PAT BASIC 1003

注意點

1、其實就是要求P之前的A的數量乘以PT之間A的數量等於T之後A的數量
2、注意PT中間等於零的情況
3、注意PT兩個字母：必須P在T之前出現，各有且僅有一次

代碼

#include<iostream>
#include<string>
using namespace std;
void checkPAT();

void checkPAT()
{
  string str;
  int pos_P = -1, pos_T = -1, state = 0;
  int i, j, len;
  int num[3];
  cin >> str;
  len = str.size();
  num[0] = 0;
  num[1] = 0;
  num[2] = 0;
  for (i = 0; i < len; i++)
  {
    if (str[i] != 'P' && str[i] != 'A' && str[i] != 'T')
    {
      cout << "NO\n";
      return;
    }
    if (str[i] == 'A')
      num[state]++;
    else if (str[i] == 'P')
    {
      if (pos_P != -1)
      {
        cout << "NO\n";
        return;
      }
      pos_P = i;
      state++;
    }
    else if (str[i] == 'T')
    {
      if (pos_T != -1)
      {
        cout << "NO\n";
        return;
      }
      pos_T = i;
      state++;
    }
  }
  if (pos_P == -1 || pos_T == -1)
  {
    cout << "NO\n";
    return;
  }
  if (pos_P >= pos_T)
  {
    cout << "NO\n";
    return;
  }
  if (num[1] == 0)
  {
    cout << "NO\n";
    return;
  }
  if (num[0] * num[1] == num[2])
    cout << "YES\n";
  else cout << "NO\n";
  return;
}

int main()
{
  int n, i;
  cin >> n;
  for (i = 0; i < n; i++)
    checkPAT();
  //while (1)
  //{ }
  return 0;
}