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HDU 3079 Vowel Counting （水）

2024-08-02 04:57:23   224人阅读

Vowel Counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1413    Accepted Submission(s): 1043


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





解题思路：水题一枚，字符大小写转化问题。






AC代码：

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

int main(){
//    freopen("in.txt", "r", stdin);
    int t;
    string s;
    while(scanf("%d", &t)!=EOF){
        cin>>s;
        for(int i=0; i<s.size(); i++){
            if(s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u' || s[i] == 'A' || s[i] == 'E' || s[i] == 'I' || s[i] == 'O' || s[i] == 'U'){
                if(s[i]>='a' && s[i]<='z')  s[i] -= 32;
                printf("%c", s[i]);    
            }
            else{
                if(s[i]>='A' && s[i]<='Z')  s[i] += 32;
                printf("%c", s[i]);    
            }
        }
        printf("\n");    
    }
    return 0;
}


HDU 3079 Vowel Counting （水）


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