poj_3371

最新推荐文章于 2018-01-28 18:51:43 发布

weixin_34235457

最新推荐文章于 2018-01-28 18:51:43 发布

阅读量85

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/vactor/p/4100004.html

本文探讨了一道涉及字符属性判断、文本结构分析的编程题，通过实现特定函数，识别输入文本中元音字母、单词边界、句点等元素，并计算元音比例、单词数量与音节数的比例，最终输出一个综合评分。该题旨在锻炼开发者对字符串操作、条件判断及数学计算的能力。

一道模拟题，写的有点麻烦

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
bool fun(char c)
{
    if(c=='a'||c=='A'||c=='e'||c=='E'||c=='i'||c=='I'||c=='o'||c=='O'||c=='u'||c=='U'||c=='y'||c=='Y')
        return true;
    else
        return false;
}
int main()
{
    int word=0,sen=0,syl=0;
    char s[1000];
    while(scanf("%s",s)!=EOF)
    {
        int len=0,y=0;
        int i;
        for(i=0; s[i]!='\0'; i++)
        {
            if((s[i]>='a'&&s[i]<='z')||(s[i]>='A'&&s[i]<='Z'))
            {
                len++;
                if(fun(s[i]))
                {
                    if(s[i]=='e')
                    {
                        if(!((s[i+1]>='a'&&s[i+1]<='z')||(s[i+1]>='A'&&s[i+1]<='Z'))&& s[i-1]=='l')
                        {
                            syl++;
                            y++;
                        }
                        else if(!fun(s[i+1])&&((s[i+2]>='a'&&s[i+2]<='z')||(s[i+2]>='A'&&s[i+2]<='Z')))
                        {
                           if((s[i+1]>='a'&&s[i+1]<='z')||(s[i+1]>='A'&&s[i+1]<='Z'))
                            {syl++;
                            y++;
                            }
                        }
                        else if(!fun(s[i+1])&&((s[i+1]>='a'&&s[i+1]<='z')||(s[i+1]>='A'&&s[i+1]<='Z'))&&(s[i+1]!='s')&&(s[i+1]!='d'))
                            {syl++;
                            y++;
                            }
                        else if(fun(s[i-1])&&!((s[i+2]>='a'&&s[i+2]<='z')||(s[i+2]>='A'&&s[i+2]<='Z'))&&((s[i+1]=='s')||(s[i+1]=='d')))
                        {
                            syl++;y++;
                        }
                        else if(fun(s[i-1])&&!((s[i+1]>='a'&&s[i+1]<='z')||(s[i+1]>='A'&&s[i+1]<='Z')))
                        {
                            syl++;y++;
                        }
                    }
                    else if(!fun(s[i+1]))
                        {syl++;y++;
                        }
                }
                if(len<=3)
                {
                    if(!((s[i+1]>='a'&&s[i+1]<='z')||(s[i+1]>='A'&&s[i+1]<='Z')))
                    {
                        syl++;
                        syl-=y;
                        y=0;
                        continue;
                    }
                }
            }
            else if(s[i]==',')
            {
                word++;
                y=0;
                len=0;
            }
            else if(s[i]=='.' || s[i]=='?'  || s[i]==':' || s[i]==';' || s[i]=='!')
            {
                word++;
                sen++;
                y=0;
                len=0;
            }
        }
        if((s[i-1]>='a'&&s[i-1]<='z')||(s[i-1]>='A'&&s[i-1]<='Z'))
        {
            word++;
            y=0;
            len=0;
        }
        memset(s,'\0',sizeof(s));

    }
    printf("%.2f\n",206.835-1.015*(double)word/(double)sen-84.6*(double)syl/(double)word);
    return 0;
}

转载于:https://www.cnblogs.com/vactor/p/4100004.html