Acdreamoj1116（Gao the string!）字符串hash+二分+矩阵快速幂

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Problem Description

give you a string, please output the result of the following function mod 1000000007

n is the length of the string

f() is the function of fibonacci, f(0) = 0, f(1) = 1...

a[i] is the total number of times any prefix appear in the suffix s[i....n-1].

(the prefix means s[0...i] )

解法：如果知道了num[i]表示i开始的后缀s[i....n]跟前缀s[1...]之间的公共的前缀，那么以i开头的后缀中就匹配了num[i]个前缀了

所以i这个后缀出现的前缀的数量实际上就是num[i] + num[i+1] + .. num[n]. 求出来之后快速幂求斐波那契数列相应项大小即可。求lcp的时候是二分+hash；字符串hash中，seed为31（java库源码中是这个数，应该是效果比较好的）

代码：

/******************************************************* author:xiefubao*******************************************************/#pragma comment(linker, "/STACK:102400000,102400000")#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <vector>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <string.h>//freopen ("in.txt" , "r" , stdin);using namespace std;#define eps 1e-8#define zero(_) (abs(_)<=eps)const double pi=acos(-1.0);typedef long long LL;const int Max=100010;const int INF=1000000007;const int hashseed=31;LL seed[Max];LL has[Max];char s[Max];LL num[Max];int len=0;struct matrix{    LL num[2][2];    matrix()    {        memset(num,0,sizeof num);    }};matrix operator*(const matrix& a,const matrix& b){    matrix ans;    for(int i=0;i<2;i++)        for(int j=0;j<2;j++)        for(int k=0;k<2;k++)        ans.num[j][k]+=(a.num[j][i]*b.num[i][k])%INF;    return ans;}matrix operator^(matrix a,LL n){    matrix ans;    ans.num[0][0]=1;    ans.num[1][1]=1;    while(n)    {        if(n&1)        {            ans=ans*a;        }        a=a*a;        n>>=1;    }    return ans;}LL getans(LL t){    if(t==0)        return 0;    if(t<=2)        return 1;    matrix tool;    tool.num[0][0]=1;    tool.num[0][1]=1;    tool.num[1][0]=1;    tool=tool^(t-1);    return tool.num[0][0]%INF;}void init(){    seed[0]=1;    for(int i=1; i<Max; i++)        seed[i]=(seed[i-1]*hashseed)%INF;}LL Hash(int i,int L){    return ((has[i]-has[i+L]*seed[L])%INF+INF)%INF;}bool OK(int i,int l){    return Hash(0,l)==Hash(i,l);}void getnum(int i){    int left=i,right=len-1;    while(left<=right)    {        int middle=(left+right)/2;        if(OK(i,middle-i+1))            left=middle+1;        else            right=middle-1;    }    num[i]=right-i+1;}void makehash(){    for(int i=len-1;i>=0;i--)    {        has[i]=(has[i+1]*hashseed+s[i])%INF;    }    num[0]=len;    for(int i=1;i<len;i++)    {        getnum(i);    }}int main(){    init();   while(scanf("%s",s)==1)   {       memset(has,0,sizeof has);       len=strlen(s);       makehash();       for(int i=len-1;i>=0;i--)        num[i]+=num[i+1];       LL ans=0;       for(int i=0;i<len;i++)        ans=(ans+getans(num[i]))%INF;       cout<<ans<<'\n';   }    return 0;}

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