Coco

1058 - Coco

Time Limit：1s Memory Limit：64MByte

Submissions：282Solved：135

DESCRIPTION

Coco just learned a math operation call mod.Now,there is an integer $a$ and $n$ integers $b_1,\dots ,b_n$. After selecting some numbers from $b_1,\dots ,b_n$ in any order, say $c_1,\dots ,c_r$, Coco want to make sure that $amod{c}_{1}mod{c}_{2}mod\dots mod{c}_r=0$\ (i.e., $a$ will become the remainder divided by $c_i$ each time, and at the end, Coco 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 $T(T\le 5)$, which represents the number of testcases. For each testcase, there are two lines:1. The first line contains two integers$n$ and $a$\ ($1\le n\le 20,1\le a\le {10}^6$).2. The second line contains $n$ integers $b_1,\dots ,b_n$\ ($\forall 1\le i\le n,1\le b_i\le {10}^6$).

OUTPUT

Print $T$ answers in $T$ lines.

SAMPLE INPUT

2

2 9

2 7

2 9

6 7

SAMPLE OUTPUT

2

-1

题意：给一个数k和其他n个数，问能否从n个数中任选几个数使k模这几个数之后结果为0。若能则输出最少选的数字个数，不能输出-1。

思路：直接从大到小排序，然后比k大的数直接跳过，比k小的数每次递归模，中间只要模到结果为0就结束，并更新最小值

#include <iostream>#include <string>#include <string.h>#include <algorithm>#include<math.h>using namespace std;int ans;int a[30],f,n;bool cmp(int a,int b){    return a>b;}int we(int a1,int s,int step){    if(a1==0){        ans=ans>step?step:ans;        return 1;    }    int i;    for(i=s;i<n;i++){        if(a[i]<=a1)        {            f=i;            break;        }    }    if(i==n)        return 0;    for(i=f;i<n;i++){        return we(a1%a[i],i+1,++step);    }}int main(){    int t;    int a1;    cin>>t;    while(t--){        cin>>n>>a1;        for(int i=0;i<n;i++){            cin>>a[i];        }        ans=10000000;        sort(a,a+n,cmp);        if(we(a1,0,0))            cout<<ans<<endl;        else        cout<<-1<<endl;    }}