PAT-1003 I would like to pass! (20 points) Python3 (knowledge: find the location of the array is the number obtained)

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:

1. String must have only P, A, T of the three characters, the characters may not contain other;
2. The string of arbitrary shape can be obtained xPATx "correct answer", or where x is the empty string, or a string consisting only of letters A;
3. 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 formats:

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.

Sample input:

8
PAT
PAAT
AAPATAA
AAPAATAAAA
xPATx
PT
Whatever
APAAATAA

Sample output:

YES
YES
YES
YES
NO
NO
NO
NO

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The key problem-solving: meet with all the elements and PAT , the determination is YESP前面的A的个数 乘以 P与T之间的A的个数 是否等于 T后面A的个数
AT > 0 and (AP*AT)==TA

AC Code:

def IsPat(s):
    Pnum=s.count("P")
    Anum=s.count("A")
    Tnum=s.count("T")
    if Pnum==1 and Tnum==1 and (Pnum+Anum+Tnum)==len(s):
        AP=s.find("P")  # 找到P的位置就相当于求得了P前面A的个数
        T=s.find("T")
        AT=T-AP-1
        TA=len(s)-T-1
        if AT > 0 and (AP*AT)==TA:
            print("YES")
        else:
            print("NO")
    else:
        print("NO")


n =int(input())
for i in range(n):
    s=input()
    IsPat(s)

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More than others a little perseverance, you will create a miracle