2019 MIST IUPC F - Palindromadness

Solution: When I write know why. Anyway, I think the question is a palindrome automaton difficult.

#include<bits/stdc++.h>
#define ls (x<<1)
#define rs (x<<1|1)
#define ll long long
#define pb push_back
#define mp make_pair
#define db double
#define pii pair<int,int>
using namespace std;
const int M=1e4+7;
const int N=1e6+7;
const int inf=1e9;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int T;
char S[N];
ll aa[N];
int n,base,mod;
	int a [N] [26], with [N] fail [N], half [N], pq [N];
Pam struct {
	int last,tot;
	ll cnt[N],val[N];
	ll ans;
	int vis[N];
	ll tmp;
	void init(){
		for(int i=0;i<=tot;i++){
			cnt[i]=len[i]=0;
		}
		for(int i=0;i<=tot;i++)for(int j=0;j<26;j++)son[i][j]=0;
		len[1]=-1;tot=1;last=0;fail[0]=fail[1]=1;
		ans=0;tmp=0;
	}
	void extend(int c,int pl,char *S){
		int p=last;int q,r;
		while(S[pl-len[p]-1]!=S[pl]){p=fail[p];}
		if(!son[p][c]){
			q=++tot,r=fail[p];
			len[q]=len[p]+2;
			while(S[pl-len[r]-1]!=S[pl])r=fail[r];
			fail[q]=son[r][c];	
			son[p][c]=q;
		}
		last=son[p][c];
		cnt[last]++;
	}
	void cal(){
		if(x>1&&view [x]) tmp = (TMP-cnt [x] * AA [n-len [x]]% v)% against, vis [x] = 0;
		
 
	while(T--){
		cin>>n>>base>>mod;
		scanf("%s",S+1);
		aa[0]=1;
		int len=strlen(S+1);
		for(int i=1;i<=n;i++)aa[i]=aa[i-1]*base%mod;
		//for(int i=1;i<=n;i++)cout<<aa[i]<<" ";cout<<"\n";
		pam.init();
		for(int i=1;i<=len;i++){
			pam.extend(S[i]-'a',i,S);
		}
		pam.cal();
		pam.dfs(0);
		//cout<<pam.tmp<<"\n";
		pam.dfs(1);
		
		//cout<<pam.cnt[0]<<pam.cnt[1]<<"\n";
		cout<<"Case "<<++cas<<": "<<(pam.ans%mod+mod)%mod<<"\n";
	}
}