【洛谷P2375】【NOI2014】—动物园（Kmp）

传送门

考虑对于$i$
就有一个为$i-nxt[i]$的循环节
那么$s[1...n]$和$s[i...n]$更短的一个相同前缀就是到$nxt[nxt[i]]$
每次暴力跳到满足$p*2<=i$即可

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
#define poly vector<int>
#define chemx(a,b) ((a)<(b)?(a)=(b):0)
#define chemn(a,b) ((a)>(b)?(a)=(b):0)
cs int mod=1e9+7;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
cs int N=1000005;
int nxt[N],fail[N],n;
char s[N];
int main(){
	int T=read();
	while(T--){
		scanf("%s",s+1);
		n=strlen(s+1);
		memset(nxt,0,sizeof(int)*(n+1));
		nxt[1]=1;
		for(int i=0,j=2;j<=n;j++){
			while(i&&s[i+1]!=s[j])i=fail[i];
			if(s[i+1]==s[j])i++;
			fail[j]=i,nxt[j]=nxt[i]+1;
		}
		int ans=1;
		for(int i=0,j=2;j<=n;j++){
			while(i&&s[i+1]!=s[j])i=fail[i];
			if(s[i+1]==s[j])i++;
			while(i*2>j)i=fail[i];
			Mul(ans,nxt[i]+1);
		}
		cout<<ans<<'\n';
	}
}