# | Submission time | Handle | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
696 | 2013-03-01T05:50:57 Z | gs12013 | 쉬운 문제 (GA3_easy) | C++ | 0 ms | 0 KB |
#include <stdio.h> int CountPair(int N, int *A) { int i, j, ans = 0; for(i=0;i<n-1;i++) for(j=0;j<n;j++) if(!(A[j]%A[i])) ans++; return ans; }