题解 | #坐标移动#

思路：暴力解法，使用strtok分割字符串，然后判断每个字符串是否有效，有效就进行移动，无效则break；

#include<stdio.h>


int main()
{
    char arr[10001] = {0};
    gets(arr);
    int x = 0;
    int y = 0;
    char* par = strtok(arr,";");
    while(par!=NULL)
    {
        switch(par[0])
        {
            case 'A':
                if(par[1]>='0'&&par[1]<='9' && par[2]>='0'&&par[2]<='9')
                {
                    int count = (par[1]-'0')*10+(par[2]-'0');
                    x-=count;
                }
                else if(par[1]>='0'&&par[1]<='9' && par[2]=='\0')
                {
                    int count = (par[1]-'0');
                    x-=count;
                }
                break;
            case 'D':
                if(par[1]>='0'&&par[1]<='9' && par[2]>='0'&&par[2]<='9')
                {
                    int count = (par[1]-'0')*10+(par[2]-'0');
                    x+=count;
                }
                else if(par[1]>='0'&&par[1]<='9' && par[2]=='\0')
                {
                    int count = (par[1]-'0');
                    x+=count;
                }
                break;
            case 'W':
                if(par[1]>='0'&&par[1]<='9' && par[2]>='0'&&par[2]<='9')
                {
                    int count = (par[1]-'0')*10+(par[2]-'0');
                    y+=count;
                }
                else if(par[1]>='0'&&par[1]<='9' && par[2]=='\0')
                {
                    int count = (par[1]-'0');
                    y+=count;
                }
                break;
            case 'S':
                if(par[1]>='0'&&par[1]<='9' && par[2]>='0'&&par[2]<='9')
                {
                    int count = (par[1]-'0')*10+(par[2]-'0');
                    y-=count;
                }
                else if(par[1]>='0'&&par[1]<='9' && par[2]=='\0')
                {
                    int count = (par[1]-'0');
                    y-=count;
                }
                break;
            default:
                break;
        }
        par = strtok(NULL,";");
    }
    printf("%d,%d\n",x,y);
    return 0;
}