Question

我在填写矩阵时遇到问题，如下图所示。矩阵必须是NxN，N是偶数。

我认为最好的方法是在for内使用两个while。像这样：

int a=0, c=1, i, j, flag=0, map[N][N] //N with a #define

//First I input my matrix with 0.

for (i=0; i<N; i++)
    for (j=0; j<N; j++)
        map[i][j]=0;

//Here I fill the matrix
while(!flag){

    for(j=a; j<N-a; j++){
        map[i][j]=c;
        c++;
    }
    for(i=a+1; i<N-a; i++){
        map[i][j]=c;
        c++;
    }
    for(j=N-a-1; j>=a; j--){
        map[i][j]=c;
        c++;
    }
    for(i=N-a-1; i>a+1; i--){
        if(map[i][j]==0){
            map[i][j]=c;
            c++;
        }
        else if(map[i][j]!=0)
            flag=1;
    }
c=1;
a++;
}

问题是id不起作用......你能帮助我吗？感谢

Answer 1

如果不修改大部分代码，最简单的方法如下：

 #define N 4
 #include<stdio.h>

  int a=0, c=1, i, j, flag=0, map[N][N]; //N with a #define


    main() {

   //First I input my matrix with 0.
    for (i=0; i<N; i++)
        for (j=0; j<N; j++)
            map[i][j]=0;

    i=0;
    //Here I fill the matrix
    while(!flag){

        for(j=a; j<N-a; j++){
            map[i][j]=c;
            c++;
        }
        j--;
        for(i=a+1; i<N-a-1; i++){
            map[i][j]=c;
            c++;
        }
        for(j=N-a-1; j>=a; j--){
            map[i][j]=c;
            c++;
        }
        j++;
        for(i=N-a-1; i>=a+1; i--){
            if(map[i][j]==0){
                map[i][j]=c;
                c++;
            }
        }
        if(map[i+1][j+1]!=0)
            flag=1;

        c=1;
        a++;
        i++;
    }
    for(i=0;i<N;i++){
        for(j=0;j<N;j++){
            printf("%d ",map[i][j]);
        }
        printf("\n");
    }
    }

此处N设置为4。所以输出如下：

Answer 2

您有两类问题：

在while循环中的4个for循环中的每个循环之前，您没有正确重置i或j。因此，在尝试填充矩阵时，您通常会超出界限。
您的flag永远不会被设置为1，因为填充矩阵后它的for循环将永远不会运行（通过纸上的代码跟踪以确切了解原因）。现在，使用for-loop边界，实际上并不需要flag来跟踪填充矩阵的时间。

考虑到这些点，您修改的代码可能如下所示：

int a=0, c=1, i, j, map[N][N] //N with a #define

// (no need to fill your matrix with zeros; your matrix will be filled when your while-loop below ends)

//Here I fill the matrix
while(a < N/2){ // <- stop filling the matrix once a == N/2

    i = a; // <- set i, otherwise the for-loop below will use i=undefined on the first iteration (or i=N if you keep the zeroing code above)

    for(j=a; j<N-a; j++){
        map[i][j]=c;
        c++;
    }

    j = N-a-1; // <- set j, otherwise the for-loop below will use j=N-a, which is out-of-bounds on the first iteration

    for(i=a+1; i<N-a; i++){
        map[i][j]=c;
        c++;
    }

    i = N-a-1; // <- set i, otherwise the for-loop below will use i=N-a, which, again, is out-of-bounds on the first iteration

    for(j=N-a-2; j>=a; j--){ // <- set j=N-a-2 instead of j=N-a-1 to avoid overwriting existing elements of the map
        map[i][j]=c;
        c++;
    }

    j = a; // <- set j, otherwise the for-loop below will use j=a-1, which is out-of-bounds on the first iteration

    for(i=N-a-2; i>a+1; i--){ // <- set i=N-a-2 instead of i=N-a-1 to avoid overwriting existing elements of the map
        // (no need to check if an element is 0)            

        map[i][j]=c;
        c++;
    }
c=1;
a++;
}

如何输入矩阵的边框？

2 个答案: