HDU 4901

dpdp，需要绕一个弯子，注意不要重复计数，然后就没什么了。

/*
ID:esxgx1
LANG:C++
PROG:hdu4901
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int mod = 1e9 + 7;

unsigned _dp[1007][1024][2], (*dp)[1024][2] = &_dp[1];
unsigned dp2[1024];
int a[1007];

int main(void)
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    #endif

    int T;
    scanf("%d", &T);
    for(int t=1; t<=T; ++t) {
        int N;
        scanf("%d", &N);
        for(int i=0; i<N; ++i)
            scanf("%u", &a[i]);
        for(int i=0; i<N; ++i) {
            for(int j=0; j<1024; ++j) {
                dp[i][j][0] = dp[i-1][j ^ a[i]][1];
                dp[i][j][1] = (dp[i][j][0] + dp[i-1][j][1]) % mod;
            }
            dp[i][a[i]][0] = (dp[i][a[i]][0] + 1) % mod;
            dp[i][a[i]][1] = (dp[i][a[i]][1] + 1) % mod;
        }
        unsigned ans = 0;
        memset(dp2,0, sizeof(dp2));
        for(int i=N; i--; ) {
            for(int j=0; j<1024; ++j) {
                ans = (ans + (((long long)dp2[j] * dp[i][j][0]) % mod)) % mod;
                dp2[j & a[i]] = (dp2[j & a[i]] + dp2[j]) % mod;
            }
            dp2[a[i]] = (dp2[a[i]] + 1) % mod;
        }
        printf("%u
", ans);
    }
    return 0;
}

2014-10-04 13:19:54

Accepted

4901

1046MS

8332K

1081 B

G++