题解：
乱搞
令x[i][j]表示(i,j)是否操作,a[i][j]表示状态
先假设都翻到0
则x[i][1]^x[i][2]^^^x[i][n]^x[1][j]^x[2][j]^^^x[n][j]^x[i][j]=a[i][j]
令d[i][j]=x[i][1]^x[i][2]^^^x[i][n]^x[1][j]^x[2][j]^^^x[n][j]^x[i][j]
则d[i][1]^d[i][2]^^^d[i][n]^d[1][j]^d[2][j]^^^d[n][j]^d[i][j]=x[i][j]
然后可以求出x[i][j]
根据式子发现翻转成0和1的步数和=n^2
然后取min
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1009;
int n;
int a[maxn][maxn];
int r[maxn],l[maxn];
char ss[maxn];
int cnt=0;
int main(){
scanf("%d",&n);
for(int i=1;i<=n;++i){
scanf("%s",ss+1);
for(int j=1;j<=n;++j)a[i][j]=ss[j]-‘0‘;
}
for(int i=1;i<=n;++i){
for(int j=1;j<=n;++j)r[i]^=a[i][j];
}
for(int j=1;j<=n;++j){
for(int i=1;i<=n;++i)l[j]^=a[i][j];
}
for(int i=1;i<=n;++i){
for(int j=1;j<=n;++j){
int tm=r[i]^l[j]^a[i][j];
// if(!tm)cout<<i<<‘ ‘<<j<<endl;
cnt+=tm;
}
}
cout<<min(cnt,n*n-cnt)<<endl;
return 0;
}