hdu 4901 The Romantic Hero (dp)

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题目要求将一个序列分成两个集合A，B，并且A中的所有元素异或的结果要等于B中所有元素与的结果，求这样的划分方案，注意的是题目要求了：这两个集合的元素必须满足对于任意的i,j ai的序号小于bj的序号。

其实就是类似背包的问题，求出正推反推对的方案数，然后枚举端点两端相乘。注意集合非空。

#include<iostream>#include<math.h>#include<stdio.h>#include<algorithm>#include<string.h>#include<string>#include<vector>#include<queue>#include<map>#include<set>#include<stack>#define B(x) (1<<(x))using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned ui;const int oo = 0x3f3f3f3f;//const ll OO = 0x3f3f3f3f3f3f3f3f;const double eps = 1e-9;#define lson rt<<1#define rson rt<<1|1void cmax(int& a, int b){ if (b > a)a = b; }void cmin(int& a, int b){ if (b < a)a = b; }void cmax(ll& a, ll b){ if (b > a)a = b; }void cmin(ll& a, ll b){ if (b < a)a = b; }void cmax(double& a, double b){ if (a - b < eps) a = b; }void cmin(double& a, double b){ if (b - a < eps) a = b; }void add(int& a, int b, int mod){ a = (a + b) % mod; }void add(ll& a, ll b, ll mod){ a = (a + b) % mod; }const ll MOD = 1000000007;const int maxn = 1100;ll f[maxn][maxn][2], g[maxn][maxn][2];int a[maxn];int main(){int n, T;scanf("%d", &T);while (T--){scanf("%d", &n);for (int i = 1; i <= n; i++)scanf("%d", &a[i]);memset(f, 0, sizeof f);memset(g, 0, sizeof g);for (int i = 1; i <= n; i++){f[i][a[i]][1] = 1;for (int j = 0; j < 1024; j++){f[i][j][0] += f[i - 1][j][0];f[i][j][0] += f[i - 1][j][1];f[i][j][0] %= MOD;f[i][j^a[i]][1] += f[i - 1][j][0];f[i][j^a[i]][1] += f[i - 1][j][1];f[i][j^a[i]][1] %= MOD;}}for (int i = n; i >= 1; i--){g[i][a[i]][1] = 1;for (int j = 0; j < 1024; j++){g[i][j][0] += g[i + 1][j][0];g[i][j][0] += g[i + 1][j][1];g[i][j][0] %= MOD;g[i][j&a[i]][1] += g[i + 1][j][0];g[i][j&a[i]][1] += g[i + 1][j][1];g[i][j&a[i]][1] %= MOD;}}ll ans = 0;for (int i = 1; i < n; i++){for (int j = 0; j < 1024; j++){ans += (f[i][j][1] * (g[i + 1][j][0] + g[i + 1][j][1]) % MOD) % MOD;ans %= MOD;}}cout << ans << endl;}return 0;}

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