题解

#include <iostream>
#include <cstdio>
using namespace std;
string a;
int x_, y_;
int curx, cury;
int main(){
	cin >> a;
	int i;
	for(i = 0; a[i] != '\0'; i++){//预处理单个命令串
		if(a[i] == 'E') x_++;
		else if(a[i] == 'S') y_--;
		else if(a[i] == 'W') x_--;
		else y_++;
	}int t, len = i;
	cin >> t;
	curx = t / len * x_, cury = t / len * y_;//整体加
	t %= len, i = 0;
	while(t--){//处理剩下的操作
		if(a[i] == 'E') curx++;
		else if(a[i] == 'S') cury--;
		else if(a[i] == 'W') curx--;
		else cury++;
		i++;
	}cout << curx << ' ' << cury;

	return 0;
}

时间复杂度：$O(|S|)$