HDU5339——Untitled

Problem Description

There is an integer a and n integers b1,…,bn. After selecting some numbers from b1,…,bn in any order, say c1,…,cr, we want to make sure that a mod c1 mod c2 mod… mod cr=0 (i.e., a will become the remainder divided by ci each time, and at the end, we want a to become 0). Please determine the minimum value of r. If the goal cannot be achieved, print −1 instead.

 

Input

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

For each testcase, there are two lines:

1. The first line contains two integers n and a (1≤n≤20,1≤a≤106).

2. The second line contains n integers b1,…,bn (∀1≤i≤n,1≤bi≤106).

 

Output

Print T answers in T lines.

 

Sample Input

22 92 72 96 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;}