机器人路径模拟器[关闭]

Question

我对编程还是很陌生，而且由于我们最近开始使用 C，因此我们的任务是在假设的无限网格上创建一个程序（机器人模拟器），当然只使用三个可能的动作：右转,左转前进。输出只需 x 和 y 坐标 (x,y)。现在的问题是，我只能通过所有四个动作（上，下，左，右 - 北，南，西，东）来理解并做到这一点（假设没有左转或右转选项，左意味着移动 1位置向左，向右意味着向右移动 1 个位置等））。所以只是它们之间的一个简单的减法给了我最终的坐标，但我根本不知道如何为那些有 2 个转弯选项和前进的人制定规则。我会把我的代码留给我已经完成的任务如果有人认为我可以升级那个或者对如何使它工作有任何建议或想法，请告诉我。

#include<stdio.h> 
static void position(int n, char* string)
{
    int left = 0, right = 0, up = 0, down = 0;
    for (int i = 0; i < n; i++) {
        if (string[i] == 'L')
            left++;
        else if (string[i] == 'R')
            right++;
        else if (string[i] == 'U')
            up++;
        else if (string[i] == 'D')
            down++;
    }
    printf("(%d,%d)", (right - left), (up - down));
}
int main()
{
    printf("Enter your movement: ");
    char string[1000];
    scanf("%s", string);
    int n;
    for (n = 0; string[n] != '\0'; ++n);
    string[100] = string[n];
    position(n, string);
}

Answer 1

解决此问题的一种方法是使用笛卡尔坐标系。

我没有在最后尝试计算所有内容，而是在处理过程中跟踪它们。

#include<stdio.h> 

enum {NORTH/*UP*/,SOUTH/*DOWN*/, EAST/*LEFT*/, WEST/*RIGHT*/};
static void position(int n, char* string)
{                           // let's use a Cartesian coordinate system
    // assume a 10x10 grid
    int X=0,Y=0;            // start at this place within the grid (the center)
    int direction = NORTH;  // we are facing NORTH
    int i;
    for (i = 0; i < n; i++) {
        if (string[i] == 'L') // TURN LEFT, no step
        {   // pivot
            switch (direction)
            {
            case NORTH: 
                direction = EAST;   // face EAST, but do not step
                break;
            case SOUTH: 
                direction = WEST;   // face WEST, but do not step
                break;
            case EAST: 
                direction = SOUTH;  // face SOUTH, but do not step
                break;
            case WEST: 
                direction = NORTH;  // face NORTH, but do not step
                break;
            }
        }
        else if (string[i] == 'R') // TURN RIGHT, no step
        {   // pivot
            switch (direction)
            {
            case NORTH: 
                direction = WEST;   // face WEST, but do not step
                break;
            case SOUTH: 
                direction = EAST;   // face EAST, but do not step
                break;
            case EAST: 
                direction = NORTH;  // face NORTH, but do not step
                break;
            case WEST: 
                direction = SOUTH;  // face SOUTH, but do not step
                break;
            }
        }
        else if (string[i] == 'A')
        {   // advance
            switch (direction)
            {
            case NORTH: 
                if (Y == 5)
                    printf("you cannot go NORTH any more.\n");
                else
                    Y++;
                break;
            case SOUTH: 
                if (Y == -5)
                    printf("you cannot go SOUTH any more.\n");
                else
                    Y--;
                break;
            case EAST: 
                if (X == -5)
                    printf("you cannot go EAST any more.\n");
                else
                    X--;
                break;
            case WEST: 
                if (X == 5)
                    printf("you cannot go WEST any more.\n");
                else
                    X++;
                break;
            }
        }
        else
        {
            printf("%c is not a valid command.\n", string[i]);
        }
    }
    printf("(%d,%d)", X, Y);
}
int main()
{
    char string[1000];
    int n;
    for(;;)
    {   // only valid movements are turn Right,turn Left and Advance. (RLA)
        printf("Enter your movement: ");
        scanf("%s", string);
        for (n = 0; string[n] != '\0'; ++n);
        string[100] = string[n];
        position(n, string);
    }
}

示例 I/O

输入你的动作：@

@ 不是有效的命令。

(0,0)

输入你的动作：LLLA

(1,0)

输入你的动作：LLLA

(1,0)

输入你的动作：LLAAA

(3,0)

输入你的动作：LLARRRAA

(2,-1)

输入你的动作：LRRLRA

(0,1)