UVA11888 Abnormal 89‘s【回文】

A palindrome is a word that can be read the same way in either direction. More formally if a string is d (d > 0) characters length and the i-th character is a_i , the string is palindrome if and only if ai equals a_(d−i+1) for 1 ≤ i ≤ d. For example “abcba” is palindrome while “aaab” is not.
It is known that everyone who gets to know palindromes, begin an emotional relationship with these beautiful strings. The harmony between the letters makes them artistic. But the 89’s (those who entered AUT at 1389) claim they love another kind of strings. It is called alindrome. Actually an alindrome is the result of concatenation of two palindromes. For example “abacc”=“aba”+“cc” is alindrome.
Now you should write a program to distinguish alindromes, palindromes and other kind of strings.
Input
The first line contains T (T ≤ 50), the number of tests. Each test that comes in a separate line contains a string to be checked. Input strings contain only lower case letters (‘a’ to ‘z’) and their length will be at most 200000.
Output
For each test output a single word in a single line. If the input string can be made by concatenating two palindromes, output ‘alindrome’. Otherwise if it’s a palindrome output ‘palindrome’. In any other case print ‘simple’. (Quotes for clarity)
Sample Input
4
aaa
aabaa
aabaaa
abc
Sample Output
alindrome
palindrome
alindrome
simple

问题链接：UVA11888 Abnormal 89’s
问题简述：（略）
问题分析：回文判定问题，这里用暴力来解决。也许用Manachar算法是正解。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA11888 Abnormal 89's */#include <bits/stdc++.h>using namespace std;const int N = 200000;
char s[N + 1];bool isPalindrome(char s[], int l, int r)
{while (l < r) {if (s[l] != s[r]) return false;l++, r--;}return true;
}int main()
{int t;scanf("%d", &t);while (t--) {scanf("%s", s);int len = strlen(s);int ans = 0;for (int i = 0; i < len - 1; i++)if (s[0] == s[i] && s[i + 1] == s[len - 1])if (isPalindrome(s, 0, i) && isPalindrome(s, i + 1, len - 1)) {ans = 1;break;}if (ans == 0) {if (isPalindrome(s, 0, len - 1)) ans = 2;else ans = 3;}if (ans == 1) puts("alindrome");else if (ans == 2) puts("palindrome");else puts("simple");}return 0;
}

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