有关直角坐标系旋转(n*90°)(1 ≤ n ≤ 4)的坐标求解

 Problem - E - Codeforces

 设初始点为(x1,y1),易知y4 = x1,再通过与点(y1,x1)横坐标之间的关系(x4 + y1 = n + 1);

可知x4 = n + 1 - y1;

再由中心对称 x2 = n + 1 - x4; y2 = n + 1 - y4;

而(x3,y3)可由(x1,y1)中心对称可得: x3 = n + 1 - x1, y3 = n + 1 - y1;

    int y4 = x1;

    int x4 = y1;

    x4 = n + 1 - x4;

    int x2 = n + 1 - x4;

    int y2 = n + 1 - y4;

    int x3 = n + 1 - x1;

    int y3 = n + 1 - y1;

 以上是求解坐标的过程

本题要求4个旋转角度得到的都是同一个矩阵，那么就要求每一个1/4矩阵中的点都与他的相关旋转点相同，可以枚举1/4矩阵中的每一个点，通过check(i,j)返回需要改变的点的最小数量

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <map>
#include <set>
using namespace std;
#define mst(x, y) memset(x, y, sizeof x);
#define X first
#define Y second
#define int long long
#define FAST ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
const int N = 1010, INF = 0x3f3f3f3f, EPS = 1e-8;
typedef pair<int, int> PII;
int t, n;
char g[N][N];
int check(int x1, int y1)
{
    int y4 = x1;
    int x4 = y1;
    x4 = n + 1 - x4;

    int x2 = n + 1 - x4;
    int y2 = n + 1 - y4;

    int x3 = n + 1 - x1;
    int y3 = n + 1 - y1;

    int res = 0;
    if (g[x1][y1] == '1')
        res++;
    if (g[x2][y2] == '1')
        res++;
    if (g[x3][y3] == '1')
        res++;
    if (g[x4][y4] == '1')
        res++;
    return min(res, 4 - res);
}
signed main()
{
    FAST;
    cin >> t;
    while (t--)
    {
        cin >> n;
        for (int i = 1; i <= n; i++)
            cin >> g[i] + 1;

        int res = 0;
        for (int i = 1; i <= (n + 1) / 2 - (n % 2); i++)
        {
            for (int j = 1; j <= (n + 1) / 2; j++)
            {
                res += check(i, j);
            }
        }
        cout << res << endl;
    }

    return 0;
}