manacher算法，求回文串

用来求字符串最长回文串或者回文串的总数量

#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<stdio.h>
#include<float.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define sf scanf
#define pf printf
#define fi first
#define se second
#define mp make_pair
#define pii pair<int,int>
#define scf(x) scanf("%d",&x)
#define prf(x) printf("%d\n",x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
#define mm(x,b) memset((x),(b),sizeof(x))
#define scfff(x,y,z) scanf("%d%d%d",&x,&y,&z)
typedef long long ll;
const ll mod=1e9+7;
using namespace std;

const double eps=1e-8;
const int inf=0x3f3f3f3f;
const double pi=acos(-1.0);
const int N=2e5+10;
int radius[N];
string pre_deal(string s)
{
    string q;
    q.append("#");
    rep(i,0,s.length() )
    {
        q.append(s,i,1);
        q.append("#");  
    } 
    return q;
}
int solve(string s)
{
    int len = s.length() ;
    int R = -1;
    int id = -1;
    int maxn = -1;
    int ans=0;
    for (int i = 0; i < len; i++) 
    {
        radius[i] = R > i ? min(radius[2*id-i],R-i+1):1;
        while(i+radius[i] < len && i - radius[i] > -1){
            if(s[i-radius[i]] == s[i+radius[i]]){
                radius[i]++;
            }else{
                break;
            }
        }
        if(i + radius[i] > R){
            R = i + radius[i]-1;
            id = i;
        }
        maxn = max(maxn,radius[i]);

        ans+=radius[i]/2;
        //cout<<radius[i]<<" ";
    }
    return ans;         //返回的是回文串的数量 
    return maxn-1;      //返回的是最大回文串的长度 
}
int main()
{
    
    string s;
    getline(cin,s);
    s=pre_deal(s);
    cout<<solve(s);
    return 0;
}