AC题解，有注释啊啊啊！！

#include<bits/stdc++.h>
using namespace std;
string str;
int dir[4][4] = {{0,0,90,-90},{0,0,-90,90},{-90,90,0,0},{90,-90,0,0}}; // 存储方向对应的角度变化

int main(){
    int n;
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> str;
        str = str + str[0]; // 将字符串首尾相接，以确保围栏闭合
        int len = str.size();
        int ang = 0;
        for (int i = 0; i <= len - 2; i++){
            int x, y;
            // 根据围栏方向字符确定对应的行和列
            if (str[i] == 'E')x = 0;
            if (str[i + 1] == 'E')y = 0; 
            if (str[i] == 'W')x = 1;
            if (str[i + 1] == 'W')y = 1; 
            if (str[i] == 'S')x = 2;
            if (str[i + 1] == 'S')y = 2; 
            if (str[i] == 'N')x = 3;
            if (str[i + 1] == 'N')y = 3; 
            // 累加角度变化
            if (dir[x][y]){
                ang += dir[x][y];
            }
        }
        // 判断角度总和，输出结果
        if (ang == -360)cout << "CCW" << endl;
        else if(ang == 360)cout << "CW" << endl;
    } 
    return 0;
}