There are n employees in Alternative Cake Manufacturing (ACM). They are now voting on some very important question and the leading world media are trying to predict the outcome of the vote.
Each of the employees belongs to one of two fractions: depublicans or remocrats, and these two fractions have opposite opinions on what should be the outcome of the vote. The voting procedure is rather complicated:
- Each of n employees makes a statement. They make statements one by one starting from employees 1 and finishing with employee n. If at the moment when it's time for the i-th employee to make a statement he no longer has the right to vote, he just skips his turn (and no longer takes part in this voting).
- When employee makes a statement, he can do nothing or declare that one of the other employees no longer has a right to vote. It's allowed to deny from voting people who already made the statement or people who are only waiting to do so. If someone is denied from voting he no longer participates in the voting till the very end.
- When all employees are done with their statements, the procedure repeats: again, each employees starting from 1 and finishing with n who are still eligible to vote make their statements.
- The process repeats until there is only one employee eligible to vote remaining and he determines the outcome of the whole voting. Of course, he votes for the decision suitable for his fraction.
You know the order employees are going to vote and that they behave optimal (and they also know the order and who belongs to which fraction). Predict the outcome of the vote.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of employees.
The next line contains n characters. The i-th character is 'D' if the i-th employee is from depublicans fraction or 'R' if he is from remocrats.
Output
Print 'D' if the outcome of the vote will be suitable for depublicans and 'R' if remocrats will win.
Examples
Input
5 DDRRR
Output
D
Input
6 DDRRRR
Output
R
题目分析
题目是说投票，但我觉得理解成打架会好一点
有两拨人打架，每个人手里都有一把枪但是只有一颗子弹，而且只能等所有人都开完枪才能装弹，
每次轮到你开枪的时候你可以打死一个人，因为每个人总是寻找着最优的解决方案，
那什么方案是最优的呐
首先我们肯定优先击杀没开过枪的，其次是排在前面的。
减小对方
对我方的杀伤，直到一方人全死光为止。
我们可以用两个队列存储两队人的位置，然后每一次双方各出队一位，位置在前面的开枪快把对方打死了
然后他去队尾等待装弹。
AC代码
#include<bits/stdc++.h>
using namespace std;
#define maxn 200005
char str[maxn];
int main()
{
int n;
queue<int> d;建立两个队列
queue<int> r;
while(cin>>n)
{
while(!d.empty())
d.pop();清空队列
while(!r.empty())
r.pop();
cin>>str;
for(int i=0;i<n;i++)
{
if(str[i]=='D')
d.push(i);用队列存储他们·的位置
else
r.push(i);
}
while(!d.empty()&&!r.empty())
{
int x = d.front();
int y = r.front();
d.pop();
r.pop();
if(x < y)两个人单挑，只有一个能活
d.push(x + n);
else
r.push(y + n);
}
if(d.empty())直到有一方死光为止
cout<<"R"<<endl;
else
cout<<"D"<<endl;
}
return 0;
}