hdu 5119 Happy Matt Friends【dp】

Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 3037    Accepted Submission(s): 1197

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亚洲区北京站-重现赛（感谢北师和上交）

题目大意：

一共有N个数，找任意个数的数使其亦或值大于等于m，问一共有多少种可行解。

思路：

1、设定dp【i】【j】表示数据选到第i个数亦或值为j的方案数。

2、

①初始化dp【0】【0】=1；

②那么不难理解其状态转移方程：dp【i】【j】=dp【i-1】【j】+dp【i-1】【j^a[i]】

③表示数据选到第i个数的亦或值为j的方案数可以从不选a【i】这个数以及选择a【i】这个数的状态转移过来。

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;long long int dp[45][(1<<20)];int a[50];int main(){    int t;    int kase=0;    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        int n,m;        scanf("%d%d",&n,&m);        int maxn=(1<<20);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        dp[0][0]=1;        for(int i=1;i<=n;i++)        {            for(int j=0;j<maxn-5;j++)            {                dp[i][j]=dp[i-1][j^a[i]]+dp[i-1][j];            }        }        long long int  output=0;        for(int i=m;i<maxn+10;i++)output+=dp[n][i];        printf("Case #%d: ",++kase);        printf("%I64d\n",output);    }}