Mutiple

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

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.

Input

There 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.

Output

For 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

Source

BestCoder Round #39 ($)

题意：求大于i（1~n-1）小于等于n的中间最小的数的和

代码：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <set>
#define LL __int64
using namespace std;
const LL Mod = 10007;
const LL M = 1e4+5;

int s[M];

int main(){
    int n;
    while(scanf("%d", &n) == 1){
        memset(s, 0, sizeof(s));
        for(int i = 1; i <= n; ++ i){
            scanf("%d", &s[i]);
        }
        int sum = 0;
        for(int i = 1; i < n; ++ i){
            int Min = 1e7;
            for(int j = i+1; j <= n; ++ j){
                if(s[j]%s[i] == 0&&Min > j){
                    Min = j;
                }
            }
            if(Min != 1e7) sum += Min;
        }
        printf("%d\n", sum);
    }
    return 0;
}