Pat B true title 1003. I'm going through! (20) (Problem solving)

Source: Internet
Author: User

" correct answer "is the most delightful reply given by the automated system. The question belongs to Pat's "answer correct" large dispatch--as long as the read-in string satisfies the following conditions, the system outputs "answer correct", otherwise the output "answer error".

The conditions for getting the answer right are:

1. The string must have only p, a, t these three characters, can not contain other characters;
2. Any string of arbitrary shapes, such as XPATX, can be "answered correctly", where x is either an empty string or a string consisting of only the letter A;
3. If the APBTC is correct, then APBATCA is also correct, where a, B, c are either empty strings, or a string consisting of only the letter A.

Now ask you to write an automatic referee program for Pat to determine which strings are " correctly answered ".

Input Format: Each test input consists of 1 test cases. The 1th line gives a natural number n (<10), which is the number of strings to be detected. Each string is followed by a single line, with a string length of no more than 100 and no spaces. output Format: The detection result of each string is one row, if the string can get "answer correct", then output yes, otherwise output No. Input Sample: 8PATPAATAAPATAAAAPAATAAAAxPATxPTWhateverAPAAATAA Output Sample: Yesyesyesyesnononono


/********* Compilation Environment **vc++ 6.0 (32-bit) *************/#include<stdio.h>#include<string.h>#include<malloc.h>#defineM 101intMain () {intn,i,j; intcp=0, ct=0, ca=0, flag=0, ce=0; Char**s; Char*p1,*P2; scanf ("%d",&N); S=(Char**)malloc(nsizeof(Char*));//defining the first column     for(i=0; i<n;i++) {S[i]=(Char*)malloc(m*sizeof(Char));//Defining Rowsscanf"%s", S[i]); Strcat (S[i]," /"); }     for(i=0; i<n;i++) {CP=0; CT=0; CA=0; Ce=0;  for(j=0; s[i][j]!=' /'; j + +)        {            if(s[i][j]=='P') CP++; Else if(s[i][j]=='T') CT++; Else if(s[i][j]=='A') CA++; ElseCE=1; }        if(cp==1&&ct==1&&ca!=0&&ce==0) {P1=STRCHR (S[i],'P'); P2=STRCHR (S[i],'T'); if((P1-s[i]) * (p2-p1-1) ==strlen (p2)-1) Flag=1; }        ElseFlag=0; if(flag==1) printf ("yes\n"); Elseprintf ("no\n"); }    return 0;}

Pat B true title 1003. I'm going through! (20) (Problem solving)

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