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 \(s_is_{i+1}s_{i+2}s_{i+3}\) 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 \(s_is_{i+1}s_{i+2}s_{i+3}\) and \(s_js_{j+1}s_{j+2}s_{j+3}\) are different, if and only if \(i \ne j\).
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.
Input
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 \(n\) (\(1 \le n \le 2\times 10^5\)), indicating the length of string \(s\).
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 \(10^6\).
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 \(s\) 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 \(s\) 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 \(s\) 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.
题目的意思就是给你一个只含“C”和P的字符串，可以在其中加一个C或P仅有一次机会，问你最多几个“CCPC”串。（CCPCCPC）算两个
我们先考虑，我们新加的C||P不能影响到已经存在的CCPC，否则我们加的就没意义了。
所以我们找到CCPC串后就可以直接跳过，不用再考虑加字符了。
我们去掉CCPC中任意一个可以得到 “CCP”，“CPC”，“CCC”。我们只要对‘CCC’这种情况特殊分析，因为这个子串的情况可能是“CCCPC”如果我们插入字符就会导致下个CCPC串被打乱，失去意义。
下面我要吐槽ZOJ的判题机，我用了sbustr就能过，为什么一个个字符看就会超时！！！
我至今还没想通，也请大佬帮我看看。
TLE代码！！！！！！
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
//999999
int main()
{
int j,i,n;
scanf("%d",&n);
while(n--)
{
char str[1000005];
int len;
scanf("%d",&len);
memset(str,0,sizeof(str));
int ans=0;
int flag=0;
scanf("%s",str);
for(i=0; i<len; i++)
{
if(str[i]=='C'&&str[i+1]=='C'&&str[i+2]=='P'&&str[i+3]=='C')
{
i=i+2;
ans++;
continue;
}
if(!flag)
{
if(str[i]=='C'&&str[i+1]=='C'&&str[i+2]=='C')
{
if(str[i+3]!='P'&&str[i+4]!='C')
flag=1;
}
if(str[i]=='C'&&str[i+1]=='C'&&str[i+2]=='P')
flag=1;
if(str[i]=='C'&&str[i+1]=='P'&&str[i+2]=='C')
flag=1;
}
}
printf("%d\n",ans+flag);
}
return 0;
}AC代码
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;
//999999
char str[1000005];
int main()
{
int j,i,n;
scanf("%d",&n);
while(n--)
{
string str;
int len;
scanf("%d",&len);
str="";
int ans=0;
int flag=0;
cin >>str;
for(i=0; i<len; i++)
{
if(str.substr(i,4)=="CCPC"){
i+=2;
ans++;
continue;
}
if(!flag){
string k=str.substr(i,3);
if(k=="CCC"||k=="CCP"||k=="CPC"){
if(k=="CCC"&&str.substr(i+1,4)=="CCPC") continue;
flag=1;
//ans++;
}
}
}
printf("%d\n",ans+flag);
}
return 0;
}
//acacacac