给你一个字符串，求字符串的最长回文串的长度．

使用 Manacher 方法来求解，能在 O(n) 内求出．

#include <bits/stdc++.h>
using namespace std;
const int N = (int) 3e5;
char s[N], t[N];
int p[N];
int main()
{
    while (~scanf("%s", s)) {
        int len_s = strlen(s), len_t = 2;
        t[0] = '$';
        t[1] = '#';
        for (int i = 0; i < len_s; i++) {
            t[len_t++] = s[i];
            t[len_t++] = '#';
        }
        t[len_t] = '\0';
        int ans = 0, id = 0, mx = 0;
        for (int i = 1; i < len_t; i++) {
            p[i] = i < mx ? min(p[id + id - i], mx - i) : 1;
            while (t[i - p[i]] == t[i + p[i]])
                p[i]++;
            if (mx < i + p[i]) {
                mx = i + p[i];
                id = i;
            }
            ans = max(ans, p[i] - 1);
        }
        printf("%d\n", ans);
    }
    return 0;
}