京东2016笔试题

题目记得不太清了，大概意思是：

小东和小C两个人在棋盘（如下：8×8）上，相互指定位置，并打印小东到小C需要走的最短步数，以及走过的最短路径，走的每一步的方向可以为：

上（U），下（D），左（L），右（R），左上（LU），左下（LD），右上（RU），右下（RD）

也就是说：当前的位置可以走到跟它相邻(有公共的点或者边)的任意一个位置

小东可以任意选择方向，一个方向就相当与一步（也就是说：走右下（RD）跟走右（R），是一样的，都是一步）



小东到达小C比较喜欢走斜角，因为他觉得走斜路将更容易到达目的地

示例：

小C为小东指定：a8                  小东为小C指定：h1

那么输入输出就是：

a8 h1        （输入）

7               （输出）

RD

RD

RD

RD

RD

RD

RD

分析：

1，因为是8×8的棋盘，所以考虑用二维数组

2，输入的是：a8，h1，b3等这一类字符，我们必须将其转化到相应的数组中去（如：将a8，转换为数组位置（0，7），将h1，转换为数组位置（7，0））

3，根据小东和小C的相应位置，调用不同的算法

      1，先考虑两个位置是不是相等，如果相等，直接退出;

      2，判断两者位置是不是在同一列或者同一列，如果是，调用相应的函数，运行完毕，退出;

      3，如果不在同一行或者同一列的话，那么判断大概的位置，左上，左下，右上，右下，调用一次相应的函数，然后继续执行1，2，3;

程序：

#include <stdio.h>

int graph[8][8]= {0};
int flag = 0;     //标志着，只打印一次总步数
int step,step1,step2;

void col(int x , int y , int x1, int y1)
{

if(x == y)   //同一列
{
if(x1 > y1) //小东在上，小C在下
{

step = x1-y1;
if(flag == 0)
printf("%d\n", step);
while(step --)
printf("D\n");
}
if(x1 < y1)
{
step = y1 - x1;
if(flag == 0)
printf("%d\n", step);
while(step--)
printf("U\n");
}
}
}

void hang(int x, int y, int x1, int y1)
{
                        if(x1 == y1)  //同一行
                        {
                                if(x < y)
                                {
                                        step = y-x;
                                        if(flag == 0)
                                                printf("%d\n", step);
                                        while(step --)
                                        {
                                                printf("R\n");  
                                        }
                                }
                                if(x > y)
                                {
                                        step = x - y;
                                        if(flag == 0)
                                                printf("%d\n", step);
                                        while(step --)
                                        {
                                                printf("L\n");  
                                        }
                                }
                        }
}

int main()
{
        int x,y,x1,y1;
        char a[10], b[10];
        scanf("%s %s", a, b);
        
                
        x = a[0] - 'a';
        x1 = a[1] - '1';
        y = b[0] - 'a';
        y1 = b[1] - '1';                      //(x,x1)为小东的坐标   (y,y1)为小C的坐标

        if(x>= 0 && x <8 && x1>= 0 && x1 < 8) //合理的小东坐标
        {
                if(y>= 0 && y< 8 && y1>= 0 && y1<8)//合理的小C坐标
                {
                        if(x == y && x1 ==y1 )
                                return 0;      //如果两者位置相同直接退出
                        col(x, y, x1, y1);     //判断在不在同一列
                        hang(x, y, x1, y1);    //判断在不在同一行
                        
                        if(x < y && x1 > y1)   //右下角的情况（即就是小c位于小东的右下角）
                        {
                                step = y- x;                                                                                                    
                                step1 = x1 - y1;
                                if(step >= step1)
                                        step2 = step;
                                else
                                        step2 = step1;
                                printf("%d\n", step2);
                                flag = 1;
                                while(x != y && x1 != y1)
                                {
                                        x++;
                                        x1--;
                                        printf("RD\n");
                                        
                                        col(x, y, x1, y1);
                                        hang(x , y,x1, y1);

                             }
                        }
                        else if(x > y && x1 < y1)   //左上角
                        {       
                                flag = 1;
                                step = x - y;
                                step1 = y1 - x1;
                                if(step >= step1)
                                        step2 = step;
                                else
                                        step2 = step1;
                                printf("%d\n", step2);
                                while(x != y && x1 != y1)
                                {
                                        x--;
                                        x1++;
                                        printf("LU\n");
                                        
                                        col(x,y, x1,y1);        
                                        hang(x,y, x1, y1);

                                }
                        }

                      else if(x < y && x1 < y1)   //右上角的情况
                        {
                                flag = 1;
                                step = y - x;
                                step1 = y1 - x1;
                                if(step >= step1)
                                        step2 = step;
                                else
                                        step2 = step1;
                                printf("%d\n", step2);
                                while(x != y && x1 != y1)
                                {
                                        x++;
                                        x1++;
                                        printf("RU\n");
                                        
                                        col(x,y,x1, y1);
                                        hang(x,y, x1, y1);      
                                }
                        }

                        else if(x > y && x1 > y1)   //左下角的情况
                        {
                                flag = 1;
                                step = x - y;
                                step1 = x1 - y1;
                                if(step >= step1)
                                        step2 = step;
                                else
                                        step2 = step1;
                                printf("%d\n", step2);
                                while(x != y && x1 != y1)
                                {
                                        x--;
                                        x1--;
                                        printf("LD\n");
                                        
                                        col(x,y,x1, y1);
                                        hang(x,y,x1, y1);       
                                }
                        }

                }
        }
        return 0;

}                                             
 

其实合理的话，也应该将所有斜角的情况写到函数调用或者递归里面去的

好多同学输入代码后，自己运行对着，但是放到指定的编译器中（赛码）去就报错

那么应该就是，我们大家写的代码不通用，只满足某些特定的例子，也就是说：我们的程序尽可能的完善一点！！！