【贪心】cf 1392 D Omkar and Bed Wars

题目

题目链接：https://codeforc.es/contest/1392/problem/D

思路

分两种情况
一.只有L/R
n/3向上取整
二.都有
每次取连续段的L/R 每次加上连续段长度/3即可

代码

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<cctype>
#include<ctime>
#include<iostream>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<iomanip>
#include<list>
#include<bitset>
#include<sstream>
#include<fstream>
#include<complex>
#include<algorithm>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#define ll long long
using namespace std;
const int INF = 0x3f3f3f3f;
int dp[100010][2];
int main(){
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		string s;
		cin>>s;
		int cnt=0;
		for(int i=n-1;i>=0;i--){
			if(s[i]==s[0]){
				cnt++;
			}
			else break;
		}
	//	cout<<cnt<<endl;
		vector<int> v;
		int cnt1=0;
		for(int i=0;i<n-cnt;i++){
			if(s[i]!=s[i+1]){
				cnt1++;
				v.push_back(cnt1);
				cnt1=0;
			}
			else cnt1++;
		}
		//for(int i=0;i<v.size();i++) cout<<v[i]<<" \n"[i==v.size()-1];
		if(cnt==n){
			cout<<n/3+(n%3==0?0:1)<<endl;
		}
		else{
			int ans=0;
			for(int i=0;i<v.size();i++){
				if(i==0){
					ans+=(v[i]+cnt)/3;
				}
				else ans+=v[i]/3;
			}
			cout<<ans<<endl;
		}
	}
    return 0;
}