c语言简单编程

Question

Input array size N (1~20)
Using Two-Dimensional array
输入数组大小氮（1 - 20）
二维数组

Answer 1

楼上站着说话不腰疼。
这道题只给思路，但是你那个最关键的思路就没给。
我以为自己十分钟可以搞定这题，结果弄了一个小时。。。
给出经过测试的C语言源码如下：

 #include "stdio.h"
#include "stdlib.h"
#define NUM 5

void right_add(int a[NUM][NUM],int line,int row,int count); //Note that it can't be written as a[][] here(with an error called "Without subscript")
void down_add(int a[NUM][NUM],int line,int row,int count);
void left_add(int a[NUM][NUM],int line,int row,int count);
void up_add(int a[NUM][NUM],int line,int row,int count);
void print(int a[NUM][NUM],int n);


int main(int argc, char* argv[])
{
    int i,j;
    int result[NUM][NUM];
    for(i = 0;i < NUM;i++)
        for(j = 0;j < NUM;j++)
            result[i][j] = 0;

    int num = NUM * NUM;
    right_add(result,0,0,1);   //The first move is always towards right

    return 0;
}

void right_add(int a[NUM][NUM],int line,int row,int count) {  //Move to right position until there are limits
    while(a[line][row] == 0) {
        a[line][row] = count;
        count++;
        if (count > NUM*NUM) {
            print(a,NUM);
            exit(0);
        }
        row++;

        if(row % NUM == 0 || a[line][row] != 0) { //When it reaches the border or its right is a valid number,goto down_add
            row--;
            line++;
            down_add(a,line,row,count);
        }
    }
}

void down_add(int a[NUM][NUM],int line,int row,int count) { //Move to downside position until there are limits
    while(a[line][row] == 0) {
        a[line][row] = count;
        count++;
        if (count > NUM*NUM) {
            print(a,NUM);
            exit(0);
        }
        line++;

        if(line % NUM == 0 || a[line][row] != 0) { //When it reaches the border or its downside is a valid number,goto left_add
            line--;
            row--;
            left_add(a,line,row,count);
        }
    }
}

void left_add(int a[NUM][NUM],int line,int row,int count) { //Move to left position until there are limits
    while(a[line][row] == 0) {
        a[line][row] = count;
        count++;
        if (count > NUM*NUM) {
            print(a,NUM);
            exit(0);
        }
        row--;

        if(row == -1 || a[line][row] != 0) { //When it reaches the border or its left is a valid number,goto up_add
            row++;
            line--;
            up_add(a,line,row,count);
        }
    }
}

void up_add(int a[NUM][NUM],int line,int row,int count) { //Move to upside position until there are limits
    while(a[line][row] == 0) {
        a[line][row] = count;
        count++;
        if (count > NUM*NUM) {
            print(a,NUM);
            exit(0);
        }
        line--;

        if(line == -1 || a[line][row] != 0) { //When it reaches the border or its left is a valid number,goto up_add
            line++;
            row++;
            right_add(a,line,row,count);
        }
    }
}

void print(int a[NUM][NUM],int n) { //Print out the result we get
    int i,j;
    for(i = 0;i < n;i++) {
        for(j = 0;j < n;j++) {
            printf("%3d",a[i][j]);   //Formatting the output
            if ((j+1) % n == 0)      //goto next line
                printf("\n");
        }
    }
}

测试结果：



再放个大招：

附：当规模大了以后，print函数的格式化输出需要调整一下。
题主看看程序思路，如果哪里不理解欢迎继续交流，我打算把它写到我的博客里了~
当然，这个代码是可以优化的，我到时再用指针重写一遍~