CodeForces 131A cAPS lOCK

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最新推荐文章于 2024-10-30 02:03:01 发布

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原文链接：http://www.cnblogs.com/cyd308/p/4771494.html

本文介绍了一个关于CAPSLOCK键误触导致输入错误的问题，并提供了一段C语言代码来自动修正这种错误。修正规则为：如果单词除了第一个字母外其余全部大写，则将所有字母转换为小写；若首字母大写则只转换首字母。

cAPS lOCK

Time Limit:500MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 131A

Description

wHAT DO WE NEED cAPS LOCK FOR?

Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.

Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:

either it only contains uppercase letters;
or all letters except for the first one are uppercase.

In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.

Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.

Input

The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.

Output

Print the result of the given word's processing.

Sample Input

Input

cAPS

Output

Caps

Input

Lock

Output

Lock

 1 #include <stdio.h>
 2 #include <string.h>
 3 int main()
 4 {
 5     char a[105];
 6     int i,j,k;
 7     while(scanf("%s",a)!=EOF)
 8     {
 9         int flg=1;
10         int l=strlen(a);
11         for(i=1;i<l;i++)
12         {
13             if('A'<=a[i] && a[i]<='Z')
14                 k=1;
15             else
16             {
17                 flg=0;
18                 break;
19             }
20         }
21         //printf("%d\n",flg);
22         if(flg!=1)
23             printf("%s\n",a);
24         else
25         {
26             if('A'<=a[0] && a[0]<='Z')
27                 printf("%c",a[0]+32);
28             else
29                 printf("%c",a[0]-32);
30             for(i=1;i<l;i++)
31                 printf("%c",a[i]+32);
32         }
33         printf("\n");
34     }
35     return 0;
36 }