hdu 4901 The Romantic Hero(dp)

题目链接：hdu 4901 The Romantic Hero

题目大意：给定一个序列，要求选出两个集合，S和T，要求S中选中的元素的下标都要小于T中元素的下标。

而且说S中元素的亦或和要等于T中元素取且的和。

解题思路:dp， 思路非常好想。就是细节比較恶心。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef __int64 ll;
const int maxn = 1024;
const ll MOD = 1e9+7;

int n, num[maxn];
ll l[maxn][maxn+10][2], r[maxn][maxn+10];

void init () {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &num[i]);
}

ll solve () {
    memset(l, 0, sizeof(l));
    memset(r, 0, sizeof(r));

    for (int i = 1; i < n; i++) {
        l[i][num[i]][1]++;
        for (int j = 0; j < maxn; j++) {
            l[i+1][j][1] = (l[i][j^num[i+1]][1] + l[i][j^num[i+1]][0])  % MOD;
            l[i+1][j][0] = (l[i][j][1] + l[i][j][0]) % MOD;
        }
    }

    for (int i = n; i; i--) {
        r[i][num[i]]++;
        for (int j = 0; j < maxn; j++) {
            r[i-1][j] = (r[i-1][j] + r[i][j]) % MOD;
            r[i-1][j&num[i-1]] = (r[i-1][j&num[i-1]] + r[i][j]) % MOD;
        }
    }

    ll ans = 0;
    for (int i = 1; i < n; i++) {

        for (int j = 0; j < maxn; j++) {
            /*
            if (l[i][j] * r[i+1][j])
                printf("%d %d %lld %lld
", i, j, l[i][j], r[i+1][j]);
                */
            ans = (ans + l[i][j][1] * r[i+1][j] % MOD) % MOD;
        }
    }
    return ans;
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        init();
        printf("%I64d
", solve());
    }
    return 0;
}