HDU

Harry And Magic Box

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.

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≤500≤n,m≤50.

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

    题意：  n 行  m 列的表格，每一行，每一列不能全为0  ，求方案数？

   思路：   容斥 + 组合数。直接假定  每一行都有  1 ，考虑每一列的情况，枚举有多少列全部是0 ，奇数减偶数加。

#pragma comment(linker, "/STACK:1024000000,1024000000")//#include <bits/stdc++.h>#include<string>#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<queue>#include<stack>#include<vector>#include<set>#include<algorithm>#define maxn 57#define INF 0x3f3f3f3f#define eps 1e-8#define mod 1000000007#define ll long longusing namespace std;ll Pow(ll a,ll b){    a=a%mod;    ll ans=1;    while(b)    {        if(b&1)  ans=ans*a%mod;        a=a*a%mod;        b>>=1;    }    return ans%mod;}ll c[maxn][maxn];void init(){    for(int i=1;i<maxn;i++)    {        c[i][0]=c[i][i]=1;        for(int j=1;j<i;j++)            c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;    }}int main(){    ll n,m;    init();    while(scanf("%lld%lld",&n,&m)!=EOF)    {        ll ans=0;        for(int i=0;i<=m;i++)        {            if(i&1)            {                ans=((ans-Pow((1ll<<(m-i))-1,n)*c[m][i]%mod)%mod+mod)%mod;            }            else                ans=((ans+Pow((1ll<<(m-i))-1,n)*c[m][i]%mod)%mod+mod)%mod;        }        printf("%lld\n",ans);    }    return 0;}