HDU Untitled（状压DP OR dfs枚举子集）

Untitled

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 325    Accepted Submission(s): 169

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 ($)

 

Recommend

hujie

 

大致题意：

20个数，求选出最少的数然后这些数的某种排列可以连续模a得到0

思路:显然先模大的数，再模小的数，否则没意义

所以状压枚举子集，1<<20 = 100w 然后dp，用了lowbit优化了一下

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
using namespace std;
typedef long long ll;
#define X first
#define Y second
typedef pair<int,int> pii;

template <class T>
inline bool RD(T &ret) {
        char c; int sgn;
        if (c = getchar(), c == EOF) return 0;
        while (c != '-' && (c<'0' || c>'9')) c = getchar();
        sgn = (c == '-') ? -1 : 1;
        ret = (c == '-') ? 0 : (c - '0');
        while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
        ret *= sgn;
        return 1;
}
template <class T>
inline void PT(T x){
        if (x < 0) {
                putchar('-');
                x = -x;
        }
        if (x > 9) pt(x / 10);
        putchar(x % 10 + '0');
}

const int N = 22;
int m[25];
int dp[1<<N];
int sum[1<<N];
int inline lowbit(int x){
        return x & -x;
}
bool cmp(int a,int b){
        return a > b;
}
int vs[1<<N];
int main(){
        for(int i = 0; i<=  20;i++) vs[1<<i] = i;
        int T;
        cin>>T;
        while(T--){
                memset(sum,0,sizeof(sum));
                int n,a;
                RD(n),RD(a);
                REP(i,n) RD(m[i-1]);
                sort(m,m+n);
                REP(i,n) {
                        dp[1<<(i-1)] = a%m[i-1];
                        sum[1<<(i-1)] = 1;
                }
                int ans = 100;
                for(int i = 1;i < (1<<n);i++){
                        int cur = i;
                        if(sum[cur] == 0){
                                sum[cur] = sum[cur-lowbit(cur)] + sum[lowbit(cur)];
                                dp[cur] = dp[cur-lowbit(cur)]%m[vs[lowbit(cur)]];
                        }
                        if( dp[cur] == 0) ans = min(ans,sum[cur]);
                }
                if(ans == 100) puts("-1");
                else printf("%d\n",ans);
        }
}