codeforces 844B Rectangles （概率）

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Rectangles

You are given n × m table. Each cell of the table is colored white or black. Find the number of non-empty sets of cells such that:

All cells in a set have the same color.
Every two cells in a set share row or column.
Input
The first line of input contains integers n and m (1 ≤ n, m ≤ 50) — the number of rows and the number of columns correspondingly.

The next n lines of input contain descriptions of rows. There are m integers, separated by spaces, in each line. The number equals 0 if the corresponding cell is colored white and equals 1 if the corresponding cell is colored black.

Output
Output single integer — the number of non-empty sets from the problem description.

Sample Input
Input
1 1
0
Output
1
Input
2 3
1 0 1
0 1 0
Output
8
Hint
In the second example, there are six one-element sets. Additionally, there are two two-element sets, the first one consists of the first and the third cells of the first row, the second one consists of the first and the third cells of the second row. To sum up, there are 8 sets.

Source
AIM Tech Round 4 (Div. 2)

mean：
给你n*m的矩阵，0，1个代表一个颜色；

同行或者同列 颜色相同就算一个set，求解set个数

ans：
同一颜色分为出现和不出现两种可能，去除全部不出现的可能行就变成了 2

的k方-1；

没法Cnm暴力求，爆int ，这题没取模，似乎只能这样搞额

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#include<bits/stdc++.h>
using namespace std;

#define ll long long
const ll mod=1e9+7;
ll work(ll  x)
{
    ll s=1,i;
    for(i=0;i<x;i++)
        s=(s*2);
    return s;
}
int main()
{
    ll n,m,i,j;
    while(cin>>n>>m)
    {
        ll mp[111][111];
        ll ans=0;
        for(i=0;i<n;i++)
        {
            ll cnt=0;
            for(j=0;j<m;j++)
            {
                scanf("%lld",&mp[i][j]);
                if(mp[i][j]) cnt++;
            }
            ans=(ans+work(cnt)-1);
            ans=(ans+work(m-cnt)-1);
        }
        for(j=0;j<m;j++)
        {
            ll cnt=0;
            for(i=0;i<n;i++)
            {
                if(mp[i][j]) cnt++;
            }
            ans=(ans+work(cnt)-1);
            ans=(ans+work(n-cnt)-1);
        }
        cout<<ans-n*m<<endl;
    }
}