Reverse It HDU HDU - 6513 Reverse It (SYSU校赛C题)(组合数学+容斥)
题目链接：
Reverse It
题目大意：给你一个n*m的01矩阵，然后你最多对这个矩阵的子矩阵进行翻转两次（0变成1，1变成0）。问你最多有多少个不同的矩阵？
学习网址：
感谢lxw的讲解，
具体思路：
AC代码：
1 #include<bits/stdc++.h>
2 using namespace std;
3 # define ll long long
4 # define inf 0x3f3f3f3f
5 const int maxn = 2e5+100;
6 ll com(ll n,ll m) {ll ret=1; for(ll i=1; i<=m; ++i)ret=ret*(n-i+1)/i; return ret;}
7 int main()
8 {
9 ll n,m;
10 while(~scanf("%lld %lld",&n,&m))
11 {
12 char u;
13 for(int i=1; i<=n; i++)
14 {
15 for(int j=1; j<=m; j++)
16 scanf(" %c",&u);
17 }
18 ll tot=com(n+1ll,2ll)*com(m+1ll,2ll)*(com(n+1ll,2ll)*(com(m+1ll,2ll))-1ll)/2ll;
19 tot-=(n-1ll+m-1ll-1ll)*com(n+1ll,2ll)*com(m+1,2ll);
20
21 tot-=2ll*2ll*com(n+1ll,3ll)*com(m+1ll,3ll);
22
23 tot-=2ll*(com(n+1ll,2ll)*com(m+1ll,4ll)+com(n+1ll,4ll)*com(m+1ll,2ll));
24
25 tot-=2ll*4ll*(com(n+1ll,3ll)*com(m+1ll,3ll));
26 tot+=1ll;
27
28 printf("%lld
",tot);
29 }
30 return 0;
31 }