Question

问题是：检查两个矩阵，如果一个是另一个矩阵的子矩阵。

我在这里遇到的问题是for循环，在代码中注释为“// problem”。当程序第一次运行时，提到的for循环不能正常工作。

#include <stdio.h>
#define N 10

int
main ()
{

  char matrix1[N][N], matrix2[N][N];
  int i, j, row1, col1, row2, col2, k, l, m, n, check = 0;

  //First Matrix
  printf ("Enter the data\n");
  printf ("Enter the size of rows\n");
  scanf ("%d", &row1);   
  printf ("Enter the size of columns\n");
  scanf ("%d", &col1);   
  printf ("Now enter the values please\n");

  //Putting Values In First Matrix
  for (i = 0; i < row1; i++)
    {
      for (j = 0; j < col1; j++)
        {
          printf ("Please enter the %dth row and %dth column\n", i + 1,
                  j + 1);
          scanf ("%s", &matrix1[i][j]);
        }
    }

  //Second Matrix
  printf ("Enter the data\n");
  printf ("Enter the size of rows\n");
  scanf ("%d", &row2);   
  printf ("Enter the size of columns\n");
  scanf ("%d", &col2);   
  printf ("Now enter the values please\n");

  //Putting Values In Second Matrix
  for (i = 0; i < row2; i++)
    {
      for (j = 0; j < col2; j++)
        {
          printf ("Please enter the %dth row and %dth column\n", i + 1,
                  j + 1);
          scanf ("%s", &matrix2[i][j]);
        }
    }

  //Checking Both Matrices
  for (i = 0; i < row1; i++)
    {
      for (j = 0; j < col1; j++)
        {
          if (matrix1[i][j] == matrix2[0][0])
            {
              k = i;
              l = j;
              for (m = 0; m < row2; m++)
                {
                  for (n = 0; n < col2; n++)
                    {           //problem
                      if (matrix1[k][l] == matrix2[m][n])
                        {
                          check++;
                          printf ("Checked\n");
                        }
                      l++;
                    }
                  l = j; 
                  k++;   
                }
            }
        }
      printf ("hello\n");
    }
  if (check == row2 * col2)
    {
      printf ("It exists\n");
    }
  else
    {
      printf ("It doesn't exist\n");
    }
}

这是输出：

 Checked
 hello
 Checked
 Checked
 Checked
 Checked
 hello
 hello
 It doesn't exist

Answer 1

在开始查找子矩阵之前，您需要将check重置为零。一旦你找到它就打破（或者用旗子来表明它是否被发现）。

从你的输出中，（假设你试图找到2x2矩阵）它连续发现它Checked打印的次数，但它的值也是第一次打印的5，这使得你的程序可以打印{ {1}}。

像：

"It does not exist"

'for'循环在多维数组中失败

1 个答案: