HDU 5155 Harry and Magic Box

Problem description

One day, Harry got a magical box. The box is made up of n*m, and there are some glittering gems in the lattice. But the top and bottom of the box were sealed with magical magic, so Harry couldn't see the inside of the box from the top and bottom. However, the four sides of the box are transparent, and Harry can see the situation from the sides of the box. Harry found that this magical box, from the left to the past, each line is shining, from the front to see the past, each column is also flashing light. Harry wondered how many of the gems in the box were distributed. The answer is likely to be large, so the output answer is 1000000007 modulo.

Enter a description

Multiple sets of input data per row of data, input two number n m for the size of the box,
    
        0≤n,m≤    
  
 
    
 

Output description

Each set of data outputs a row, an integer, representing the number of scenarios

Idea: A checkerboard type of DP, expanded by line, to each row must have a pawn, so enumerate each line of the number of pieces filled,

DP[I][J] indicates that it has been extended to line I with the number of schemes in which J has a pawn.

  Dp[i+1][j+t] + = dp[i][j] * c[m-j][t] * C[j][k-t]  (k is the number of pieces of the i+1 line, T is the number of new columns)

Code:

#include <iostream> #include <cmath> #include <cstdio> #include <cstring> #include < algorithm>using namespace Std;const int mod = 1000000007;const int N = 100+10;typedef long long ll;ll c[n][n];ll dp[n][ N];int n,m;void init () {    memset (c,0,sizeof (C));    for (int i=0;i<=60;i++) c[i][0]=1;    for (int i=1;i<=50;i++) for        (int j=1;j<=i;j++)            c[i][j] = (C[i-1][j] + c[i-1][j-1])% MoD;} ll solve () {    memset (dp,0,sizeof (DP));    for (int i=1;i<=m;i++)    {        Dp[1][i] = C[m][i];    }    for (int i=1;i<=n;i++) for (int        j=1;j<=m;j++) for (            int. k=1;k<=m;k++) for (int                t=0;t<=k& &t<=m-j;t++)                dp[i+1][j+t] = (Dp[i+1][j+t] + ((dp[i][j]*c[m-j][t)%mod*c[j][k-t])%mod)%mod;    return dp[n][m];} int main () {    init ();    while (scanf ("%d%d", &n,&m)!=eof)    {        printf ("%d\n", Solve ());    }    return 0;}

HDU 5155 Harry and Magic Box