dp[i][j] 表示 长宽为i,j的矩形的可能的总数

dp[i][j+1] 可由 dp[i][j] 推过来,枚举dp[i][j]所保留的行数(1...i)即可

Harry And Magic Box

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

One day, Harry got a magical box. The box is made of n*m grids. There are sparking jewel in some grids. But the top and bottom of the box is locked by amazing magic, so Harry can’t see the inside from the top or bottom. However, four sides of the box are transparent,
so Harry can see the inside from the four sides. Seeing from the left of the box, Harry finds each row is shining(it means each row has at least one jewel). And seeing from the front of the box, each column is shining(it means each column has at least one
jewel). Harry wants to know how many kinds of jewel’s distribution are there in the box.And the answer may be too large, you should output the answer mod 1000000007.

Input

There are several test cases.

For each test case,there are two integers n and m indicating the size of the box. 0≤n,m≤50.

Output

For each test case, just output one line that contains an integer indicating the answer.

Sample Input

1 1
2 2
2 3

Sample Output

1
7
25

Hint

There are 7 possible arrangements for the second test case.
They are:
11
11

11
10

11
01

10
11

01
11

01
10

10
01

Assume that a grids is ‘1‘ when it contains a jewel otherwise not.

Source

BestCoder Round #25

/* ***********************************************
Author        :CKboss
Created Time  :2015年02月15日 星期日 22时41分33秒
File Name     :HDOJ5155.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

const LL mod=1000000007LL; 

LL dp[100][100];
LL C[100][100];
LL two[100];

void init()
{
	for(int i=1;i<=50;i++)
	{
		dp[1][i]=dp[i][1]=1LL;
		two[i]=(1LL<<i)%mod;
	}

	for(int i=1;i<=50;i++)
		C[i][0]=C[i][i]=1LL;
	for(int i=2;i<=50;i++)
	{
		for(int j=1;j<i;j++)
		{
			C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
		}
	}

	for(int i=1;i<=50;i++)
	{
		for(int j=1;j<=50;j++)
		{
			if(dp[i][j]==0)
			{
				LL temp=dp[i][j-1]*(two[i]+mod-1)%mod;
				for(int k=1;k<i;k++)
				{
					temp=(temp+((C[i][k]*dp[i-k][j-1])%mod*two[i-k])%mod)%mod;
				}
				dp[i][j]=dp[j][i]=temp;
			}
		}
	}
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int a,b;
	init();
	while(scanf("%d%d",&a,&b)!=EOF)
	{
		cout<<dp[a][b]<<endl;
	}

    return 0;
}