206 lines
5.7 KiB
C++
206 lines
5.7 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 15;
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char a[N][N];//原始图案
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char c[N][N];//目标图案
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char t[N][N];//转换后的图案
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char t2[N][N];
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int n;
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//判断两个二维数组是不是一致
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bool equal() {
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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if (t[i][j] != c[i][j]) return false;
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return true;
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}
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int main() {
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cin >> n;
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//读入原始图案
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for (int i = 1; i <= n; ++i)
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for (int j = 1; j <= n; ++j)
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cin >> a[i][j];
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//读入目标图案
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for (int i = 1; i <= n; ++i)
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for (int j = 1; j <= n; ++j)
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cin >> c[i][j];
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//1、顺时针转90度
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//通过实际例子，找规律，得i,j--->j,n-i+1
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/**
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1 2 3 7 4 1
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4 5 6 --> 8 5 2
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7 8 9 9 6 3
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以数字1为例，原来的坐标(1,1),现在的坐标(1,3)
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以数字2为例，原来的坐标(1,2),现在的坐标(2,3)
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以数字3为例，原来的坐标(1,3),现在的坐标(3,3)
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成功发现列坐标变成了行坐标。那么行坐标呢?它是怎么变化的呢?
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以7，8，9为例，它们原来在不同的列，现在都是第1列，看来后期的
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列与行有关，当然也得与n有关，所以猜测试为n-i,结果呢，3-3=0,不对啊，
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再加个1试试，就是n-i+1,再验算一下，完全正确！
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*/
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[j][n - i + 1] = a[i][j];
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//如果通过，输出1，返回
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if (equal()) {
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cout << 1 << endl;
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return 0;
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}
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//2、顺时针转180度
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/**
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1 2 3 9 8 7
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4 5 6 --> 6 5 4
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7 8 9 3 2 1
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用纸画出来模拟一下
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找规律
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1,1--->n,n
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1,2--->n,n-1
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1,3--->n,n-2
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2,1--->n-1,n
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2,2--->n-1,n-1
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总结 ： i,j---> n-i+1,n-j+1
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*/
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[n - i + 1][n - j + 1] = a[i][j];
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//如果通过，输出2，返回
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if (equal()) {
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cout << 2 << endl;
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return 0;
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}
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//3、顺时针转270度
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//通过实际例子，找规律，得i,j---> n-j+1,i
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/**
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1 2 3 3 6 9
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4 5 6 --> 2 5 8
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7 8 9 1 4 7
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用纸画出来模拟一下
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找规律
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1,1--->n,1
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1,2--->n-1,1
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1,3--->n-2,1
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2,1--->n,2
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2,2--->n-1,2
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总结 ： i,j---> n-j+1,i
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*/
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[n - j + 1][i] = a[i][j];
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//如果通过，输出3，返回
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if (equal()) {
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cout << 3 << endl;
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return 0;
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}
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//4、图案在水平方向翻转
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//通过实际例子，
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/**
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1 2 3 3 2 1
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4 5 6 --> 6 5 4
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7 8 9 9 8 7
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用纸画出来模拟一下
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找规律
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1,1--->1,n
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1,2--->1,n-1
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1,3--->1,n-2
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2,1--->2,n
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2,2--->2,n-1
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总结 ： i,j---> i,n-j+1
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*/
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[i][n - j + 1] = a[i][j];
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//如果通过，输出4，返回
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if (equal()) {
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cout << 4 << endl;
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return 0;
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}
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//5、组合 图案在水平方向翻转，然后再按照 1 ∼3 之间的一种再次转换。
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//***********************************************************************************************
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//5.1 水平+顺时针90度
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//水平
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[i][n - j + 1] = a[i][j];
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//复制出来
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t2[i][j] = t[i][j];
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//顺时针90度
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[j][n - i + 1] = t2[i][j];
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//如果通过，输出5，返回
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if (equal()) {
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cout << 5 << endl;
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return 0;
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}
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//***********************************************************************************************
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//***********************************************************************************************
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//5.2 水平+180度
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[i][n - j + 1] = a[i][j];
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t2[i][j] = t[i][j];
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[n - i + 1][n - j + 1] = t2[i][j];
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//如果通过，输出5，返回
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if (equal()) {
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cout << 5 << endl;
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return 0;
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}
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//***********************************************************************************************
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//***********************************************************************************************
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//5.3 水平+270度
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[i][n - j + 1] = a[i][j];
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t2[i][j] = t[i][j];
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[n - j + 1][i] = t2[i][j];
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//如果通过，输出5，返回
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if (equal()) {
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cout << 5 << endl;
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return 0;
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}
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//***********************************************************************************************
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//6、不改变
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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t[i][j] = a[i][j];
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//如果通过，输出6，返回
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if (equal()) {
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cout << 6 << endl;
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return 0;
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}
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//7、无效转换
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cout << 7 << endl;
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return 0;
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} |