Vowel Counting

最新推荐文章于 2025-08-15 23:10:13 发布

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最新推荐文章于 2025-08-15 23:10:13 发布

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文章标签： java

原文链接：http://www.cnblogs.com/Wurq/p/3750263.html

本文介绍了如何通过编程解决Vowel-Counting-Word问题，包括输入样例、输出样例及实现代码。重点在于字符转换逻辑，即将大写元音转换为小写，保持小写字母不变。

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Vowel Counting

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 58 Accepted Submission(s) : 18

Problem Description

The "Vowel-Counting-Word"(VCW), complies with the following conditions. Each vowel in the word must be uppercase. Each consonant (the letters except the vowels) must be lowercase. For example, "ApplE" is the VCW of "aPPle", "jUhUA" is the VCW of "Juhua". Give you some words; your task is to get the "Vowel-Counting-Word" of each word.

Input

The first line of the input contains an integer T (T<=20) which means the number of test cases. For each case, there is a line contains the word (only contains uppercase and lowercase). The length of the word is not greater than 50.

Output

For each case, output its Vowel-Counting-Word.

Sample Input

4 XYz application qwcvb aeioOa

Sample Output

xyz ApplIcAtIOn qwcvb AEIOOA

Author

AppleMan

Source

HDU 2nd “Vegetable-Birds Cup” Programming Open Contest

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 
 5 int main()
 6 {
 7   int T,L,i;
 8   char a[50];
 9   scanf("%d",&T);
10   while(T--)
11   {
12       scanf("%s",a);
13       L=strlen(a);
14       for(i=0;i<L;i++)
15       {
16         if(a[i]=='a'||a[i]=='e'||a[i]=='i'||a[i]=='o'||a[i]=='u')
17         {
18             a[i]-=32;
19             continue;
20         }
21         else if(a[i]=='A'||a[i]=='E'||a[i]=='I'||a[i]=='O'||a[i]=='U')
22         {
23             continue;
24         }
25         else
26         {
27             if(a[i]>='A'&&a[i]<='Z')
28             {
29                 a[i]+=32;
30                 continue;
31             }
32         }
33       }
34       printf("%s\n",a);
35   }
36   return 0;
37 }