Question

我有一个C代码来生成对角线单位矩阵。代码如下：

#include<stdio.h>

void main()
{
    int n,m,a,b,c,d,i,j;
    int ar1[100][100];

    printf("Enter value of n to create a square matrix of order n (odd no.): ");
    scanf("%d", &n);

    printf(" matrix in diagonal direction is - \n");
    for (i=0; i<n; i++) {
        for (j=0; j<n; j++) {
            if (ar1[j]==ar1[i]) {
                ar1[j][i] = 2;
                printf("%d, ", ar1[j][i]);
            } else {
                ar1[j][i] = -1;
                printf("%d, ", ar1[j][i]);
            }
        }
        printf("\n");
    }
}

给我一个这样的结果：

 2, -1, -1, -1, -1, 
-1,  2, -1, -1, -1, 
-1, -1,  2, -1, -1, 
-1, -1, -1,  2, -1, 
-1, -1, -1, -1,  2

但我想在相反方向打印对角矩阵，如下所示：

-1, -1, -1, -1,  2, 
-1, -1, -1,  2, -1, 
-1, -1,  2, -1, -1, 
-1,  2, -1, -1, -1, 
 2, -1, -1, -1, -1

有人可以帮我解决一下如何在相反方向打印对角矩阵吗？我想改变实际矩阵而不仅仅是输出

我已经像这样更新了我的代码（如下所示），但我仍然收到错误（如下所示）

#include<stdio.h>

void main()
{
     int n,m,a,b,c,d,i,j,s;
         int ar1[100][100];

     printf("Enter value of n to create a square matrix of order n (odd no.): ");
     scanf("%d",&n);

         printf("Enter m: (odd no.) ");
     scanf("%d",&m );


         if(n>m)
          { 
     for(i=0;i<n;i++)
     {
           for(j=0;j<n;j++)
           {

                       for(a=1;a<=(m-1)/2;a++){
                       if(ar1[i-a]==ar1[j]){
                       ar1[i][j] = 2;
                       printf("%d, ",ar1[i][j]);
                      }
                      }

                      for(a=(m+1)/2;a<=(n-1);a++){
                       if(ar1[i-a]==ar1[j]){
                       ar1[i][j] = -1;
                       printf("%d, ",ar1[i][j]);
                      }
                      }

                      if(ar1[i]==ar1[j]){
                       ar1[i][j] = 2;
                       printf("%d, ",ar1[i][j]);
                      }

                     for(a=1;a<=(m-1)/2;a++){
                       if(ar1[i+a]==ar1[j]){
                       ar1[i][j] = 2;
                       printf("%d, ",ar1[i][j]);

                      }
                      }

        for(a=(m+1)/2;a<=(n-1);a++){
                       if(ar1[i+a]==ar1[j]){
                       ar1[i][j] = -1;
                       printf("%d, ",ar1[i][j]);
                      }
                      }
s = flip (ar1[100][100], n);

             }
           printf("\n");
     }

void flip(int a[100][100], int n) {
 int t;
 for (i=0; i<n; i++) {
     for (j=0; j<n/2; j++) {
         t=a[i][j];
         a[i][j] = a[i][n-1-j];
         a[i][n-1-j] = t;
       }
   }
  }
    }

}

我遇到了这样的错误

      /tmp/cc72v9hd.o: In function `main':
     faltu2.c:(.text+0x629): undefined reference to `flip'
     collect2: ld returned 1 exit status

Answer 1

只需撤消任何循环，即从n-1迭代应该这样做。下面是我的版本，

#include<stdio.h>

void main()
{
    int n,m,a,b,c,d,i,j;
    int ar1[100][100];

    printf("Enter value of n to create a square matrix of order n (odd no.): ");
    scanf("%d",&n);

    printf(" matrix in diagonal direction is - \n");

    for(i=0;i<n;i++)
    {
            for(j=n-1;j>=0;j--)
            {
                    //if(ar1[j]==ar1[i]){       // any two garbage values can be equal
                    if(j==i)
                    {
                            ar1[j][i] = 2;
                            printf("%d, ",ar1[j][i]);
                    }
                    else
                    {
                            ar1[j][i] = -1;
                            printf("%d, ",ar1[j][i]);
                    }

            }
            printf("\n");
    }
}

Answer 2

如果您只想更改输出而不是实际矩阵，只需替换此外循环：

for (i=0; i<n; i++) {
    ...

这一个：

for (i=n-1; i>=0; i--) {
    ...

<强>更新即可。如果你需要翻转矩阵，只需使用这样的函数或代码：

 void flip(int a[][], int n) {
     int i,j,t;
     for (i=0; i<n; i++) {
         for (j=0; j<n/2; i++) {
             t=a[i][j];
             a[i][j] = a[i][n-1-j];
             a[i][n-1-j] = t;
         }
     }
 }

C中相反方向的对角矩阵

2 个答案: