HDU 5339 Untitled (狀態壓縮列舉)

_TCgogogo_發表於2015-08-02
原文網址 : https://blog.csdn.net/tc_to_top/article/details/47206995


Untitled

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 570    Accepted Submission(s): 291

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
2
2 9
2 7
2 9
6 7




 

Sample Output



2
-1




 

Source 

BestCoder Round #49 ($)


題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=5339


題目大意：重排列bi，問a對重排列的數不斷取模最快能為0的取模次數


題目分析：其實是水題，首先對數字小的取完餘後再對數字大的取餘等於沒取，所以先對大數字取餘，從大到小排序，因為n很小，DFS隨意搜，也可以用狀態壓縮做，把每個數字選或不選的狀態二進位制壓縮

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 1 << 22;
int const INF = 0x3fffffff;
int b[25], sta[MAX];

int lowbit(int x)
{
    return x & (-x);
}

bool cmp(int a, int b)
{
    return a > b;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T --)
    {
        memset(sta, 0, sizeof(sta));
        int n, a;
        scanf("%d %d", &n, &a);
        for(int i = 1; i <= n; i++)
            scanf("%d", &b[i]);
        sort(b + 1, b + n + 1, cmp);
        for(int i = 1; i <= n; i++)
            sta[1 << (i - 1)] = b[i];
        int cnt, ans = INF;
        for(int i = 1; i < (1 << n); i++)
        {
            int tmp = a;
            cnt = 0;
            for(int j = i; j > 0; j -= lowbit(j))
            {
                tmp %= sta[lowbit(j)];
                cnt ++;
            }
            if(tmp == 0)
                ans = min(ans, cnt);
        }
        if(ans == INF)
            printf("-1\n");
        else
            printf("%d\n", ans);
    }
}




 





            




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