HDU5339——Untitled

 

Problem Description

There is an integer 1 instead.

 

 

Input

The first line contains one integer 5, which represents the number of testcases. 

For each testcase, there are two lines:

1. The first line contains two integers 6).

2. The second line contains 6).

 

 

Output

Print T answers in T lines.

 

 

Sample Input

2 2 9 2 7 2 9 6 7

 

 

Sample Output

2 -1

 

 

用a去模b中的数，问最少需要多少个数才能使a变为0

感觉模的数字大，需要的数可能就会少一点，所以排个序，再暴力解决

#include <iostream>
#include <cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int a[30];

bool cmp(int a,int b)
{
    return a > b;
}

int work(int q,int n)
{
    int minx = 0x3f3f3f3f;
    int i,j;
    for(i = 1; i <= n; i++)
    {
        int temp = q;
        for( j = i; j <= n; j++)
        {
            if(temp == 0)
                break;
            else
                temp %= a[j];
        }
        if(temp == 0)
            minx = min(minx,j-i);
    }
    return minx;
}

int main()
{
    int T;
    int n,all;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&all);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        sort(a+1,a+n+1,cmp);
        int p = work(all,n);
        if(p != 0x3f3f3f3f)
            printf("%d\n",p);
        else
            printf("-1\n");
    }
    return 0;
}