标签:turn not rac ons red procedure font imp moment
Problem Description
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:
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
5
DDRRR
D
6
DDRRRR
R
Note
Consider one of the voting scenarios for the first sample:
这题的思路用使用队列模拟。
用两个队列来维护。一个表示D,一个表示R.
分别存入他们所在的位置后,比较两个队列首的元素。将较小的那一个+n后放入该队列的尾部(+n代表下一轮循环)。
然后再同时将队首元素pop。直到有一个队列为空。不为空的那个队列所表示是的就是答案。
AC代码如下
#include<cstdio> #include<algorithm> #include<queue> using namespace std; char v[200005]; int n; queue<int>r,d; int main(){ scanf("%d%s",&n,v); for(int i=0;i<n;i++){ if(v[i]==‘R‘) r.push(i); else d.push(i); } while(!r.empty()&&!d.empty()){ if(r.front()<d.front()) r.push(r.front()+n); else d.push(d.front()+n); d.pop(); r.pop(); } if(r.empty()) printf("D"); else printf("R"); return 0; }
2020 BIT冬训-贪心 G - Voting CodeForces - 749C
标签:turn not rac ons red procedure font imp moment
原文地址:https://www.cnblogs.com/mikku39/p/14393828.html