Initial commit: C language learning code

2025-07-20 16:30:56 +08:00
commit 06e24173a6
139 changed files with 9303 additions and 0 deletions
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+#include <stdio.h>
+#include <stdlib.h>
+#include <time.h>
+#include <math.h>
+#include <stdbool.h>
+
+// !二维数组
+/*
+    int a[3][5];
+    通常理解为a是一个3行5列的矩阵
+*/
+
+// !二维数组的遍历
+/*
+    for ( i=0; i<3; i++)
+    {
+        for ( j=0; j<5; j++ )
+        {
+            a[i][j] = i*j;
+        }
+    }
+    a[i][i]是一个int
+    表示第i行第j列上的单元
+    但a[i,j]是数学上表示a是一个i行j列的矩阵，在C语言是a[i][j]
+*/
+
+// !二维数组的初始化
+/*
+    nt a[] [5] = 
+    { 
+        {0,1,2,3,4}, 
+        {2,3,4,5,6
+    };
+    列数是必须给出的，行数可以由编译器来数每行一个，逗号分隔
+    最后的逗号可以存在，有古老的传统
+    如果省略，表示补零
+    也可以用定位(*C99 标准)
+*/
+
+int check_symmetry(int matrix[][200], int size);
+
+int main()
+{
+    int n;
+    scanf("%d", &n);
+    int matrix1[100][100], matrix2[100][100], matrix3[100][100], matrix4[100][100];
+
+    // 读取四个n×n的矩阵
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < n; j++)
+        {
+            scanf("%d", &matrix1[i][j]);
+        }
+    }
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < n; j++)
+        {
+            scanf("%d", &matrix2[i][j]);
+        }
+    }
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < n; j++)
+        {
+            scanf("%d", &matrix3[i][j]);
+        }
+    }
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < n; j++)
+        {
+            scanf("%d", &matrix4[i][j]);
+        }
+    }
+
+    int big_matrix[200][200];
+    // 方案一：常规左上、右上、左下、右下拼接
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < n; j++)
+        {
+            big_matrix[i][j] = matrix1[i][j];
+            big_matrix[i][j + n] = matrix2[i][j];
+            big_matrix[i + n][j] = matrix3[i][j];
+            big_matrix[i + n][j + n] = matrix4[i][j];
+        }
+    }
+    if (check_symmetry(big_matrix, 2 * n))
+    {
+        printf("YES\n");
+        return 0;
+    }
+
+    // 方案二：交换右上和左下
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < n; j++)
+        {
+            big_matrix[i][j] = matrix1[i][j];
+            big_matrix[i][j + n] = matrix3[i][j];
+            big_matrix[i + n][j] = matrix2[i][j];
+            big_matrix[i + n][j + n] = matrix4[i][j];
+        }
+    }
+    if (check_symmetry(big_matrix, 2 * n))
+    {
+        printf("YES\n");
+        return 0;
+    }
+
+    // 方案三：交换左上和右下
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < n; j++)
+        {
+            big_matrix[i][j] = matrix4[i][j];
+            big_matrix[i][j + n] = matrix2[i][j];
+            big_matrix[i + n][j] = matrix3[i][j];
+            big_matrix[i + n][j + n] = matrix1[i][j];
+        }
+    }
+    if (check_symmetry(big_matrix, 2 * n))
+    {
+        printf("YES\n");
+        return 0;
+    }
+
+    // 方案四：交换左上和右上，左下和右下
+    for (int i = 0; i < n; i++)
+    {
+        for (int j = 0; j < n; j++)
+        {
+            big_matrix[i][j] = matrix2[i][j];
+            big_matrix[i][j + n] = matrix1[i][j];
+            big_matrix[i + n][j] = matrix4[i][j];
+            big_matrix[i + n][j + n] = matrix3[i][j];
+        }
+    }
+    if (check_symmetry(big_matrix, 2 * n))
+    {
+        printf("YES\n");
+        return 0;
+    }
+
+    printf("NO\n");
+    return 0;
+}
+
+// !检查矩阵是否为对称矩阵
+int check_symmetry(int matrix[][200], int size)
+{
+    for (int i = 0; i < size; i++)
+    {
+        for (int j = 0; j < size; j++)
+        {
+            if (matrix[i][j] != matrix[j][i])
+            {
+                return 0;
+            }
+        }
+    }
+    return 1;
+}