HDU 5155 Harry And Magic Box（组合数学+容斥）

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

容斥+组合知识：感谢凡神写的题解点击打开链接

我们首先要保证的是每行都有1，这是前提。然后要想得到答案我们可以枚举i

列全为0（共有C(m,i)选法)，然后就是考虑每行还剩的(m-i)个位置，我们

此时可以随意放置01(2^(m-i)方法)，但是为了保证每行都有1故要减1，此时

为(2^(m-i)-1).然后有n行，所以sum[i]=(2^(m-i)-1)^n,然后我们要从

里面剔除一些东西，就是仅仅第i列全为0（注意刚刚i列全为不是仅仅）

所以要容斥。ans=sum[0]-sum[1]+sum[2]..
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int maxn=55;
const int MOD=1000000007;
LL c[maxn][maxn];//2^(m-i)-1;
int n,m;
void init()
{
    REPF(i,1,50)
    {
        c[i][0]=c[i][i]=1;
        REPF(j,1,i-1)
           c[i][j]=(c[i-1][j]+c[i-1][j-1])%MOD;
    }
}
LL pow_mod(LL a,int b)
{
    a%=MOD;
    LL ans=1;
    while(b)
    {
        if(b&1)  ans=ans*a%MOD;
        a=a*a%MOD;
        b>>=1;
    }
    return ans;
}
int main()
{
    init();
    while(~scanf("%d%d",&n,&m))
    {
        LL ans=0;
        REPF(i,0,m)
        {
            if(i&1)
                ans=(ans-(pow_mod((1LL<<(m-i))-1,n)*c[m][i])%MOD+MOD)%MOD;
            else
                ans=(ans+(pow_mod((1LL<<(m-i))-1,n)*c[m][i])%MOD)%MOD;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}