HDU-5119 Happy Matt Friends （背包DP，递推枚举）2017寒假集训

题意：n个物品，第i个物品的权值为ki，选出一些物品使它们的异或不小于m，求有多少种方案

数据范围：1 <= n <= 40，0 <= m <= 1e6

思路：其实就是换了一种要求的背包，MX要开得比1e6大一些，不滚动数组也能过去。

老套路设dp[i][j]为前i个物品异或为j时的方案，dp[i][j] = dp[i-1][j] + dp[i-1][j^a[i]]，最后把dp[n][m以上]的值求和

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define LL long long
 5 using namespace std;
 6 
 7 const int mx = 1<<20;
 8 int a[50];
 9 LL dp[50][mx+10];
10 
11 int main(){
12     int t, kase = 0;
13     scanf("%d", &t);
14     while (t--){
15         int n, m;
16         memset(dp, 0, sizeof dp);
17         LL ans = 0;
18         scanf("%d%d", &n, &m);
19         for (int i = 1; i <= n; i++)
20             scanf("%lld", &a[i]);
21         dp[0][0] = 1;
22         for (int i = 1; i <= n; i++)
23             for (int j = 0; j <= mx; j++)
24                 dp[i][j] = dp[i-1][j] + dp[i-1][j^a[i]];
25         for (int i = m; i < mx; i++)
26             ans += dp[n][i];
27         printf("Case #%d: %lld\n", ++kase, ans);
28     }
29     return 0;
30 }