poj 1575 Easier Done Than Said?

最新推荐文章于 2018-04-04 22:41:51 发布

yzl_rex

最新推荐文章于 2018-04-04 22:41:51 发布

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分类专栏： POJ 文章标签： include system string

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本文探讨了一段用于评估程序代码质量的C++代码片段，重点在于通过一系列逻辑判断来确定代码中字母字符的使用情况，从而评估其是否符合特定的可接受标准。文章详细解释了代码中实现的三种不同条件判断逻辑，并最终决定代码是否符合要求。

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#include <iostream>
#include <string>
using namespace std;

int main()
{
    int i, len;
    string str;
    bool flag1, flag2, flag3, flag4;
    while (cin >> str)
    {
          if (str == "end")  break;
          len = str.length();
          flag1 = flag2 = flag3 = flag4 = false;
          //第一种情况的判断 
          for (i = 0; i < len; i++)
          {
              if (str[i] == 'a' || str[i] == 'e' || str[i] == 'o' || str[i] == 'i' || str[i] == 'u')
              {
                 flag1 = true;
                 break;
              }
          }
          //第二种情况的判断 
          for (i = 0; i < len - 2; i++)
          {
              if (str[i] == 'a' || str[i] == 'e' || str[i] == 'o' || str[i] == 'i' || str[i] == 'u')
              {
                  if (str[i+1] == 'a' || str[i+1] == 'e' || str[i+1] == 'o' || str[i+1] == 'i' || str[i+1] == 'u')
                  {
                       if (str[i+2] == 'a' || str[i+2] == 'e' || str[i+2] == 'o' || str[i+2] == 'i' || str[i+2] == 'u')
                       {
                            flag2 = true;
                            break;
                       }
                  }
              }
          }
          for (i = 0; i < len - 2; i++)
          {
              if (str[i] != 'a' && str[i] != 'e' && str[i] != 'o' && str[i] != 'i' && str[i] != 'u')
              {
                  if (str[i+1] != 'a' && str[i+1] != 'e' && str[i+1] != 'o' && str[i+1] != 'i' && str[i+1] != 'u')
                  {
                      if (str[i+2] != 'a' && str[i+2] != 'e' && str[i+2] != 'o' && str[i+2] != 'i' && str[i+2] != 'u')
                      {
                          flag3 = true;
                          break;
                      }
                  }
              }
          }
          //第三种情况的判断 
          for (i = 0; i < len - 1; i++)
          {
              if (str[i] == str[i+1] && str[i] != 'e' && str[i] != 'o')
              {
                  flag4 = true;
                  break;
              }
          }
          
          if (flag1 && !flag2 && !flag3 && !flag4)
             cout << "<" << str << ">" << " is acceptable." << endl;
          else
             cout << "<" << str << ">" << " is not acceptable." << endl;
    }
    
    system("pause");
}