hdu 5119__Happy Matt Friends

Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 1642    Accepted Submission(s): 646

Problem Description

Matt has N friends. They are playing a game together.

Each of Matt’s friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends’magic numbers is no less than M , Matt wins.

Matt wants to know the number of ways to win.

 

Input

The first line contains only one integer T , which indicates the number of test cases.

For each test case, the first line contains two integers N, M (1 ≤ N ≤ 40, 0 ≤ M ≤ 10⁶).

In the second line, there are N integers ki (0 ≤ k_i ≤ 10⁶), indicating the i-th friend’s magic number.

 

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y indicates the number of ways where Matt can win.

 

Sample Input

2 3 2 1 2 3 3 3 1 2 3

 

Sample Output

Case #1: 4 Case #2: 2

Hint

In the first sample, Matt can win by selecting: friend with number 1 and friend with number 2. The xor sum is 3. friend with number 1 and friend with number 3. The xor sum is 2. friend with number 2. The xor sum is 2. friend with number 3. The xor sum is 3. Hence, the answer is 4.

 

Source

2014ACM/ICPC亞洲區北京站-重現賽（感謝北師和上交）

 

題意：給出n個數要求任意數量的數的異或和大於m的方案數。

想法：類似於揹包問題。對於第i個數，可以選擇取或不取，取就是dp[i-1][j^a[i]]，不取就是dp[i-1][j]；所以狀態轉移方程是dp[i][j]=dp[i-1][j]+dp[i-1][j^a[i]]；

代碼如下：

#include<cstdio>
#include<cstring>
int t,n,m,ch[44],dp[44][1<<20];
int main()
{
    scanf("%d",&t);
    int cas=1;
    while(t--) {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) {
            scanf("%d",&ch[i]);
        }
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(int i=1;i<=n;i++) {
            for(int j=0;j<=(1<<20);j++) {
                dp[i][j]=dp[i-1][j]+dp[i-1][j^ch[i]];
            }
        }
        long long ans=0;
        for(int i=m;i<(1<<20);i++) {
            ans+=dp[n][i];
        }
        printf("Case #%d: %lld\n",cas++,ans);
    }
    return 0;
}