算法1:(暴力枚举) $O(n)$
//字符串操作 数字操作都可以
#include <bits/stdc++.h>
using namespace std;
int main() {
string s, last;
while (cin >> s, s != "99999") {
int x = s[0]-'0' + s[1]-'0';
if (x % 2 == 1) {
cout << "left ";
last = "left "; //记录上一次的操作
} else if (x%2 == 0 && x!=0) {
cout << "right ";
last = "right ";
}else {
cout << last; //输出上一次的操作
}
cout << s.substr(2) << endl;
}
return 0;
}