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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#853781 | #55. 欧几里得距离之和 | sxyzc233 | 0 | 0ms | 0kb | C++14 | 357b | 2025-01-11 19:13:53 | 2025-01-11 19:14:05 |
answer
#include <bits/stdc++.h>
int main(){
int n;
scanf("%d",&n);
int a[n];
int b[n];
for(int i=0;i<n;i++){
scanf("%d %d",&a[i],&b[i]);
}
double sum=0;
double tmp;
for(int i=0;i<n;i++){
for(int j=i+1;i<n;j++){
tmp=sqrt(1.0*(a[i]-a[j])*(a[i]-a[j])+1.0*(b[i]-b[j])*(b[i]-b[j]));
sum+=tmp;
}
}
printf("%f",sum);
return 0;
}
Details
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Subtask #1:
score: 0
Time Limit Exceeded
Test #1:
score: 0
Time Limit Exceeded
input:
3000 -802420 -321989 227507 956314 -460698 -819834 -479809 -341770 191520 109304 712327 -189558 -578326 -41090 282566 982266 -859119 686756 209058 -23298 -884994 -349898 -11358 182915 -507706 -81622 745434 575941 -374809 139274 810223 367608 960234 -197223 439081 573568 -275182 999306 -583036 -61808...
output:
result:
Subtask #2:
score: 0
Skipped
Dependency #1:
0%
Subtask #3:
score: 0
Skipped
Dependency #1:
0%
Subtask #4:
score: 0
Skipped
Dependency #1:
0%
Subtask #5:
score: 0
Skipped
Dependency #1:
0%