BestCoder Round #39 1002——降低复杂度——Mutiple

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[10005];
int b[10005];
int main()
{
    int n;
    while(~scanf("%d",&n)){
        int sum = 0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i  = 1; i <= n ; i++){
            scanf("%d",&a[i]);
        }

        for(int i = n ; i >= 1;  i--){
            sum += b[a[i]];
            for(int j = 1; j*j <= a[i]; j++){
               if(a[i] % j == 0){
                   b[j] = b[a[i]/j] = i;
               }
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}