首页 > 其他 > 详细

HDU 5339 Untitled (状态压缩枚举)

时间:2015-08-02 15:13:58      阅读:286      评论:0      收藏:0      [点我收藏+]


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 ($)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5339

题目大意:重排列bi,问a对重排列的数不断取模最快能为0的取模次数

题目分析:其实是水题,首先对数字小的取完余后再对数字大的取余等于没取,所以先对大数字取余,从大到小排序,因为n很小,DFS随意搜,也可以用状态压缩做,把每个数字选或不选的状态二进制压缩

#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);
    }
}


 

版权声明:本文为博主原创文章,未经博主允许不得转载。

HDU 5339 Untitled (状态压缩枚举)

原文:http://blog.csdn.net/tc_to_top/article/details/47206995

踩
(0)
赞
(0)
   
举报
评论 一句话评论(0)
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!