[5211]:Mutiple

a sequence a[1..N]. // 数组判断从1开始，而不是0

sequence of N integers. // n为每个测试组的测试量

For every index i, WLD wants to find the smallest index F(i) ( if exists ), that i**大于F(i)小于等于**n // 这里说明一遇到整除即停止，它只要知道整除的数是第几个

and aF(i) mod ai = 0. If there is no such an index F(i), we set F(i) as 0.

5.There are Multiple Cases.(At MOST 10) // 输入的测试组数至少为10

6.The first line contains one integers N(1≤N≤10000). // 测试组的测试量的范围

/*
author : Yangchengfeng
*/

#include<stdio.h>
#define N 10000

int main()
{
int n;
while(scanf("%d", &n)!=EOF){
int i = 1, j, a
, sum = 0;
for(; i<=n; i++){
scanf("%d", &a[i]);
}
for(i = 1; i<=n; i++){
for(j=i+1; j<=n; j++){
if (a[j] % a[i] == 0){
sum += j;
break;
}
}
}
printf("%d\n", sum);
}
return 0;
}