题解 | #坐标移动#

自己一开始想得复杂了，预期要靠考虑输入不对的情况，不如反过来仅考虑输入正确的情况。反过来考虑那么可能出现的case就更少了

还是太菜了，多看看吧！

#include <bits/stdc++.h>
using namespace std;

struct position {
    int x;
    int y;
    position(int _x, int _y): x(_x), y(_y) {}
};

void movePo(string s1) {
    position po = position(0,0);
    int lens1 = s1.size();
    //s2用来存储子串
    vector<string> s2;
    //统计子串长度
    int sublen = 0;
    for(int i = 0; i < lens1; i++) {
        if(s1[i] != ';') {
            sublen++;
            //结束本轮循环
            continue;
        }
        //获取子串，压入s2 vector向量组中
        s2.push_back(s1.substr(i-sublen,sublen));
        sublen = 0;
    }
    //先计算出某个方向移动的数值
    for(int i = 0; i < s2.size(); i++) {
        int posiNum = 0;
        if(s2[i].size() == 3 && s2[i][1] >= '0' && s2[i][1] <= '9' && s2[i][2] >= '0' && s2[i][2] <= '9') {
            posiNum = (s2[i][1] - '0') * 10 + (s2[i][2] - '0');
        }
        if(s2[i].size() == 2 && s2[i][1] >= '0' && s2[i][1] <= '9') {
            posiNum = s2[i][1] - '0';
        }
        if(s2[i].size() == 1) {
            posiNum = 0;
        }
    //确定移动的方向
    switch(s2[i][0]) {
        case 'A': po.x -= posiNum;
        break;
        case 'D': po.x += posiNum;
        break;
        case 'S': po.y -= posiNum;
        break;
        case 'W': po.y += posiNum;
        break;
        default:
            break;
        }

    }
    cout << po.x << "," << po.y <<endl;
}


int main() {
    string s;
    getline(cin, s);
    movePo(s);
    return 0;
}