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 .
<h4< dd="">Output
For each test case output one line containing one integer, indicating the maximum final value BaoBao can obtain.
<h4< dd="">Sample Input
3 3 CCC 5 CCCCP 4 CPCP
<h4< dd="">Sample Output
1 1 1
<h4< dd="">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.
题意：每次可以插入一个C或者P，花费是0，1，2，3…如此递增的。每个“CCPC”可视作一分。问可能出现的最高分数是多少。
题解：我们增加一个“P”至多可能增加一个新的“CCPC”。我们增加一个“C”，最多可以增加两个“CCPC”的同时减少一个“CCPC”，也相当于就加了一个。
所以，我们只需要考虑第一次插入操作就可以。
- 注意边界控制。
参考代码：
1 #include<bits/stdc++.h> 2 #define rep(i,a,n) for(int i=a;i<n;++i) 3 #define per(i,a,n) for(int i=n-1;i>=a;--i) 4 #define fi first 5 #define se second 6 #define mp make_pair 7 #define pb push_back 8 #define np next_permutation 9 #define INF 0x3f3f3f3f 10 #define EPS 1e-8 11 #define mod 1000000007 12 using namespace std; 13 typedef long long ll; 14 typedef unsigned long long ull; 15 typedef long double ld; 16 typedef vector<int,int > vi; 17 typedef pair<int,int> pii; 18 typedef pair<string,string> pss; 19 20 typedef map<string,string> mss; 21 typedef map<string,int> msi; 22 23 string s; 24 int m,t,n; 25 //加P 26 //CCCC 稳+1 27 //CCC 后两个C不被用就 +1 否则破坏 28 //加C 29 //PCC 非CCPC 30 //CPC 非CCPC 31 //CCP 非CCPC 32 33 int main(){ 34 ios::sync_with_stdio(false ); 35 cin>>t; 36 while(t--){ 37 cin>>n; 38 cin>>s; 39 if(n<=3){ 40 if(s=="CCC" || s=="CCP" || s=="CPC") cout<<1<<endl; 41 else cout<<0<<endl; 42 continue; 43 } 44 int flag=0,cnt1=0; 45 rep(i,0,n-2){ 46 if(s[i]=='C' && s[i+1]=='C' && s[i+2]=='P'){ 47 // cout<<"#1"<<endl; 48 if(i+3==n){ 49 flag=1; 50 break; 51 } 52 else if(s[i+3]=='C'){ 53 cnt1++; 54 continue; 55 } 56 else { 57 flag=1; 58 continue; 59 } 60 } 61 else if(s[i]=='C' && s[i+1]=='P' && s[i+2]=='C'){ 62 // cout<<"#2"<<endl; 63 if(i==0){ 64 flag=1; 65 continue; 66 } 67 else if(s[i-1]=='C') continue; 68 else { 69 flag=1; 70 continue; 71 } 72 } 73 //CCCPP 74 else if(s[i]=='C' && s[i+1]=='C' && s[i+2]=='C'){ 75 // cout<<"#3"<<endl; 76 if(i+3==n){ 77 flag=1; 78 break; 79 } 80 else if(s[i+3]=='C'){ 81 flag=1; 82 continue; 83 } 84 else { 85 86 if(i+4==n){ 87 flag=1; 88 break; 89 } 90 else if(s[i+4]=='C') continue; 91 else { 92 flag=1; 93 continue; 94 } 95 } 96 } 97 98 99 100 } 101 if(flag) cnt1++; 102 cout<<cnt1<<endl; 103 } 104 105 }