Happy Matt Friends

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

Total Submission(s): 0 Accepted Submission(s): 0

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

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.

解题：dp计数

dp弱到渣啊！在码代码的猿猿巨巨指导下才搞定！

1 #include 
         
           2 #include 
          
            3 #include 
           
             4 #include 
            
              5 #include 
             
               6 #include 
              
                7 #include 
               
                 8 #include 
                
                  9 #include 
                 
                  10 #include 
                  
                   11 #include 
                   
                    12 #include 
                    13 #include 
                     
                      14 #define LL long long15 #define INF 0x3f3f3f3f16 #define pii pair
                      
                       17 using namespace std;18 const int maxn = 1<<21;19 int dp[2][maxn],d[41],n,m;20 int main() {21 int T,cs=1;22 scanf("%d",&T);23 while(T--) {24 scanf("%d %d",&n,&m);25 memset(dp,0,sizeof(dp));26 for(int i = 0; i < n; ++i)27 scanf("%d",d+i);28 LL ans = 0;29 dp[0][0] = 1;30 int cur = 0;31 for(int i = 0; i < n; ++i) {32 memset(dp[cur^1],0,sizeof(dp[cur^1]));33 for(int j = 0; j < maxn; ++j) {34 int tmp = j^d[i];35 dp[cur^1][tmp] += dp[cur][j];36 dp[cur^1][j] += dp[cur][j];37 }38 cur ^= 1;39 }40 for(int i = m; i < maxn; ++i)41 if(dp[cur][i]) ans += dp[cur][i];42 printf("Case #%d: %I64d\n",cs++,ans);43 }44 return 0;45 }

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转载于:https://www.cnblogs.com/crackpotisback/p/4130776.html

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