Hdoj 5211 Mutiple 【水】

Mutiple

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 65    Accepted Submission(s): 45

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

41 3 2 4

 

Sample Output

6
Hint
F(1)=2F(2)=0F(3)=4F(4)=0
 

 

Source

BestCoder Round #39 ($)

 

题意：求大于i（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 __int64using 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;}