HDU 5339 Untitled (状态压缩枚举)

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 integerT≤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

 

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);    }}