标签：水题

Mutiple

4
1 3 2 4

6

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

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;
}

Hdoj 5211 Mutiple 【水】

标签：水题

原文地址：http://blog.csdn.net/shengweisong/article/details/45293413