acdream1116 Gao the string!(hash二分 or 后缀数组)

问题套了一个斐波那契数，归根结底就是要求对于所有后缀s[i...n-1]，所有前缀在其中出现的总次数。我一开始做的时候想了好久，后来看了别人的解法才恍然大悟。对于一个后缀来说 s[i...n-1]来说，所有与它匹配的前缀必然是和 s[i+1...n-1]  s[i+2...n-1] ....s[n-1..n-1]里的前缀匹配的，因而如果我们定义一个num[i]表示的是后缀s[i...n-1]与前缀的总长公共前缀，那么num[i]+num[i+1]+..num[n-1]就是前缀在后缀i里出现的次数总和，所以问题转化为求s[i...n-1]跟前缀的最长公共前缀，也可以理解成是  s[0...n-1]和s[i...n-1]求最长公共前缀。

求后缀的最长公共前缀，一个做法是后缀数组，然后建lcp,然后通过rmq询问，理论的复杂度nlogn,我自己套的模板用的是快排，所以后缀数组本身的复杂度就已经是nlog^2n，然后我就愉快的TLE了。

失败之后就只能转而用hash＋二分，毕竟二分还是慢，所以我卡着时限过了这题，心有余悸。

#pragma warning(disable:4996)

#include <iostream>

#include <cstring>

#include <cstdio>

#include <vector>

#include <cmath>

#include <string>

#include <algorithm>

using namespace std;



#define maxn 110000

#define ll long long

#define mod 1000000007

#define step 31



/*int rk[maxn], sa[maxn], lcp[maxn];

int tmp[maxn];

int d[maxn + 50][25];*/



int n;



/*bool cmp_sa(int i, int j)

{

	if (rk[i] != rk[j]) return rk[i] < rk[j];

	else{

		int ri = i + k <= n ? rk[i + k] : -1;

		int rj = j + k <= n ? rk[j + k] : -1;

		return ri < rj;

	}

}



void construct_sa(char *s, int *sa)

{

	n = strlen(s);

	for (int i = 0; i <= n; i++){

		sa[i] = i;

		rk[i] = i < n ? s[i] : -1;

	}

	for (k = 1; k <= n; k <<= 1){

		sort(sa, sa + n + 1, cmp_sa);

		tmp[sa[0]] = 0;

		for (int i = 1; i <= n; i++){

			tmp[sa[i]] = tmp[sa[i - 1]] + (cmp_sa(sa[i - 1], sa[i]) ? 1 : 0);

		}

		for (int i = 0; i <= n; i++){

			rk[i] = tmp[i];

		}

	}

}



void construct_lcp(char *s, int *sa, int *lcp)

{

	n = strlen(s);

	for (int i = 0; i <= n; i++) rk[sa[i]] = i;

	int h = 0;

	lcp[0] = 0;

	for (int i = 0; i < n; i++){

		int j = sa[rk[i] - 1];

		for (h ? h-- : 0; i + h < n&&j + h < n&&s[i + h] == s[j + h]; h++);

		lcp[rk[i] - 1] = h;

	}

}



void construct_rmq(int *lcp, int sizen)

{

	for (int i = 0; i <= sizen; i++) d[i][0] = lcp[i];

	for (int j = 1; (1 << j) <= sizen; j++){

		for (int i = 0; (i + (1 << j) - 1) <= sizen; i++){

			d[i][j] = min(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);

		}

	}

}



int rmq_query(int l, int r)

{

	if (l > r) swap(l, r); r -= 1;

	int k = 0; int len = r - l + 1;

	while ((1 << (k + 1)) < len) k++;

	return min(d[l][k], d[r - (1 << k) + 1][k]);

}



*/



char s[maxn];

ll num[maxn];



struct Matrix

{

	ll a[2][2];

	Matrix(){ memset(a, 0, sizeof(a)); }

}m;



Matrix operator * (const Matrix &a,const Matrix &b){

	Matrix ret;

	for (int i = 0; i < 2; i++){

		for (int j = 0; j < 2; j++){

			for (int k = 0; k < 2; k++){

				ret.a[i][j] += (a.a[i][k] * b.a[k][j]) % mod;

				ret.a[i][j] %= mod;

			}

		}

	}

	return ret;

}

Matrix operator ^ (Matrix a,ll n){

	Matrix ret;

	for (int i = 0; i < 2; i++) ret.a[i][i] = 1;

	while (n){

		if (n & 1) ret = ret*a;

		n >>= 1;

		a = a*a;

	}

	return ret;

}



ll cal(ll n)

{

	m.a[0][0] = 0; m.a[0][1] = 1;

	m.a[1][0] = 1; m.a[1][1] = 1;

	m = m^n;

	return m.a[0][1];

}



ll seed[maxn];

ll h[maxn];



ll get(int l, int r){

	return ((h[r] - h[l - 1] * seed[r - l + 1]%mod) + mod) % mod;

}





int getlcp(int x)

{

	if (get(1, 1) != get(x, x)) return 0;

	int l = 1,r = n - x + 1;

	while (l <= r){

		int m = (l + r) >> 1;

		if (get(1, 1 + m - 1) == get(x, x + m - 1)) l = m+1;

		else r = m-1;

	}

	return r;

}





int main()

{

	seed[0] = 1;

	for (int i = 1; i < maxn; i++){

		seed[i] = seed[i - 1] * step%mod;

	}

	while (~scanf("%s", s + 1))

	{

		n = strlen(s + 1); h[0] = 0;

		for (int i = 1; i <= n; i++){

			h[i] = (h[i - 1] * step + s[i]) % mod;

		}

		num[1] = n;

		for (int i = 2; i <= n; i++){

			num[i] = getlcp(i);

		}

		ll ans = 0;

		num[n + 1] = 0;

		for (int i = n; i >= 1; i--){

			num[i] += num[i + 1];

			ans += cal(num[i]);

			ans %= mod;

		}

		printf("%lld\n", ans);

	}

	return 0;

}