One day two PAT (1)

In fact, since yesterday brush, preparing for what PAT (B, chicken dishes do not explain, I hope a few months can grow in it), do a bit and found his poor arithmetic skills good ah. . . . .

First on the subject

1. The string must only P, A, T three character, other characters can not contain;
2 xPATx arbitrary shape such as a string can get "correct answer", or where x is an empty string, or a string is composed only of letters;
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 format: Each test comprises a test input. Line 1 shows a natural number 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


Again on my code

#include<iostream>

#include<string>

using namespace std;

int main ()

{

int n;

cin>>n;

string *a;

int j,n1=0,n2=0,n3=0;

a = new string[n];

for(int i=0;i<n;i++)

{

cin>>a[i];

}

for(int i=0;i<n;i++)

{ j=0;

n1=0;

n2=0;

n3=0;

if(a[i][j]=='P')

{ j++;

if(a[i][j]!='A')

{

cout<<"NO"<<endl;

continue;

}

while(a[i][j]=='A'&&j<a[i].length())

{

j++;

}

if(j==a[i].length())

{

cout<<"NO"<<endl;

continue;

}

else if(a[i][j]=='T'&&(j+1)==a[i].length())

{

cout<<"YES"<<endl;

continue;

}

else

{

cout<<"NO"<<endl;

continue;

}

}

while(j<a[i].length()&&a[i][j]=='A')

{

//cout<<j<<n1<<endl;

j++;

n1++;

}

if(j==a[i].length())

{

cout<<"NO"<<endl;

continue;

}

if(a[i][j]!='P')

{

cout<<"NO"<<endl;

continue;

}

j++;

while(j<a[i].length()&&a[i][j]=='A')

{

j++;

n2 ++;

}

if(j==a[i].length())

{

cout<<"NO"<<endl;

continue;

}

if(a[i][j]!='T')

{

cout<<"NO"<<endl;

continue;

}

n3 = n1 * n2;

j++;

if((n1+n2+n3+2)!= a[i].length())

{

cout<<"NO"<<endl;

continue;

}

while(j<a[i].length())

{

if(a[i][j]!='A')

{

cout<<"NO"<<endl;

break;

}

j++;

}

if(j==a[i].length())

{

cout<<"YES"<<endl;

}

}

delete []a;

}

Good food ah, checked the others, get 28 rows. . . . . .