Idea: Introduce enough correct strings to find rules,
1. PAT ->>> PAAT ->>> PAAAT
2. APATA ->>> APAATAA ->>> APAAAPAAA
3.AAPATAA ->>> AAPAATAAAA ->>> AAPAAATAAAAAA
需要注意的是，要用初始正确的去推导。
发现了一个规律 ，P左边A的数量 c1 ,P T 中间A的数量 c2, T右边A的数量 c3 之间；存在数量关系 c1*c2==c3;
import java.util.ArrayList;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int flagSum = 0;
ArrayList<String> stringList = new ArrayList<String>();
Scanner sc = new Scanner(System.in);
flagSum = Integer.parseInt(sc.nextLine());
for (int i = 0; i < flagSum; ++i) {
stringList.add(sc.nextLine());
}
// 判断
for (String x : stringList) {
judge(x);
}
}
public static void judge(String s) {
String pattern = "A*PA+TA*";//推导式
String pattern1 = "PA+T";//对应A数量的第一项正确项
if (s.matches(pattern)) {
if (s.matches(pattern1)) {
System.out.println("YES");
} else {
String temp[] = s.split("P|T");
int aLength = temp[0].length();
int bLength = temp[1].length();
int cLength = temp[2].length();
if (aLength*bLength==cLength) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
} else {
System.out.println("NO");
}
}
}