2017 CCPC 秦皇岛 & ZOJ 3985 - String of CCPC （子串）

String of CCPC
Time Limit: 1 Second      Memory Limit: 65536 KB

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.
Author: WENG, Caizhi
Source: The 2017 China Collegiate Programming Contest, Qinhuangdao Site

题意：

可以加字符。字符有代价 0 1 2 3.

POINT：

容易得知只能加一次。那么直接遍历加的位置，和加的p还是c。

然后往前走5个长度，往后从5个长度。比较这10或11个长度里有几个ccpc，增加了说明就成功了。

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>

using namespace std;
const int maxn = 2e5+10;
char s[5]="CCPC";
char str[maxn];
char a[22];
int main()
{
int T;
scanf("%d",&T);
while(T--){
int n;
scanf("%d",&n);
scanf("%s",str);
int tmp=0;
for(int i=0;i<=n;i++)
{
int cs=0;
int ans1=0,ans2=0,ans3=0;
for(int k=max((i-5),0);k<min(n,i+5);k++)
{
a[cs++]=str[k];
}
for(int j=0;j<cs-3;j++)
{
bool flag=1;
for(int k=0;k<4;k++)
{
if(a[j+k]!=s[k])
flag=0;
}
ans1+=(int)flag;
}
cs=0;
for(int k=max(i-5,0);k<i;k++)
{
a[cs++]=str[k];
}
a[cs++]='P';
for(int k=i;k<min(n,i+5);k++)
{
a[cs++]=str[k];
}
for(int j=0;j<cs-3;j++)
{
bool flag=1;
for(int k=0;k<4;k++)
{
if(a[j+k]!=s[k])
flag=0;
}
ans2+=(int)flag;
}
cs=0;
for(int k=max(i-5,0);k<i;k++)
{
a[cs++]=str[k];
}
a[cs++]='C';
for(int k=i;k<min(n,i+5);k++)
{
a[cs++]=str[k];
}
for(int j=0;j<cs-3;j++)
{
bool flag=1;
for(int k=0;k<4;k++)
{
if(a[j+k]!=s[k])
flag=0;
}
ans3+=(int)flag;
}
if((ans2>ans1)||(ans3>ans1))
{
tmp=1;
break;
}
}
int ans=0;
for(int i=0;i<n-3;i++)
{
int flag=1;
for(int j=0;j<4;j++)
{
if(str[i+j]!=s[j])
flag=0;

}
ans+=flag;
}
printf("%d\n",ans+tmp);
}

return 0;
}