Description
Given a string S, which consists of lowercase characters, you need to find the longest palindromic sub-string.
A sub-string of a string S is another string S' that occurs "in" S. For example, "abst" is a sub-string of "abaabsta". A palindrome is a sequence of characters which reads the same backward as forward.
Input
There are several test cases.
Each test case consists of one line with a single string S (1 ≤ |S | ≤ 50).
Output
For each test case, output the length of the longest palindromic sub-string.
Sample Input
sasadasa
bxabx
zhuyuan
Sample Output
7
1
3
分析：
题意：给出一个字符串，求这个字符串中的最长回文串的长度。
首先要一个一个看字符串中的字符：
对于第 i 个字符 str[i] ，假如回文子串是奇数个字符，那么考虑以i为中心，同时向左向右扩展，直到发现对称位置字符不相等，假如此时共扫过x个字符，则当前回文串长度为2*x+1。加上的1就是加上i本身。如下图所示：
由第 i 个字符a[i]构成回文偶数串。i在最接近对称轴的左侧。
代码：
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#define INF 999999
using namespace std;
int main()
{
int maxlen,i,j,len; ///最长对称子串长度。
char str[1003];
while(gets(str))
{
maxlen = 0;
len = strlen(str);
for(i = 0; i < len; i++)
{
///考虑是i是奇数串的中心,以i为中心，同时往左往右扩展
for(j = 0; i-j>=0&&i+j<=len; j++)
{
if(str[i-j]!=str[i+j])
break;
if(2*j+1>maxlen)
maxlen = 2*j+1;
}
///i是偶数串的中心
for(j = 0; i-j>=0&&i+j+1<len;j++)
{
if(str[i-j]!=str[i+j+1])
break;
if(2*j+2>maxlen)
maxlen = 2*j+2;
}
}
printf("%d
",maxlen);
}
return 0;
}