UVA10945 Mother bear【Ad Hoc】

Unforunately for our lazy “heroes”, the nuts were planted by an evil bear known as… Dave, and they’ve fallen right into his trap. Dave is not just any bear, he’s a talking bear, but he can only understand sentences that are palindromes.
While Larry was dazed and confused, Ryan figured this out, but need a way to make sure his sentences are palindromic. So he pulled out his trustly iPod, which thankfully have this program you wrote just for this purpose… or did you?
Input
You’ll be given many sentences. You have to determine if they are palindromes or not, ignoring case and punctuations. Every sentence will only contain the letters A-Z, a-z, ‘.’, ‘,’, ‘!’, ‘?’. The end of input will be a line containing the word ‘DONE’, which should not be processed.
Output
On each input, output ‘You won’t be eaten!’ if it is a palindrome, and ‘Uh oh…’ if it is not a palindrome.
Sample Input
Madam, Im adam!
Roma tibi subito motibus ibit amor.
Me so hungry!
Si nummi immunis
DONE
Sample Output
You won’t be eaten!
You won’t be eaten!
Uh oh…
You won’t be eaten!

问题链接：UVA10945 Mother bear
问题简述：（略）
问题分析：
简单回文判定题不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10945 Mother bear */#include <bits/stdc++.h>using namespace std;int main()
{string s;while(getline(cin, s)) {if(s == "DONE") break;int k = 0;for(int i = 0; s[i]; i++)if(isupper(s[i]))s[k++] = s[i] - 'A' + 'a';else if(islower(s[i]))s[k++] = s[i];// 判定回文int flag = 1, l = 0, r = k - 1;while(l < r)if(s[r] != s[l]) {flag = 0;break;} elsel++, r--;printf(flag ? "You won't be eaten!\n" : "Uh oh..\n");}return 0;
}

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