介绍三种做法。
设字符串长度为 \(n\)。
（一）暴力枚举左右端点，再判断中间是否是回文串，可以得到 \(O(n^3)\) 解法，足以通过本题。
（二）动态规划，设 \(f_{l,r}\) 表示子串 \([l,r]\) 是不是回文串，则初值为 \(f_{i,i}=1\)、\(f_{i,i+1}=[s_i=s_{i+1}]\)（\([\textrm{cond}]\) 是艾弗森括号，若 \(\textrm{cond}\) 为真，其值为 \(1\)，否则为 \(0\)）。对于 \(r-l+1\ge 3\)，有 \(f_{l,r}=[s_l=s_r]\land f_{l+1,r-1}\)。DP 出来之后求 \(r-l+1\) 的最大值，使得 \(f_{l,r}=1\) 即可。时间复杂度 \(O(n^2)\)。
（三）使用 Manacher 算法求出以每个位置为中心的回文串长度，取最大值即可，时间复杂度 \(O(n)\)。鉴于此做法难度显著高于本题要求，Manacher 略过不讲，有兴趣的读者可以前往 P3805 【模板】Manacher 学习。
代码是第二种做法：
// Problem: P15520 [CCC 2016 J3] Hidden Palindrome
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P15520
// Memory Limit: 512 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(int x = (y); x <= (z); ++x)
#define per(x, y, z) for(int x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {uniform_int_distribution<int> dist(L, R);return dist(rnd);
}template<typename T> void chkmin(T& x, T y) {if(y < x) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}template<int mod>
inline unsigned int down(unsigned int x) {return x >= mod ? x - mod : x;
}template<int mod>
struct Modint {unsigned int x;Modint() = default;Modint(unsigned int x) : x(x) {}friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}friend Modint operator/(Modint a, Modint b) {return a * ~b;}friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}friend Modint operator~(Modint a) {return a ^ (mod - 2);}friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}friend Modint& operator++(Modint& a) {return a += 1;}friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}friend Modint& operator--(Modint& a) {return a -= 1;}friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}friend bool operator==(Modint a, Modint b) {return a.x == b.x;}friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};const int N = 45;int n, f[N][N], ans;
string s;int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);cin >> s;n = s.size();s = " " + s + " ";rep(i, 1, n) f[i][i] = 1;rep(i, 1, n - 1) if(s[i] == s[i + 1]) f[i][i + 1] = 1;rep(len, 3, n) {rep(l, 1, n - len + 1) {int r = l + len - 1;if(s[l] == s[r]) f[l][r] = f[l + 1][r - 1];}}rep(l, 1, n) rep(r, l, n) if(f[l][r]) chkmax(ans, r - l + 1);cout << ans << endl;return 0;
}