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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.

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).

Print T answers in T lines.

2
2 9
2 7
2 9
6 7

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 1000000000000000007
int 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;
}

hdu 5339 Untitled

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原文地址：http://blog.csdn.net/mengzhengnan/article/details/47191085