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
5DDRRR
output
D
input
6DDRRRR
output
R
Note
Consider one of the voting scenarios for the first sample:
- Employee 1 denies employee 5 to vote.
- Employee 2 denies employee 3 to vote.
- Employee 3 has no right to vote and skips his turn (he was denied by employee 2).
- Employee 4 denies employee 2 to vote.
- Employee 5 has no right to vote and skips his turn (he was denied by employee 1).
- Employee 1 denies employee 4.
- Only employee 1 now has the right to vote so the voting ends with the victory of depublicans.
题意：可能有点难懂，说的是有两个分派R和D
样列：DDRRR
第一个是D表态说第五个人出局（R），那么出局人没有表态权利
第二个是D表态说第三个人出局（R），同上
那么剩下4（R），说第二个人出局（D）
一轮结束
第二局第一个人说第四个人出局，好，R全军覆没，D保留
问的也是最后保留的是谁
解法：
1 自己人当然不能让自己人出局啊
2 有机会轮到自己时，一定是让对方出局，最容易想到的是让离自己最近的对方出局，例如第一个D让第三个R出局
3 直到有一个全军覆没，保留的胜利
4 那么就算一个模拟贪心，实际上最多也就玩两次也有结果了，用vector超时了，改set
1 #include<bits/stdc++.h>
2 typedef long long LL;
3 typedef unsigned long long ULL;
4 typedef long double LD;
5 using namespace std;
6 set<int>Md,Mr;
7 int flag[400000];
8 int main(){
9 int n;
10 string s;
11 cin>>n>>s;
12 for(int i=0;i<n;i++){
13 if(s[i]==‘D‘){
14 Md.insert(i);
15 }else{
16 Mr.insert(i);
17 }
18 }
19 int i=-1;
20 while(Md.size()&&Mr.size()){
21 i=(i+1)%n;
22 if(flag[i]) continue;
23 if(s[i]==‘D‘){
24 auto it=Mr.upper_bound(i);
25 if(it==Mr.end()){
26 it=Mr.begin();
27 }
28 flag[*it]=1;
29 Mr.erase(it);
30 }else{
31 auto it=Md.upper_bound(i);
32 if(it==Md.end()){
33 it=Md.begin();
34 }
35 flag[*it]=1;
36 Md.erase(it);
37 }
38 }
39 if(Md.size()){
40 cout<<"D"<<endl;
41 }else{
42 cout<<"R"<<endl;
43 }
44 return 0;
45 }