ACM(白银组题)problem O

75 阅读 0 评论 50 点赞

我是靠谱客的博主高兴云朵，最近开发中收集的这篇文章主要介绍ACM(白银组题)problem O，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

O - cAPS lOCK
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.

Examples

Input

cAPS

Output

Caps

Input

Lock

Output

Lock

题目简述：
输入一串字母，如果这串字母全身大写就将其变成小写，如果这串字母开头是小写，后面都是大写，那么将开头字母变为大写，后面的字母变为小写。其它情况不做处理。

题目分析：
理解了题意后，我们只用考虑两种情况，那么将其抽象成判断条件即可。大写字母与小写字母的区分就在于ASCLL码值的不同，转变也是通过ASCLL码的对应值来改变。

AC代码：

#include<iostream>
using namespace std;
int main()
{
 char a[210];
 int i;
 while (cin >> a)
 {
  int count = 0,count1=0;
  for (i = 0; i < strlen(a); i++)
  {
   if (a[i] >= 'A'&&a[i] <= 'Z')
    count++;
  }
  if (count == strlen(a))
  {
   for (i = 0; i < strlen(a); i++)
    a[i] = a[i] + 32;
   cout << a << endl;
  }
  else
  {
   if (a[0] >= 'a'&&a[0] <= 'z')
   {
    for (i = 1; i < strlen(a); i++)
     if(a[i]>='A'&&a[i]<='Z')
     count1++;
   }
   if (count1 == strlen(a) - 1)
   {
    a[0] = a[0] - 32;
    for (i = 1; i < strlen(a); i++)
     a[i] = a[i] + 32;
    cout << a << endl;
   }
   else
    cout << a << endl;
  }
 }
 return 0;
}