【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+;
 __int64 dp[MAXN][MAXN];
 __int64 C[MAXN][MAXN];
 __int64 two[MAXN];

 void init() {
     int i, j, k, tmp;

     two[] = ;
     for (i=; i<MAXN; ++i) {
         two[i] = (two[i-] << );
         two[i] %= MOD;
     }

     C[][] = C[][] = ;
     for (i=; i<MAXN; ++i) {
         C[i][] = C[i][i] = ;
         for (j=; j<i; ++j) {
             C[i][j] = C[i-][j] + C[i-][j-];
             C[i][j] %= MOD;
         }
     }

     for (i=; i<MAXN; ++i)
         dp[i][] = dp[][i] = ;

     for (i=; i<MAXN; ++i) {
         for (j=; j<MAXN; ++j) {
             dp[i][j] = dp[i-][j]*(two[j]-)%MOD;
             for (k=; k<j; ++k) {
                 dp[i][j] += dp[i-][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\n", dp[n][m]);

     return ;
 }