dpdp，需要绕一个弯子，注意不要重复计数，然后就没什么了。
1 /* 2 ID:esxgx1 3 LANG:C++ 4 PROG:hdu4901 5 */ 6 #include <cstdio> 7 #include <cstring> 8 #include <iostream> 9 #include <algorithm> 10 using namespace std; 11 12 const int mod = 1e9 + 7; 13 14 unsigned _dp[1007][1024][2], (*dp)[1024][2] = &_dp[1]; 15 unsigned dp2[1024]; 16 int a[1007]; 17 18 int main(void) 19 { 20 #ifndef ONLINE_JUDGE 21 freopen("in.txt", "r", stdin); 22 #endif 23 24 int T; 25 scanf("%d", &T); 26 for(int t=1; t<=T; ++t) { 27 int N; 28 scanf("%d", &N); 29 for(int i=0; i<N; ++i) 30 scanf("%u", &a[i]); 31 for(int i=0; i<N; ++i) { 32 for(int j=0; j<1024; ++j) { 33 dp[i][j][0] = dp[i-1][j ^ a[i]][1]; 34 dp[i][j][1] = (dp[i][j][0] + dp[i-1][j][1]) % mod; 35 } 36 dp[i][a[i]][0] = (dp[i][a[i]][0] + 1) % mod; 37 dp[i][a[i]][1] = (dp[i][a[i]][1] + 1) % mod; 38 } 39 unsigned ans = 0; 40 memset(dp2,0, sizeof(dp2)); 41 for(int i=N; i--; ) { 42 for(int j=0; j<1024; ++j) { 43 ans = (ans + (((long long)dp2[j] * dp[i][j][0]) % mod)) % mod; 44 dp2[j & a[i]] = (dp2[j & a[i]] + dp2[j]) % mod; 45 } 46 dp2[a[i]] = (dp2[a[i]] + 1) % mod; 47 } 48 printf("%u ", ans); 49 } 50 return 0; 51 }
| 2014-10-04 13:19:54 | Accepted | 4901 | 1046MS | 8332K | 1081 B | G++ |