回文串动态规划：
s[i] == s[j]：dp[i][j]=dp[i+1][j-1]
s[i]!=s[j]:dp[i][j]=0
边界条件：dp[i][i]=1 dp[i][i+1]=(s[i]==s[i+1])?1:0
注意ans初始值为1
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
const int maxn = 1e3 + 10;
int dp[maxn][maxn];
int main()
{
string s;
getline(cin, s);
int len = s.length();
int ans = 1;
for (int i = 0; i < len; i++)
{
dp[i][i] = 1;
if (i < len - 1 && s[i] == s[i + 1])
{
dp[i][i + 1] = 1;
ans = 2;
}
}
for (int L = 3; L <= len; L++)
{
for (int i = 0; i + L - 1 < len; i++)
{
int j = i + L - 1;
if (s[i] == s[j] && dp[i + 1][j - 1] == 1)
{
dp[i][j] = 1;
ans = L;
}
}
}
printf("%d", ans);
return 0;
}
