adong的acm作业1

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我是靠谱客的博主可靠巨人，最近开发中收集的这篇文章主要介绍adong的acm作业1，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

                           ACM第一次训练题1

Time limit
2000 ms

                            Memory limit
                            262144 kB
                        
                            Source
                            Codeforces Beta Round #89 (Div. 2)
                        
                            Tags
                            implementation*1100

Problem Description
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:

deletes all the vowels,
inserts a character “.” before each consonant,
replaces all uppercase consonants with corresponding lowercase ones.

Vowels are letters “A”, “O”, “Y”, “E”, “U”, “I”, and the rest are consonants. The program’s input is exactly one string, it should return the output as a single string, resulting after the program’s processing the initial string.
Help Petya cope with this easy task.

    Input
    The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
    Output
    Print the resulting string. It is guaranteed that this string is not empty.

问题描述:
Petya开始参加编程课程。第一课他的任务是写一个简单的程序。该程序应该做以下工作:在给定的字符串中，包含大写和小写的拉丁字母，它:删除所有的元音，在每个辅音前插入一个字符“。”将所有大写辅音替换为相应的小写辅音。元音是字母“A”、“O”、“Y”、“E”、“U”、“I”，其余都是辅音。程序的输入恰好是一个字符串，它应该以单个字符串的形式返回输出，这是在程序处理初始字符串之后产生的。帮助彼佳处理这个简单的任务。输入第一行表示Petya程序的输入字符串。该字符串仅由大写和小写拉丁字母组成，其长度从1到100(包括在内)。输出打印结果字符串。可以保证这个字符串不是空的。

问题分析：
首先将输入的字符串存入一个数组，然后用for循环逐个判断每个字符是否为小写，如果不是则将字符转化为小写。然后用bool类型的表达式逐个判断是否为“a",“o”,“y”,“e”,“u”,"i"中的一个。不是输出.和字符。
代码如下：



#include <stdio.h>

 

#include<string.h>

 

 

 

int main()

{

    char a[2000];

    

    scanf("%s", a);

    for(int i=0;a[i]!='';i++)

    {

         

         if (a[i] >= 'A'&&a[i] <=
'Z')

         {

             a[i] +=
32;

         }

         

         

             if (a[i] == 'a' || a[i] == 'o' || a[i] == 'y' || a[i] == 'e' || a[i] == 'u' || a[i] == 'i')

                  a[i]
= 0;

         

         

             if (a[i] != 0)

             {

                  printf(".");

                  printf("%c", a[i]);

             }

 

         

         

    }

    printf("n");

    return 0;

}