hdu 5339 Untitled (dfs)

最新推荐文章于 2021-03-28 21:19:39 发布

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本篇介绍了一个整数问题：给定一个整数a及一系列整数b1至bn，需选取部分bi进行连续模运算使a最终变为0，目标是最小化选择的bi数量。若无法实现则返回-1。文中提供了具体的输入输出格式、示例及一种基于排序和深度优先搜索的解决方案。

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Untitled

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1294 Accepted Submission(s): 759

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

T answers in

T lines.

Sample Input

Sample Output

2
-1
solution:
我们知道一个大数模小数大数会变小，而小数模大数不变，因此我们把b[i]从大到小排序，然后dfs即可
#include<cstdio>
#include<algorithm>
using namespace std;
int b[30],n;
bool cmp(int a, int b)
{
    return a > b;
}
int dfs(int cnt, int pos, int remain)
{
    if (remain == 0)return cnt;
    if (pos >= n)return n + 1;
    int x = dfs(cnt, pos + 1, remain);
    int y = dfs(cnt + 1, pos + 1, remain%b[pos]);
    return min(x, y);
}
int main()
{
    int t,a;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d", &n, &a);
        for (int i = 0; i < n; i++)
            scanf("%d", &b[i]);
        sort(b, b + n,cmp);
        int x = dfs(0, 0, a);
        if (x == n + 1)printf("-1\n");
        else printf("%d\n", x);
    }
    return 0;
}