String of CCPC

NC14367. String of CCPC

描述

BaoBao has just found a string s of length n consisting of 'C' and 'P' in his pocket. As a big fan of the China Collegiate Programming Contest, BaoBao thinks a substring s_is_i+1s_i+2s_i+3 of s is "good", if and only if s_i=s_i+1=s_i+3= 'C' ,and s_i+2='P', where s_i denotes the i-th character in string s. The value of s is the number of different "good" substrings in s. Two "good" substrings s_is_i+1s_i+2s_i+3 and s_js_j+1s_j+2s_j+3 are different ,if and only i ≠ j .

To make this string more valuable, BaoBao decides to buy some characters from a character store. Each time he can buy one 'C' or one 'P' from the store, and insert the character into any position in s.But everything comes with a cost. If it's the i-th time for BaoBao to buy a character, he will have to spend i - 1 units of value.

The final value BaoBao obtains is the final value of s minus the total cost of all the characters bought from the store. Please help BaoBao maximize the final value.

输入描述

There are multiple test cases. The first line of the input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1 ≤ n ≤ 2 ×10⁵)indicating the length of strings.

The second line contains the string s (|s| = n)consisting of 'C' and 'P'.

It's guaranteed that the sum of n over all test cases will not exceed 10⁶.

输出描述

For each test case output one line containing one integer, indicating the maximum final value BaoBao can obtain.

示例1

输入:

3
3
CCC
5
CCCCP
4
CPCP

输出:

1
1
1

说明:

For the first sample test case, BaoBao can buy one 'P' (cost 0 value) and change to "CCPC". So the final value is 1 - 0 = 1.
For the second sample test case, BaoBao can buy one 'C' and one 'P' (cost 0 + 1 = 1 value) and change to "CCPCCPC". So the final value is 2 - 1 = 1.
For the third sample test case, BaoBao can buy one 'C' (cost 0 value) and change to "CCPCP". So the final value is 1 - 0 = 1.
It's easy to prove that no strategies of buying and inserting characters can achieve a better result for the sample test cases.

上次编辑到这里，代码来自缓存点击恢复默认模板

C++14(g++5.4) 解法, 执行用时: 159ms, 内存消耗: 776K, 提交时间: 2020-02-10 16:25:37

#include<iostream>
#include<cstdio>
using namespace std;
int t,n;
string s;
int ans,flag;
int main(void){
	cin>>t;
	while(t--){
		ans = flag = 0;
		cin>>n;
		cin>>s;
		for(int i = 0;i<n;++i){
			if(s.substr(i,4) == "CCPC"){
				++ans;
				i+=2;
				continue;
			}
			if(!flag && (s.substr(i,3)=="CCC" || s.substr(i,3) == "CCP" || s.substr(i,3) == "CPC")){
				if(s.substr(i,3) == "CCC" && s.substr(i+1,4) == "CCPC") continue;
				++ans;
				flag = 1; 
			}
		}
		cout<<ans<<endl;
	} 
	
	return 0;
}

C++(clang++ 11.0.1) 解法, 执行用时: 74ms, 内存消耗: 876K, 提交时间: 2022-10-16 14:15:52

#include<bits/stdc++.h>
using namespace std;

void slove(){
	int n;
	cin>>n;
	string s;
	cin>>s;
	int cnt=0;
	for(int i=0;i<n-3;i++){
		string temp=s.substr(i,4);
        if(temp=="CCPC"){
            cnt++;
            s[i+1]=s[i+2]='x';
        }
	}
    if(s.find("CCC")!=-1||s.find("CPC")!=-1||s.find("CCP")!=-1)cnt++;
	cout<<cnt<<endl;
	return;
}
int main(){
	int t;
	cin>>t;
	while(t--){
		slove();
	}
return 0;
}