hdu 5339 Untitled

Untitled

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 166    Accepted Submission(s): 83

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

 

对于一组可能的答案$c$，如果先对一个觉小的cic_ic​i​​取模，再对一个较大的cjc_jc​j​​取模，那么这个较大的cjc_jc​j​​肯定是没有用的。因此最终的答案序列中的$c$肯定是不增的。那么就枚举选哪些数字，并从大到小取模看看结果是否是$0$就可以了。时间复杂度O(2n)O(2^n)O(2​n​​).从小到大枚举，就可以了,。复杂度o(2^n)

#define N 205#define M 100005#define maxn 205#define MOD 1000000000000000007int T,n,a,pri[N],ans;int Dfs(int x,int m,int num){    if(m == 0){        ans = min(ans,num);        return 0;    }    for(int i = x - 1;i>=0;i--){        if(m >= pri[i])            Dfs(i,m%pri[i],num+1);    }}int main(){    while(S(T)!=EOF)    {        while(T--){            S2(n,a);            FI(n) S(pri[i]);            sort(pri,pri+n);            ans = INF;            Dfs(n,a,0);            if(ans == INF) ans = -1;            printf("%d\n",ans);        }    }    return 0;}

Source

BestCoder Round #49 ($)

 

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