HDU 5119 Happy Matt Friends （01背包）

题意:

N<=40个人，每个人有一个权值ki<=106，你可以挑任意多的人并将他们的权值异或(也可以不选)
求最后得到的值大于等于M的取法有多少种

分析:

由于ki<=106我们可以知道xor sum<=(220−1)
根据时限和空间完全足够,无非就是每个数选还是不选,01背包就好了

代码:

////  Created by TaoSama on 2015-10-11//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1 << 20, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;int n, m, a[45];LL dp[45][N];LL dfs(int i, int w) {    LL& ret = dp[i][w];    if(ret >= 0) return ret;    if(i == n + 1) return w >= m;    ret = 0;    ret += dfs(i + 1, w ^ a[i]);    ret += dfs(i + 1, w);    return ret;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d%d", &n, &m);        for(int i = 1; i <= n; ++i) scanf("%d", a + i);        memset(dp, -1, sizeof dp);        printf("Case #%d: %I64d\n", ++kase, dfs(1, 0));    }    return 0;}