首页 > 代码库 > Acdreamoj1116(Gao the string!)字符串hash+二分+矩阵快速幂
Acdreamoj1116(Gao the string!)字符串hash+二分+矩阵快速幂
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] )
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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