下面是一个5阶的螺旋方阵，具体情况如下：

Question

试编程打印出此形式的n(n<10)阶的方阵(顺时针方向旋进).1 2 3 4 516 17 18 19 615 24 25 20 714 23 22 21 813 12 11 10 9

Answer 1

// It is so easy. Just use five FOR sentences with high efficiency and simplest.
void fun(int **p, unsigned char n) // create a screw matrix with n
{ // return by pointer p
unsigned char k, i, l = (n+1)/2;
int c = 1;
p[l-1][l-1] = n * n;
for(k=0; k<l; k++, n-=2)
{
for(i=0;i<n-1;i++) p[k][k+i]=c++;
for(i=0;i<n-1;i++) p[k+i][k+n-1]=c++;
for(i=n-1;i>0;i--) p[k+n-1][k+i]=c++;
for(i=n-1;i>0;i--) p[k+i][k]=c++;
}
} // end of fun
// You can write a main function to test it, which is so simple even without using one IF sentence.

Answer 2

#include <stdio.h>
//Up,Down,Left,Right
#define U 'U'
#define D 'D'
#define L 'L'
#define R 'R'

const int MAX = 10;

int main(int argc, char *argv[]){
    int matrix[MAX][MAX];
    int n, x, y, k;
    char direction; //Up,Down,Left,Right

    // get input
    printf("Please input n(n<%d): ", MAX+1);
    scanf("%d", &n);

    // invalid input
    if(n>MAX || n<1) return -1;

    // initializations
    for (y=0;y<n;y++)
        for (x=0;x<n;x++)
            matrix[x][y]=0;
    for (y=n;y<MAX;y++)
        for (x=n;x<MAX;x++)
            matrix[x][y]=-1;

    x=y=k=0;
    direction = R;

    // build the matrix
    while(1){
        if (matrix[x][y]==0){
            matrix[x][y]=++k;
            switch (direction){
                case U:y--;break;
                case D:y++;break;
                case L:x--;break;
                case R:x++;break;
                default:break;
            }
        }else{
            switch (direction){
                case U:direction=R;y++;x++;break;
                case D:direction=L;y--;x--;break;
                case L:direction=U;y--;x++;break;
                case R:direction=D;y++;x--;break;
                default:break;
            }
        }

        // finished
        if (k==n*n) break;
    }

    // output
    for (y=0;y<n;y++){
        for (x=0;x<n;x++)
            printf("%3d", matrix[x][y]);
        printf("\n");
    }

    // exit with no error
    return 0;
}
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