旋转矩阵

这类题关键在于寻找下标之间的变换关系，在纸上写两个找一下规律比较合适
旋转矩阵是不靠谱的，嗯

  1. #include <time.h>
  2. #include <iostream>
  3. #include <cstring>
  5. using namespace std;
  7. const int N = 110;
  9. int a[N][N], tmp[N][N];
  10. int n, m, k;
  12. void rotate1() {    // 方阵顺时针旋转90°
  13.     for (int i = 1; i <= n; i++) {
  14.         for (int j = 1; j <= m; j++) {
  15.             tmp[j][n - i + 1] = a[i][j];
  16.         }
  17.     }
  19.     memcpy(a, tmp, sizeof tmp);
  20.     swap(m, n);    // 注意交换m, n
  21. }
  23. void rotate2() {    // 方阵逆时针旋转90°
  24.     for (int i = 1; i <= n; i++) {
  25.         for (int j = 1; j <= m; j++) {
  26.             tmp[m - j + 1][i] = a[i][j];
  27.         }
  28.     }
  30.     memcpy(a, tmp, sizeof tmp);
  31.     swap(m, n);    // 注意交换m, n
  32. }
  34. void mirror() {    // 将方阵沿纵向对称轴翻折
  35.     for (int i = 1; i <= n; i++) {
  36.         for (int j = 1; j <= m; j++) {
  37.             tmp[i][m - j + 1] = a[i][j];
  38.         }
  39.     }
  41.     memcpy(a, tmp, sizeof tmp);
  42. }
  44. int main() {
  45.     #ifdef SUBMIT
  46.     freopen("in.txt", "r", stdin);
  47.     freopen("out.txt", "w", stdout);
  48.     long _begin_time = clock();
  49.     #endif
  51.     while (cin >> n >> m >> k) {
  52.         for (int i = 1; i <= n; i++)
  53.             for (int j = 1; j <= m; j++)
  54.                 scanf("%d", &a[i][j]);
  56.         while (k--) {
  57.             int op;
  58.             cin >> op;
  60.             if (op == 1)
  61.                 rotate1();
  62.             else if (op == 2)
  63.                 mirror();
  64.             else
  65.                 rotate2();
  66.         }
  68.         for (int i = 1; i <= n; i++) {
  69.             for (int j = 1; j <= m; j++)
  70.                 printf("%d ", a[i][j]);
  71.             putchar('\n');
  72.         }
  73.     }
  75.     #ifdef SUBMIT
  76.     long _end_time = clock();
  77.     printf("\n\ntime = %ld ms", _end_time - _begin_time);
  78.     #endif
  80.     return 0;
  81. }

类似题：旋转矩阵 - 北航

旋转方阵

大一时候觉得很难的题…关键是想好每一圈怎么填比较合适，这种限制边界，填充的数做全局变量的做法还不错

  1. #include <time.h>
  2. #include <iostream>
  4. using namespace std;
  6. const int N = 25;
  8. int ans[N][N];
  9. int cnt = 1;
  11. void fill(int a, int b) {
  12.     for (int i = a; i < b; i++) {
  13.         ans[i][a] = cnt++;
  14.     }
  15.     for (int j = a; j < b; j++) {
  16.         ans[b][j] = cnt++;
  17.     }
  18.     for (int i = b; i > a; i--) {
  19.         ans[i][b] = cnt++;
  20.     }
  21.     for (int j = b; j > a; j--) {
  22.         ans[a][j] = cnt++;
  23.     }
  24. }
  26. int main() {
  27.     #ifdef SUBMIT
  28.     freopen("in.txt", "r", stdin);
  29.     freopen("out.txt", "w", stdout);
  30.     long _begin_time = clock();
  31.     #endif
  33.     int n;
  34.     cin >> n;
  36.     int i, j;
  37.     for (i = 1, j = n; i < j; i++, j--)
  38.         fill(i, j);
  39.     if (i == j)
  40.         ans[i][j] = cnt;
  42.     for (int i = 1; i <= n; i++) {
  43.         for (int j = 1; j <= n; j++)
  44.             printf("%-4d", ans[i][j]);
  45.         putchar('\n');
  46.     }
  49.     #ifdef SUBMIT
  50.     long _end_time = clock();
  51.     printf("\n\ntime = %ld ms", _end_time - _begin_time);
  52.     #endif
  54.     return 0;
  55. }