HDU-5211 (aF(i) mod ai = 0)

WLD likes playing with a sequence

a[1..N].
One day he is playing with a sequence of
N
integers. For every index i, WLD wants to find the smallest index
F(i)
( if exists ), that
i<F(i)≤n,
and
aF(i)
mod
ai
= 0. If there is no such an index
F(i),
we set
F(i)

as 0.

InputThere are Multiple Cases.(At MOST

10)

For each case:

The first line contains one integers
N(1≤N≤10000).

The second line contains
N
integers
a1,a2,...,aN(1≤ai≤10000),denoting the sequence WLD plays with. You can assume that all ai is distinct.OutputFor each case:

Print one integer.It denotes the sum of all
F(i)
for all
1≤i<n
Sample Input

4
1 3 2 4

Sample Output

6

Hint

F(1)=2
F(2)=0
F(3)=4
F(4)=0

aF(i)
mod
ai
= 0是关键，思路：从这个数开始向后找它的公倍数，如果是它的公倍数，总数加上公倍数的下表，就是第几个数，如果没有这个数的公倍数，就加0；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[10010];
main()
{
int n;
while(~scanf("%d",&n))
{
int sum=0;
memset(a,0,sizeof(a));
for(int i=1; i<=n; i++)
scanf("%d",&a[i]);
for(int i=1;i<=n-1;i++)
{
for(int j=i+1;j<=n;j++)
{
if(a[j]%a[i]==0)
{
sum+=j;
break;
}
}
}
printf("%d\n",sum);
}
}