hdu--5155Harry And Magic Box（组合数+容斥原理）

Harry And Magic Box

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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Appoint description:  System Crawler  (2015-01-07)

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. $0 \leq n, m \leq 50$.

 

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.

题目大意在n*m的矩阵内每一行每一列都有钻石，问钻石分布的种类？

在n*m的矩阵中，假设每一行都存在，对于每一行设i为有i列不存在钻石，那么共有C(m,i)种排列。

对于其他的m-i列中可以放也可以不放，但是要排除全都不放的情况，得到2^(m-i) - 1种，再加上n行得到(2^(m-i)-1)^n

得到f(i) = C(m,i) * (2^(m-i)-1)^n ;

容斥原理排除多余的  f(0) - f(1) + f(2)....f(n) ;

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define MOD 1000000007
#define LL long long
LL c[60][60] , num[60][60] , k[60] ;
int main()
{
    int i , j , n , m ;
    memset(num,0,sizeof(num)) ;
    for(i = 0 ; i <= 50 ; i++)
        num[i][0] = num[0][i] = 1 ;
    k[0] = 1 ;
    for(i = 0 ; i <= 50 ; i++)
        c[i][0] = 1 ;
    for(i = 1 ; i <= 50 ; i++)
    {
        k[i] = k[i-1] * 2 ;
        k[i] %= MOD ;
    }
    for(i = 1 ; i <= 50 ; i++)
    {
        for(j = 1 ; j < i ; j++)
        {
            c[i][j] = c[i-1][j-1] + c[i-1][j] ;
            c[i][j] %= MOD ;
        }
        c[i][i] = 1 ;
    }
    while(scanf("%d %d", &n, &m) != EOF )
    {
        if( num[n][m] == 0 )
        {
            int temp = 1 ;
            LL ans = 0 , s ;
            for(i = 0 ; i <= m ; i++)
            {
                s = c[m][i] ;
                for(j = 1 ; j <= n ; j++)
                {
                    s *= ( k[m-i]-1 ) ;
                    s %= MOD ;
                }
                ans += temp * s ;
                ans %= MOD ;
                temp = -temp ;
            }
            if( ans < 0 )
                ans += MOD ;
            num[n][m] = num[m][n] = ans ;
        }
        printf("%lld\n", num[n][m]) ;
    }
    return 0;
}