){
scanf("%d",&checker);
if(checker==1){
a[n1]=j;
n1+=3;
j++;
ones++;
}
else if(checker==2){
a[n2]=j;
n2+=3;
j++;
twoes++;
}
else{
a[n3]=j;
... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | ad47f8056a6a7e98b4dfe67b64f36a0e | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
#include<stdlib.h>
int main(void)
{
unsigned n,i,j,k,l,num,num1,num2,num3,*a,*b,*c,*t;
scanf("%u",&n);
t=(unsigned*)malloc(n*sizeof(unsigned));
a=(unsigned*)malloc((n)*sizeof(unsigned));
b=(unsigned*)malloc((n)*sizeof(unsigned));
c=(unsigned*)malloc((n)*sizeof(unsigned));
f... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | e77a1d5f0a5a379fc9dd0f0ae97547ad | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | // Codeforces A. Team Olympiad
// Created by Abdulrahman Elsayed on 04/07/2020
#include <stdio.h>
int min(int *a, int *b, int *c);
int main()
{
int n, c1 = 0, c2 = 0, c3 = 0, teams = 0;
scanf("%d", &n);
int array[n];
int count1[n], count2[n], count3[n];
for (int i = 0; i < n; i++)
... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 7f892cdfea0f0bb71ab92a61c88c5946 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
main(){
int n, i, j, input;
scanf("%d", &n);
int prog[n], math[n], sports[n], p=0, m=0, s=0, teams;
for(i=0; i<n; i++){
scanf("%d", &input);
// store the index
if(input==1){
prog[p] = i+1;
p++;
}else if(input==2){
math[m] = i+1;
m++;
}else if(input==3){
sports[s] =... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 87db91479654336c44cdc297d1d2f91a | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
main(){
int n, i, input;
scanf("%d", &n);
int skills[3][n], p=0, m=0, s=0;
for(i=0; i<n; i++){
scanf("%d", &input);
// store the index
if(input==1){
skills[0][p] = i+1;
p++;
}else if(input==2){
skills[1][m] = i+1;
m++;
}else if(input==3){
skills[2][s] = i+1;
s++... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | e2ee8d99e2f0db9d6762c89d729fe026 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
main(){
int n, i, input;
scanf("%d", &n);
int skills[3][n], p=0, m=0, s=0, teams;
for(i=0; i<n; i++){
scanf("%d", &input);
// store the index
if(input==1){
skills[0][p] = i+1;
p++;
}else if(input==2){
skills[1][m] = i+1;
m++;
}else if(input==3){
skills[2][s] = i+1;... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 6b8d078876733abf6be038e03d00ca18 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
main(){
int i,j,k,a[4],b[6000],n1,n,l,p,j1,j2,j3;
scanf("%d",&n);
a[1]=0;a[2]=0;a[3]=0;
for(i=0;i<n;i++){
scanf("%d",&k);
b[i]=k;
a[k]=a[k]+1;
}
if(a[1]<=a[2] && a[1]<=a[3])
n1=a[1];
if(a[2]<=a[1] && a[2]<=a[3])
n1=a[2];
if(a[3]<=a[2] ... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 2559d5acfdabd05d64d49220ad8ea035 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
int main()
{
int a,b,c,d=0,e=0,f=0,j=0,k,x=0,y=0,z=0,w=0,r=0,t=0;
scanf("%d",&a);
int s[a];
int i;
for(i=1;i<=a;i++)
{
scanf("%d",&s[i]);
if(s[i]==1)
{
d++;
}
else if(s[i]==2)
{
e++;
}else
{
f++;
}
}
if(d==0||e==0||f==0)
{
printf("0");
}
else
{
... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 7310f1cbd987b810feb4258514316d09 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
int main()
{
int i,j,k,n,a[10000],u=0,o=0,p=0,q[10000],w[10000],e[10000],count,ct=0;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n;i++)
{
count=0;
if(a[i]==1)
{
for(j=0;j<n;j++)
{
if(a[j]==... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 7890ed9a13f13953dac54e766c81b2ec | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
int main() {
int n;
scanf("%d", &n);
int a[n], i, b[4]={0}, min=5010;
for(i=0; i<n; i++) {
scanf("%d", &a[i]);
b[a[i]]++;
}
for(i=1; i<4; i++) {
if(b[i]<min) min = b[i];
}
printf("%d\n", min);
int p=0, q=0, r=0;
for(i=0; i<min; i++) {
while(a[p]!=1) p++;
printf("%d ", ++p)... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 8d11fc7f24429840e0a3f04018938c4f | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
#include <stdlib.h>
int cmp(void const* a,void const* b)
{
return (*(int *)a - *(int *)b);
}
int main()
{
int n;
scanf("%d",&n);
int i,o=0,t=0,p=0;
int q;
int cal[4]={0};
int one[5002]={0},two[5002]={0},three[5002]={0};
for(i=1;i<=n;i++)
{
scanf("%d",&q);
cal[q]++;
if(q==1)
{
on... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | c431977125f4d8fb0566bc864e03e3cc | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
int main()
{
int n,i,min;
scanf("%d",&n);
int a[n];
int freq[4]={0};
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(a[i]==1)
{
freq[1]=freq[1]+1;
}
else if(a[i]==2)
{
freq[2]=freq[2]+1;
}
else
... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 585b3e0161320024bb691112911cc320 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
int main()
{
int n;
scanf("%d",&n);
int a[5005];
int q1[5000],q2[5000],q3[5000];
int x=0,y=0,z=0;
int min;
for(int i=1; i<=n ; ++i){
scanf("%d",&a[i]);
if(a[i]==1) {
q1[x] = i;
x=x+1;
}
if(a[i]==2) {
... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | d6d76e99f66d46461273e7e964c6a3f9 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
int mini (int a, int b) {
return a > b ? b : a;
}
void teams(void) {
int n;
scanf("%d", &n);
int arr[n];
int o = 0, t = 0, th = 0;
for (int i = 0; i < n; i++) {
scanf("%d", &arr[i]);
if (arr[i] == 1) {
o++;
} else if (arr[i] == 2) {
t++;
} else if (arr[i] == ... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 2d7800ad3f194308e701582a3dd55a25 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
int a[5000],b[5000],c[5000],a1=0,b1=0,c1=0,n,m,i,j=0,k=0,l=0;
int main(){
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%d",&m);
if(m==1) a1++,a[j]=i,j++;
else if(m==2) b1++,b[k]=i,k++;
else if(m==3) c1++,c[l]=i,l++;
}
... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 57ae667eb7455cdccacc632c082a0135 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
int min(int a,int b,int c){
if(a<=b && a<=c) return(a);
else if(b<=a && b<=c) return(b);
else if(c<=a && c<=b) return(c);
}
int a[5000],b[5000],c[5000],a1=0,b1=0,c1=0,n,m,i,j=0,k=0,l=0;
int main(){
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%d",&... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | b96547662726fb8985008064623c5da7 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
int main()
{
int n,a[5005],k,i,d,j,one=0,two=0,th=0;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(a[i]==1) one++;
else if(a[i]==2) two++;
else if(a[i]==3) th++;
}
if(one<two)
{
d=one;
one=two;
two=d;
}i... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | abd62fadfecc833c3dfa8814f49ddd24 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
#include <stdlib.h>
int nearestOne(int * childrenArray,int numberOfChildren);
int nearestTwo(int * childrenArray,int numberOfChildren);
int nearestThree(int * childrenArray,int numberOfChildren);
int numberOfTeams(int * childrenArray,int numberOfChildren);
void printIndex(int * childrenArray,int numb... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | e353d71fff4a3ed17f0aee67d42008b9 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int nearestOne(int*, int);
int nearestTwo(int*, int);
int nearestThree(int*, int);
int main() {
int n;
int* children;
int one, two, three,numberOfTeams=0;
int outputArray[5000];
scanf("%d", &n);
children = (int*)malloc(n * sizeof(int));
... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 82aeddcd7549f56e9b1753d665ea9924 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
#include <stdlib.h>
int nearestOne(int * childrenArray,int numberOfChildren);
int nearestTwo(int * childrenArray,int numberOfChildren);
int nearestThree(int * childrenArray,int numberOfChildren);
int numberOfTeams(int * childrenArray,int numberOfChildren);
void printIndex(int * childrenArray,int numb... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | c3b29f62ebf35c385b2fd1292f139102 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
#include<math.h>
int main()
{
int n,i,x=0,y=0,z=0,l,m=0,j,p=0,o=0;
scanf("%d",&n);
int a[n];
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
{
if(a[i]==1)
x++;
else if(a[i]==2)
y++;
else if(a[i]==3)
z++;
}
if(x==0||y==0||z==0)
printf("0");
else
{
if(x<y)... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | fc5a185011665182d7e5b53ecbb945e9 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
int min(int i, int j, int k)
{
int x;
x=i;
if(j<x)
x=j;
if(k<x)
x=k;
return x;
}
int main()
{
int a[5000];
int p[5000],q[5000],r[5000];
int n,i;
int x=0,y=0,z=0;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | a9b70553dcefd8aaa2237c1b2b49e168 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
int main(void)
{
int n;
scanf("%d", &n);
int students[3][5000];
int cnts[3] = {0, 0, 0};
for(int i = 0; i < n; i++)
{
int a;
scanf("%d", &a);
students[a-1][cnts[a-1]] = i+1;
cnts[a-1]++;
}
int smallest = cnts[0];
for(int i = 1; ... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | db28aa183f5f7c078c71e6aed96d7a0e | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
#include <stdlib.h>
int min(int k, int l, int m){
if(m<=k&&m<=l){
return m;
}
else if(k<=m&&k<=l){
return k;
}
else{
return l;
}
}
int main()
{
int n,team[5001],i,onepost[5001],twopostion[5001],threepostion[5001],k=0,l=0,m=0,teamnumber;
scanf("%d",&n);... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 46d1018a83deb2e3c7d70161937c2c65 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include <stdio.h>
#include <stdlib.h>
int main()
{
int num,i,j=0,k=0,l=0,array[5001],array1[5001],array2[5001],array3[5001],team;
scanf("%d",&num);
for(i=0;i<num;++i){
scanf("%d",&array[i]);
}
for(i=0;i<num;++i){
if(array[i]==1){
array1[j]=i;
j++;
}
else if(array[i]==2){
array2... | |
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 ti: ti = 1, if the i-th child is good at programming, ti = 2,... | 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... | C | c014861f27edf35990cc065399697b10 | 195b48cbf3b97fcd137f8969e99e0c97 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"sortings",
"greedy"
] | 1416733800 | ["7\n1 3 1 3 2 1 2", "4\n2 1 1 2"] | null | PASSED | 800 | standard input | 1 second | The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 3), where ti describes the skill of the i-th child. | ["2\n3 5 2\n6 7 4", "0"] | #include<stdio.h>
int main()
{
int n,t[10000],a=0,b=0,c=0,i,p,e[10000],f[10000],g[10000];
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&t[i]);
if(t[i]==1)
{
e[a]=i;
a++;
}
else if(t[i]==2)
{
f[b]=i;
b++;
... | |
There is a country with $$$n$$$ citizens. The $$$i$$$-th of them initially has $$$a_{i}$$$ money. The government strictly controls the wealth of its citizens. Whenever a citizen makes a purchase or earns some money, they must send a receipt to the social services mentioning the amount of money they currently have.Somet... | Print $$$n$$$ integers — the balances of all citizens after all events. | C | 7b788c660fb8ca703af0030f4c84ce96 | 852ee8d505f20c21fe2e2d8041546906 | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"data structures",
"binary search",
"sortings",
"brute force"
] | 1564497300 | ["4\n1 2 3 4\n3\n2 3\n1 2 2\n2 1", "5\n3 50 2 1 10\n3\n1 2 0\n2 8\n1 3 20"] | NoteIn the first example the balances change as follows: 1 2 3 4 $$$\rightarrow$$$ 3 3 3 4 $$$\rightarrow$$$ 3 2 3 4 $$$\rightarrow$$$ 3 2 3 4In the second example the balances change as follows: 3 50 2 1 10 $$$\rightarrow$$$ 3 0 2 1 10 $$$\rightarrow$$$ 8 8 8 8 10 $$$\rightarrow$$$ 8 8 20 8 10 | PASSED | 1,600 | standard input | 2 seconds | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{5}$$$) — the numer of citizens. The next line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$0 \le a_{i} \le 10^{9}$$$) — the initial balances of citizens. The next line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \... | ["3 2 3 4", "8 8 20 8 10"] | #include<stdio.h>
#define MAX(a,b) a>b?a:b
int tc[200000],x[200000];
int main()
{
int n;
scanf("%d",&n);
int current[n],i;
for(i=0; i<n; i++)
scanf("%d",¤t[i]);
int q;
scanf("%d",&q);
int serial,max=0,t,tx;
for(i=0; i<q; i++)
{
scanf("%d",&serial);
if(se... | |
There is a country with $$$n$$$ citizens. The $$$i$$$-th of them initially has $$$a_{i}$$$ money. The government strictly controls the wealth of its citizens. Whenever a citizen makes a purchase or earns some money, they must send a receipt to the social services mentioning the amount of money they currently have.Somet... | Print $$$n$$$ integers — the balances of all citizens after all events. | C | 7b788c660fb8ca703af0030f4c84ce96 | 1fe8af35dbbc4c82f2a871becd0f835b | GNU C11 | standard output | 256 megabytes | train_000.jsonl | [
"data structures",
"binary search",
"sortings",
"brute force"
] | 1564497300 | ["4\n1 2 3 4\n3\n2 3\n1 2 2\n2 1", "5\n3 50 2 1 10\n3\n1 2 0\n2 8\n1 3 20"] | NoteIn the first example the balances change as follows: 1 2 3 4 $$$\rightarrow$$$ 3 3 3 4 $$$\rightarrow$$$ 3 2 3 4 $$$\rightarrow$$$ 3 2 3 4In the second example the balances change as follows: 3 50 2 1 10 $$$\rightarrow$$$ 3 0 2 1 10 $$$\rightarrow$$$ 8 8 8 8 10 $$$\rightarrow$$$ 8 8 20 8 10 | PASSED | 1,600 | standard input | 2 seconds | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^{5}$$$) — the numer of citizens. The next line contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$0 \le a_{i} \le 10^{9}$$$) — the initial balances of citizens. The next line contains a single integer $$$q$$$ ($$$1 \le q \le 2 \... | ["3 2 3 4", "8 8 20 8 10"] | #include<stdio.h>
#define MAX(a,b) a>b?a:b
int tc[200000],x[200000];
int main()
{
int n;
scanf("%d",&n);
int current[n],i;
for(i=0; i<n; i++)
scanf("%d",¤t[i]);
int q;
scanf("%d",&q);
int serial,max=0,t,tx;
for(i=0; i<q; i++)
{
scanf("%d",&serial);
if(se... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 9af15124a2eda317f02851efadfb158f | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
#include<string.h>
int main()
{
int n, i, j, count = 0;
scanf ("%d", &n);
char str[n][102];
for (i = 0; i < n; i++) {
scanf ("%s", str[i]);
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if ((j + 1) < n && str[i][j + 1]=='o')
... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 18190a2e4bc0845f4d7473974f862bd9 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
int main(){
int n;
scanf("%d", &n);
int i, j;
char a[102][102];
for (i = 1; i <= n; i++)
scanf("%s", a[i]);
for (i = 1; i <= n; i++)
for (j = n; j > 0; j--)
a[i][j] = a[i][j-1];
for (i = 1; i <= n; i++){
a[0][i] = 'a';
a[n+1][i]... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | bcd09fbf55eaa5d57283e5dd185ea73a | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
int main(){
int n;
scanf("%d", &n);
int i, j;
char a[1002][1002];
for (i = 1; i <= n; i++)
scanf("%s", a[i]);
for (i = 1; i <= n; i++)
for (j = n; j > 0; j--)
a[i][j] = a[i][j-1];
for (i = 1; i <= n; i++){
a[0][i] = 'a';
a[n+1][... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | bf29f3150eeb29803f093b532a3d9f8d | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<Stdio.h>
int main()
{
int i,j,n,c=0;
scanf("%d",&n);
char s[105],a[n][n];
for(i=0;i<n;i++)
{
scanf("%s",s);
for(j=0;j<n;j++)
a[i][j]=s[j];
}
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
c=0;
if(a[i-1][j]=='o'&&i>0)
... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | c689be621e26d936dcc6e2fd8b98e065 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
int main()
{
int n, i, j;
char a[105][105];
scanf("%d", &n);
for(i=0; i<n; i++)
scanf("%s", &a[i]);
int flag=0, sum;
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
{
sum=0;
if(a[i][j+1]=='o')
sum++;
if(a[... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 4990d1f5558fb7041f5cac14bdbafed3 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int main()
{ int n,i,j,count=0;
scanf("%d",&n);
char a[n+1][n+1];
for(i=0;i<n;i++)
{
scanf("%s",&a[i]);
}
int flag =0;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{ count =0;
if(i-1>=0)
{
if(a[i-1][j]=='o')
... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 83c85093139c0233d629a54b5ea997c4 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
int main()
{
int n, k, i, j, flag, count;
char a[111][111];
scanf ("%d", &n);
for (i = 0; i < n; i++) {
scanf ("%c", &k);
for (j = 0; j < n; j++) {
scanf ("%c", &a[i][j]);
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 98c8ac51cb3f1ce37d8693b11cf96f03 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int i,j,n,c=0;
char a[101][101];
int main()
{
scanf("%d",&n);
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
scanf(" %c",&a[i][j]);
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(a[i-1][j]=='o' && i-1>=0 )
c++;
if(a[i][j-1]=='o' && j-1>=0)
c+... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 3944af2d1e289f1cc6e6ecd84b3df9e9 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int i,j,n,c=0;
char a[101][101];
int main()
{
scanf("%d",&n);
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
scanf(" %c",&a[i][j]);
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(a[i-1][j]=='o' && i>=1 )
c++;
if(a[i][j-1]=='o' && j>=1)
c++;
... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | ac6f9542b22e1d2c387028a1b707c142 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
#include <stdlib.h>
int main()
{
int n,i,j,ocount=0;
char c[100][100];
scanf("%d ",&n);
for(i=0;i<n;i++)
{
scanf("%s",c[i]);
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if((i < (n-1)) && (c[i+1][j] == 'o'))
{
... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 286302edcedf6b820df881b305ae7585 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] |
#include <stdio.h>
int main() {
int n,i,j;
scanf("%d",&n);
int a[n][n];
getchar();
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
char temp;
scanf("%c",&temp);
if (temp == 'x')
a[i][j] = 1;
else
a[i][j] = 0;
}
getchar();
}
int ... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 60d0511909705592fa99616de30a8519 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
char arr[250][250];
int main(){
int a;
scanf("%d",&a);
getchar();
int i,b;
for(i=0;i<a;i++){
for(b=0;b<a;b++){
scanf("%c",&arr[i][b]);
}
getchar();
}
for(i=0;i<a;i++){
int cnt=0;
for(b=0;b<a;b++... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 5b3ee389e2c5aaa57345560660f70437 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int main()
{
int i,j,c=0,flag=0;
int n;
scanf("%d",&n);
getchar();
char a[n][n];
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
scanf("%c",&a[i][j]);
// printf("%c",a[i][j]);
}
getchar();
// printf("\n");
}
for(... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | a768afb53a263801c0c8e694621e176f | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | /*
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?
Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | adb188a63d791fd6e6de2c598dcb7477 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int main()
{
int n,i,p,q,r,s,sum,j;
char a[101][101];
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%s",a[i]);
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
sum=0;
p=j-1;
q=j+1;
r=i-1;
s=i+1;
if(p... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | b0f633037732103f26bb153751712ffe | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
int n, i, j, sum;
char a[105][105];
int main () {
scanf("%d",&n);
for(i = 1 ; i <= n ; i++)
for(j = 1 ; j <= n ; j++)
scanf(" %c",&a[i][j]);
for(i = 1 ; i <= n ; i++)
for(j = 1 ; j <= n ; j++, sum = 0){
if(a[i-1][j] == 'o')sum++;
if(a[i+1][j] == 'o')sum++;
if(a[i][j-1] == 'o')sum... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 76710c1f30404acca17ab6fa65eca4c3 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
#define MAXN 100
typedef enum bool {FALSE,TRUE} bool;
int n;
int a [MAXN+2][MAXN+2];
int main (void) {
scanf("%d",&n);
int i, j;
char s [MAXN+1];
for (i=1; i<=n; i++) {
scanf("%s",s);
for (j=1; j<=n; j++) {
if (s[j-1]=='x')
a[i][j] = 0;
else
a[i][j] = 1;
}
}
for (i=0; i<... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 3f5320140f4138039c95c7dac5015260 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
char c[101][101];
int main(){
int n, sum = 0, i, j;
scanf("%d", &n);
for(i=0; i<n; i++){
for(j=0; j<n; j++){
scanf("%c", &c[i][j]);
if(c[i][j] == '\n') scanf("%c", &c[i][j]);
}
}
for(i=0; i<n; i++){
for(j=0; j<n; j++)... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 825c47e354a4fc0322aebf7642fa6226 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int main()
{
int n,i,j;
scanf("%d",&n);
char a[101][101];
for(i=0;i<n;i++)
{
scanf(" %s",a[i]);
}
int count;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
count=0;
if(i-1>=0 && a[i-1][j]=='o')count++;
if(i+1<n && a[i+1][j]=='o')count++;
if(j-1>=0 && a[i][j-1]=='o')count++;
if(j... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | f7f5f867957e4c52b1bada7af385c87f | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int main()
{
char ch;
int i,j,n,cnt,flag=0;
char mat[101][101];
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%s",mat[i]);
}
/*for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%c",mat[i][j]);
}
printf("\n");
}*/
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
cnt=0;
if((i-1)>=0 && mat[i-1][j]=='o')
{
cnt++;
}
if((j-1)... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | e17a2657588577204d8d81b0cfb17226 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int main()
{
int n,i,j,p=0,q=0,count=0,count1=0;
char A[101][101];
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("\n");
for(j=0;j<n;j++)
{
scanf("%c",&A[i][j]);
}
}
if(n==1)
{
printf("YES");
return 0;
}
for(... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | b07191d0669a983060881560c52a2819 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int main()
{
int n,i,j,p=0,q=0,count=0,count1=0;
char A[101][101];
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("\n");
for(j=0;j<n;j++)
{
scanf("%c",&A[i][j]);
}
}
if(n==1)
{
printf("YES");
return 0;
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
count=0;
if(i>0)
{
if(A... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 9be99d7ff59694e43fefda2b642d15c7 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
#include<string.h>
#define loop(i,a,b) for(int i=a;i<b;i++)
int main(){
int n,t;
scanf("%d",&n);
char cb[n][n];
loop(i,0,n){
scanf("%s",cb[i]);
}
loop(i,0,n){
loop(j,0,n){
t=0;
if(i-1>=0){
if(cb[i-1][j]=='o'){
t++;
}
}
if(i<n-1){
... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | a068ab2ef37d4d686202fc7f37a77fbb | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
char ar[100][100];
int check_horizontal(int a, int b,int n)
{
int counter=0;
if((b-1)>=0 && ar[a][b-1]=='o')
counter++;
if((a-1)>=0 && ar[a-1][b]=='o')
counter++;
if((b+1)<=(n-1) && ar[a][b+1]=='o')
counter++;
if((a+1)<=(n-1) && ar[a+1][b]=='o')
counter++;
return counter;
}
int check_possibl... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 4cf2ff28149330a44938685f560b626b | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int main () {
int i ,n , j;
char c[100][101] ;
scanf ("%d" , &n) ;
int counter ;
for(i = 0; i < n; ++i)
scanf ("%s" , c[i] ) ;
for(i = 0; i < n; ++i){
counter= 0;
for(j = 0; j < n; ++j){
if ( i -1 >= 0 ... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 24074e193bafcd966a12d804557f30b5 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
#include <string.h>
int main()
{
int i, j, k, l, m, n;
scanf("%d", &n);
char a[n][n + 1];
for(i = 0; i < n; i++)
{
scanf("%s", a[i]);
}
for(i = 0; i < n; i++)
{
for(j = 0; j < n; j++)
{
k = 0;
if(i+1 < n && a[i + 1][j] =... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 8a9ed61bd4a26b133697f3945e1a4c77 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
#include <string.h>
int main(){
char p[101][101] ;
int i , j, k=0 , n ;
scanf("%d", &n) ;
for(i=0;i<n;i++){
scanf("%s", &p[i]) ;
}
for(i=0;i < n;i++){
k= 0 ;
for(j=0;j<n;j++){
if(p[i+1][j] == 'o'){
k++ ;
// p... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 6a53db393e0242b7d8440dda84011c24 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
int main (int argc, char * argv[]) {
int n;
scanf ("%d\n", &n);
char black_hole;
char ** board = (char **) malloc (sizeof(char *) * n);
for (int i = 0; i < n; i++) {
board[i] = (char *) malloc (sizeof(char) * n);
for (int ... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 14871fa7b5531bc8bc28e4c6abd2da0b | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
char mat[101][101];
int main()
{
int n;
scanf("%d", &n);
int i, j, a=0;
for(i=0; i<n; i++)
{
scanf("%s", &mat[i]);
}
for(i=0; i<n-1; i++)
{
a=0;
for(j=0; j<n; j++)
{
if(mat[i+1][j]=='o')
a++;
... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 7e84ce58351dd1c23a2654a251a98421 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
int main() {
char tempc;
int a[104][104] = {0};
int i, flag = 0, n, j, temp;
// STDIN
scanf("%d\n", &n);
for(i=1; i<=n; i++) {
for(j=1; j<=n; j++) {
scanf("%c", &tempc);
if(tempc == 'o') a[i][j] = 1;
}
scanf("%c", &temp);
}
// GO ON
for(i=1; i<=n; i++)
for(j=1; j<=n; j++) ... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | a9594c7f6463237337972a5680699283 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
int main() {
int i, j, n, cnt, yes;
static char a[100][101];
scanf("%d", &n);
for (i = 0; i < n; i++)
scanf("%s", a[i]);
yes = 1;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++) {
cnt = 0;
if (i > 0 && a[i - 1][j] == 'o')
cnt++;
if (i + 1 < n && a[i + 1][j] == 'o')
cnt+... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 96aded186d02618af37be46a664ec299 | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include<stdio.h>
int main()
{
int n,i,j,c=0,count,u;
char chk[100][100];
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%s",&chk[i]);
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
count=0;
if((i-1>=0)&&(chk[i-1][j]=='o'))
count++;... | |
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the ... | Print "YES" or "NO" (without the quotes) depending on the answer to the problem. | C | 03fcf7402397b94edd1d1837e429185d | 83742a6ad175a930904295cc6b29380f | GNU C | standard output | 256 megabytes | train_000.jsonl | [
"implementation",
"brute force"
] | 1409061600 | ["3\nxxo\nxox\noxx", "4\nxxxo\nxoxo\noxox\nxxxx"] | null | PASSED | 1,000 | standard input | 1 second | The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces. | ["YES", "NO"] | #include <stdio.h>
#include <stdlib.h>
int main() {
int n, i, j, k;
scanf("%d", &n);
char c[n][n+1];
for (i = 0; i < n; ++i) {
scanf("%s", &c[i]);
}
for (i = 0; i < n; ++i) {
for (j = 0; j < n; ++j) {
k = 0;
if (i == 0) {
if (j == 0) {
if (c[i][j+1] == 'o') {
k++;
}... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | 92c1e21e52118048d37bb1090c9ce58d | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | #include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#define int long long int
struct adjacencylist
{
int nodeindex;
int weight;
struct adjacencylist* adj;
};
struct minheap
{
int mindist;
int vertex;
};
struct adjacencylist *graph[100002],*traverse;
int swap(int *a,int *b){... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | 76f03ac09949f74f113ff97e51c71140 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | #include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#define int long long int
struct adjacencylist
{
int nodeindex;
int weight;
struct adjacencylist* adj;
};
struct minheap
{
int mindist;
int vertex;
};
struct minheap* heap;
struct adjacencylist *graph[100002],*traverse;
int ... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | 935b226cde59743816be7a4a5d1eced8 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | #include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#define int long long int
struct adjacencylist
{
int nodeindex;
int weight;
struct adjacencylist* adj;
};
struct minheap
{
int mindist;
int vertex;
};
struct adjacencylist *graph[100002],*traverse;
int swap(int *a,int *b){ ... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | 05b55ec6e547a14ad38bd63c0a5b60b0 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | #include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#define int long long int
struct adjacencylist
{
int nodeindex;
int weight;
struct adjacencylist* adj;
};
struct minheap
{
int mindist;
int vertex;
};
struct adjacencylist *graph[100002],*traverse;
int swap(int *a,int *b){ ... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | 21eb9c734b08ea7f87f0cce1aa2423c1 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | #include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#define int long long int
struct adjacencylist
{
int nodeindex;
int weight;
struct adjacencylist* adj;
};
struct minheap
{
int mindist;
int vertex;
};
struct adjacencylist *graph[100002],*traverse;
int swap(int *a,int *b){ ... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | 39beac03da2c4cce02fecd8e083b7fa0 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | #include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#define int long long int
struct adjacencylist
{
int nodeindex;
int weight;
struct adjacencylist* adj;
};
struct minheap
{
int mindist;
int vertex;
};
struct minheap* heap;
struct adjacencylist *graph[100002],*traverse;
int ... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | 5f9ad3f2486546fc964ca06b2a128e28 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | //OLD CODE WITHOUT FUNCTION
#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#define int long long int
struct adjacencylist
{
int nodeindex;
int weight;
struct adjacencylist* adj;
};
struct minheap
{
int mindist;
int vertex;
};
struct adjacencylist *graph[100002],*travers... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | b603f3dd3a7c304e963721c0fe8d50c2 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | //Testing somebody's code
#include <stdio.h>
#include <stdlib.h>
#define MAX 100005
#define ll long long
#define M 0x7f7f7f7f7f7f7f7f
// 有向边
typedef struct node{
// 终点
int e;
// 权值
int w;
// 指向下一个边的下标
int next;
}Edge;
// 用来保存边的链表
Edge edges[MAX<<2];
// 记录图上的每一个点能到边的链表的头位置
int first[MAX];
// 最短... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | 9d3b57e97511d1d745311b376f46c933 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | //Testing Somebody's code
#include <limits.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 222222
typedef long long ll;
typedef struct {ll v, w;} pair;
typedef struct _list{
pair p;
struct _list *next;
} list;
list *g[N];
ll n, m, u, v, w, size, dist[N], prev[N];
pai... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | a1b6861ce734515b888c58ba7bd935e8 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | //OLD CODE WITHOUT FUNCTION
#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#define int long long int
struct adjacencylist
{
int nodeindex;
int weight;
struct adjacencylist* adj;
};
struct minheap
{
int mindist;
int vertex;
};
struct adjacencylist *graph[100002],*travers... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | b840a1c9c6434e7cefc4480a343c0111 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct {
int to;
int len;
} edge;
typedef struct {
edge* vec;
int size;
int max_size;
} vector;
vector* vector_new() {
vector* vec = calloc(1, sizeof(*vec));
return vec;
}
void vector_free(vector* vec) {
free(vec->vec);
free(vec);
... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | 470a561ee26ec3fb2700d4cea4f8c742 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | #include <limits.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 222222
typedef long long ll;
typedef struct {ll v, w;} pair;
typedef struct _list{
pair p;
struct _list *next;
} list;
list *g[N];
ll n, m, u, v, w, size, dist[N], prev[N];
pair pq[N];
void min_heapify(... | |
You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | C | bda2ca1fd65084bb9d8659c0a591743d | 19ddbe733c809c4ff497e55871028504 | GNU C11 | standard output | 64 megabytes | train_000.jsonl | [
"shortest paths",
"graphs"
] | 1276875000 | ["5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1", "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1"] | null | PASSED | 1,900 | standard input | 1 second | The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi is the length of the edge. It is possible that the... | ["1 4 3 5", "1 4 3 5"] | #include <limits.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 222222
typedef long long ll;
typedef struct {ll v, w;} pair;
typedef struct _list{
pair p;
struct _list *next;
} list;
list *g[N];
ll n, m, a, b, w, size, dist[N], prev[N], ans[N], ind;
pair pq[N];
void... |
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