B1003 me through! (20 points)

By 1003 I want to! (20 points)

"Correct answer" is the most joy automatic reply sentence topic given system. This title belongs to the PAT of "correct answer" big delivery - just read the string following conditions are satisfied, the system will output "correct answer", otherwise outputs "Wrong Answer."

Get the "correct answer" conditions are:

String must have only P, A, T three character, other characters can not contain;
arbitrary shape such as a string can be obtained xPATx "correct answer", or where x is the empty string, or only by the letter a string thereof;
if aPbTc is correct, then aPbATca is correct, wherein a, b, c or all the empty string, or a string consisting only of letters a.
Now ask you to write a PAT referee program automatically determines which strings can get "the answer right".

Input format:
Each test comprises a test input. Line 1 is given a positive integer n (<10), is the number of strings to be detected. Next, one row for each string, the string length of not more than 100, no spaces.

Output formats:
the detection result row for each string, if the string can get "correct answer", the output YES, otherwise output NO.

Yang_li shu_ru:
8
PAT at
PAAT
AAPATAA
AAPAATAAAA
xPATx
the PT
Whatever the
APAAATAA

Output Sample:
YES
YES
YES
YES
NO
NO
NO
NO

analysis

Topic analysis

Two conditions are required to answer correctly

1. PAT only three character classes
2. xPATx, aPbTc, aPbATca of x, a, b, c may be empty or may be a string of A or A.
3. Sample heard a summary to get in front of P between a * P and T = a subsequent T (summary of the law, I did not find the reference of willow God's blog)

---

Learn to master

Use map map of the C ++ STL

#include <iostream>
#include <map>
using namespace std;
int main(){
    int n,p = 0,q = 0;
    cin >> n;
    while (n--!=0) {
        string s;
        cin >> s;
        map<char,int> m;
        for (int i=0; i<s.size();i++) {
            m[s[i]] ++;
            if(s[i]=='P') p = i;
            if(s[i]=='T') q = i;
            
        }
if(m['P']==1&&m['T']==1&&m['A']!=0&&m.size()==3&&q-p!=1&&((s.size()-q-1)==(q-p-1)*(p)))
            printf("YES\n");
else printf("NO\n");
    }

    return 0;
}

java version

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n =sc.nextInt();
        sc.nextLine();
        while(n--!=-1){
            String str = sc.nextLine();
            pat_test(str);
        }sc.close();
    }
    static void pat_test(String str){

        String s = str;
        if(s.contains("P")&&s.contains("A")&&s.contains("T")){
            s=s.replace("P","");
            s=s.replace("A","");
            s=s.replace("T","");
            s=s.replace("\\s","");
            if(s.isEmpty()){
                int a =str.indexOf("P");
                int c =str.indexOf("T");
                int b =c-a-1;
                int l=str.length()-c-1;
                if(a*b==l){
                    System.out.println("YES");
                }else System.out.println("NO");
            }else System.out.println("NO");
        }else System.out.println("NO");

    }
}

reference

Liu God blog