笨方法求解
迷路的牛牛
http://www.nowcoder.com/questionTerminal/fc72d3493d7e4be883e931d507352a4a
我用的是笨办法，先遍历字符串，统计出向左和向右的次数，如果两次相同的话，则仍是面向北。如果向左或者向右次数多的话，则多出的部分对4求余就可以了。
#include <bits/stdc++.h>
using namespace std;
int main(){
int n;
string s;
int count=0;
int sount=0;
char left[] = {"NWSE"};
char right[] = {"NESW"};
int a=0;
int b=0;
cin >> n;
cin >> s;
for(int i=0; i<n; i++){
if(s[i]=='L'){
count++;
}
else if(s[i]=='R'){
sount++;
}
}
if(count==sount){
cout << 'N' << endl;
}
else if(count>sount){
int ccount = count-sount;
a = ccount%4;
cout << left[a] << endl;
}
else if(count < sount){
int ssount = sount-count;
b = ssount%4;
cout << right[b] << endl;
}
} 
