YTU2356: The mixed letters

最新推荐文章于 2023-06-01 12:49:10 发布

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最新推荐文章于 2023-06-01 12:49:10 发布

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本文介绍了一个算法，用于将单词中的字母统一转换为全大写或全小写，以解决网络上常见的大小写混用问题。该算法通过比较单词中大写字母和小写字母的数量，决定最终的转换方式，确保转换后的单词保持原有的形式，但只使用一种大小写格式。

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2356: The mixed letters

时间限制: 1 Sec 内存限制: 128 MB
提交: 189 解决: 102
[提交][状态][讨论版][命题人:外部导入]

题目描述

Mike is very upset that many people on the Internet usually mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' case in every word so that it either only consisted of lowercase letters or only consisted of uppercase ones. And he wants to change as few letters as possible in the word.
For example, the word "HoUse" must be changed to "house", and the word "ViP", to "VIP".
If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, "maTRIx" should be changed to "matrix".
You task is to use the given method to change the given word.

输入

The first line contains a single integer n (n<=30), indicating the number of test cases.
Then following n lines, each line contains a word s, it consists of uppercase and lowercase
Latin letters and its length is between 1 and 100, inclusive.

输出

Print the word s after change. If the given word s has strictly more uppercase letters, make the word written in the uppercase register, otherwise, in the lowercase one.

样例输入

3
HoUse
ViP
maTRIx

样例输出

house
VIP
matrix

#include<stdio.h>
#include<string.h>
int main()
{
    int num;
    scanf("%d",&num);
    while(num--)
    {
        int maxnum=0,minnum=0,i,l;
        char a[110];
        scanf("%s",a);
        l=strlen(a);
        for(i=0;i<l;i++)
        {
            if(a[i]>='A'&&a[i]<='Z')
                maxnum++;
            if(a[i]>='a'&&a[i]<='z')
                minnum++;
        }
        if(minnum>=maxnum)
        {
            for(i=0;i<l;i++)
            {
                if(a[i]>='A'&&a[i]<='Z')
                    a[i]+=32;
            }
        }
        else
        {
            for(i=0;i<l;i++)
            {
                if(a[i]>='a'&&a[i]<='z')
                    a[i]-=32;
            }
        }
        printf("%s\n",a);
    }
    return 0;
}