HDU-5119 Happy Matt Friends(背包)

Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)

Total Submission(s): 3341    Accepted Submission(s): 1301

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 ≤ 10 6).

In the second line, there are N integers ki (0 ≤ k i ≤ 10 6), 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-重制賽（感謝北師和上交） 題意:有n個數,随意取數使得異或和不小于m,問有多少種取法

題解:由于題意說明每個數都小于1e6,最多隻有40個數,我們可以用背包來做,背包容量就是1e6,物品數量是n,dp[i][j]表示當取到第i個數的時候,能使背包容量為j的方法數,

dp[i][j]=dp[i-1][j]+dp[i-1][j^a[i]],dp[i-1][j]表示不取a[i],dp[i-1][j^a[i]]表示取a[i]

#include<cstdio>
#include<vector>
#include<algorithm>
#include<functional>
#include<string.h>
using namespace std;
typedef long long LL;
LL dp[41][1<<20];
int a[41];
int main(){
    int T,n,m;
  //  freopen("in.txt","r",stdin);
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        int mx=1<<20;
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(int i=1;i<=n;i++){
            for(int j=0;j<mx;j++){
                dp[i][j]=dp[i-1][j^a[i]]+dp[i-1][j];
            }
        }
        LL ans=0;
        for(int i=m;i<mx;i++) ans+=dp[n][i];
        printf("Case #%d: %I64d\n",cas,ans);
    }
    return 0;
}