【HDOJ】5155 Harry And Magic Box

DP。dp[i][j]可以表示i行j列满足要求的组合个数，考虑dp[i-1][k]满足条件，那么第i行的那k列可以为任意排列（2^k），其余的j-k列必须全为1，因此dp[i][j] += dp[i-1][k]*(2^k)*C(j, k)。

/* 5155 */
#include <cstdio>
#include <cstring>
#include <cstdlib>

#define MAXN 51

const __int64 MOD = 1e9+7;
__int64 dp[MAXN][MAXN];
__int64 C[MAXN][MAXN];
__int64 two[MAXN];

void init() {
    int i, j, k, tmp;
    
    two[0] = 1;
    for (i=1; i<MAXN; ++i) {
        two[i] = (two[i-1] << 1);
        two[i] %= MOD;
    }
    
    C[1][0] = C[1][1] = 1;
    for (i=2; i<MAXN; ++i) {
        C[i][0] = C[i][i] = 1;
        for (j=1; j<i; ++j) {
            C[i][j] = C[i-1][j] + C[i-1][j-1];
            C[i][j] %= MOD;
        }
    }
    
    for (i=0; i<MAXN; ++i)
        dp[i][1] = dp[1][i] = 1;
    
    for (i=2; i<MAXN; ++i) {
        for (j=1; j<MAXN; ++j) {
            dp[i][j] = dp[i-1][j]*(two[j]-1)%MOD;
            for (k=1; k<j; ++k) {
                dp[i][j] += dp[i-1][k]*C[j][k]%MOD*two[k];
                dp[i][j] %= MOD;
            }
        }
    }
}

int main() {
    int n, m;
    
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
        freopen("data.out", "w", stdout);
    #endif
    
    init();
    while (scanf("%d %d", &n, &m) != EOF)
        printf("%I64d
", dp[n][m]);
    
    return 0;
}