String of CCPC 思维，字符串，细节

最新推荐文章于 2024-08-31 18:11:57 发布

Cworld2017

最新推荐文章于 2024-08-31 18:11:57 发布

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分类专栏：字符串思维

本文链接：https://blog.youkuaiyun.com/Cworld2017/article/details/81939423

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本文探讨了一个字符串问题，通过在字符串中插入字符以最大化其价值。详细介绍了如何通过算法找到最佳插入位置，实现字符串价值的最大化。

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BaoBao has just found a string of length consisting of 'C' and 'P' in his pocket. As a big fan of the China Collegiate Programming Contest, BaoBao thinks a substring of is "good", if and only if 'C', and 'P', where denotes the -th character in string . The value of is the number of different "good" substrings in . Two "good" substrings and are different, if and only if .

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 . But everything comes with a cost. If it's the -th time for BaoBao to buy a character, he will have to spend units of value.

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

Input

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

The first line contains an integer (), indicating the length of string .

The second line contains the string () consisting of 'C' and 'P'.

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

Output

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

Sample Input

3
3
CCC
5
CCCCP
4
CPCP

Sample Output

1
1
1

Hint

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.

有一个只有包含C和P两种字符的字符串，问只在该字符串某一个位置插入一个'C'或'P'，最多可使该字符串中含有多少个"CCPC"，可以两个CCPC共用同一个C，如CCPCCPC算是有两个

题目中说可以插入多个字符，但一旦插入的字符大于1，那么反而会亏本。所以在判断题目给的串有多少个ccpc后，只要

再考虑插入一个c或者p后是否能形成ccpc这一种情况就行。而能插入一个字符就成功的只有ccc,cpc,ccp三种情况，而且

在ccc的检查中国后面不能是pc两个字符，否则就重复计算了

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
int main()
{
    int t;cin>>t;
    while(t--)
    {
        int L;cin>>L;string a;cin>>a;
        int ans=0;
        for(int i=0;i<L;i++)
        {
            if(a[i]=='C'&&a[i+1]=='C'&&a[i+2]=='P'&&a[i+3]=='C'){
                ans++;
            }
        }
        for(int i=0;i<L;i++)
        {
            int flag=0;
            if(a[i]=='C'&&a[i+1]=='C'&&a[i+2]=='C'){
                if(a[i+3]!='P'||a[i+4]!='C')flag++;
            }
            if(i-1<0&&a[i]=='C'&&a[i+1]=='P'&&a[i+2]=='C'||a[i]=='C'&&a[i+1]=='P'&&a[i+2]=='C'&&a[i-1]!='C'){
                flag=1;
            }
            if(a[i]=='C'&&a[i+1]=='C'&&a[i+2]=='P'&&a[i+3]!='C'){
                flag=1;
            }
            if(flag){
                ans++;break;
            }
        }
        cout<<ans<<endl;

    }
}