HDOJ 5119 Happy Matt Friends DP
N*M暴力DP....
Happy Matt Friends
Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others)Total Submission(s): 82 Accepted Submission(s): 34
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.
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 ≤ ki ≤ 106), indicating the i-th friend’s magic number.
2 3 2 1 2 3 3 3 1 2 3
Case #1: 4 Case #2: 2HintIn the ?rst 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.
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long int LL; LL dp[2][1<<20+1]; int a[100]; int n,m; int main() { int T_T,cas=1; scanf("%d",&T_T); while(T_T--) { scanf("%d%d",&n,&m); memset(a,0,sizeof(a)); memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) scanf("%d",a+i); int now=1,pre=0; dp[0][0]=1LL; for(int i=1;i<=n;i++) { memset(dp[now],0,sizeof(dp[now])); for(int j=0;j<=(1<<20);j++) { if(dp[pre][j]) { int x=j^a[i]; dp[now][x]+=dp[pre][j]; dp[now][j]+=dp[pre][j]; } } swap(now,pre); } LL ans=0; for(int i=m;i<=(1<<20);i++) ans+=dp[pre][i]; printf("Case #%d: %I64d\n",cas++,ans); } return 0; }
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