hdu 5155 Harry And Magic Box（递推）

Harry And Magic Box

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

Total Submission(s): 625    Accepted Submission(s): 294


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
HintThere 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.

题意：有一个n*m的盒子，每行每列至少有一个1，问一共有多少种可能

思路：参考此人博客。

点击打开链接

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define mod 1000000007
#define N 55
long long dp

;
long long c

;
void init()
{
for(int i=1;i<N;i++)
{
c[i][0]=c[i][i]=1;
for(int j=1;j<i;j++)
{
c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
}
}
}
long long pow_mod(long long a,long long n)
{
long long ans=1;
while(n)
{
if(n&1) ans=ans*a%mod;
a=a*a%mod;
n>>=1;
}
return ans;
}
int main()
{
int n,m;
init();
while(~scanf("%d %d",&n,&m))
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=m;i++)
dp[1][i]=1;
for(int i=1;i<=n;i++)
dp[i][1]=1;
for(int i=2;i<=n;i++)
{
for(int j=2;j<=m;j++)
{
dp[i][j]=dp[i][j-1]*(pow_mod(2,i)-1)%mod;
for(int k=1;k<=i;k++)
{
dp[i][j]=(dp[i][j]+dp[i-k][j-1]*c[i][k]%mod*pow_mod(2,i-k)%mod)%mod;
}
}
}
printf("%lld\n",dp
[m]);
}
return 0;
}