hdu 4549 M斐波那契数列 (矩阵快速幂+费马小定理+指数循环节)

Posted by 111qqz on Wednesday, October 26, 2016

TOC

题意:M斐波那契数列F[n]是一种整数数列,它的定义如下:

F[0] = a F[1] = b F[n] = F[n-1] * F[n-2] ( n > 1 )

现在给出a, b, n,你能求出F[n]的值吗?

思路:观察发现。。。F[n] = a^(fib(n-1)) * b ^ (fib(n))

此处要用到指数循环节的知识:

111qqz_指数循环节学习笔记

a^n ≡ a^(n % Phi(M) + Phi(M)) (mod M) (n >= Phi(M))

然后 因为1000000007是质数,对于任意的x,有gcd(x,1000000007) = 1,所以可以结合费马小定理化简上式:

a^n ≡ a^(n%(m-1)) * a^(m-1)≡ a^(n%(m-1)) (mod m)

记得特判一下n为0和1的情况。

xiaodingli

/* ***********************************************
Author :111qqz
Created Time :Wed 26 Oct 2016 09:16:22 AM CST
File Name :code/hdu/4549.cpp
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const LL mod = 1E9+7;
LL a,b,n;
struct Mat
{
    LL mat[2][2];
    void clear()
    {
    ms(mat,0);
    }
}M,M1;
Mat operator * ( Mat a,Mat b)
{
    Mat res;
    res.clear();
    for ( int i = 0 ; i < 2 ; i ++)
    for ( int j = 0 ; j < 2 ; j++)
        for ( int k = 0 ; k < 2 ; k++)
        res.mat[i][j] = (res.mat[i][j] + a.mat[i][k] * b.mat[k][j] ) % (mod - 1);
    return res;
}
Mat operator ^ (Mat a,LL b)
{
    Mat res;
    res.clear();
    for ( int i = 0 ; i < 2 ; i++) res.mat[i][i] = 1;
    while (b>0)
    {
    if (b&1) res = res * a;
    b = b >> 1LL;
    a = a * a ;
    }
    return res;
}
LL ksm( LL a,LL b)
{
    LL res = 1LL;
    while (b>0)
    {
    if (b&1){
        res = (res * a) %mod;
    }
    b = b >> 1LL;
    a = ( a * a ) % mod;
    }
    return res;
}
void init()
{
    M.clear();
    M.mat[0][1] = 1;
    M.mat[1][0] = 1;
    M.mat[1][1] = 1;
    M1.clear();
    M1.mat[0][0] = 0;
    M1.mat[1][0] = 1;
}
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif
    while (~scanf("%lld %lld %lld",&a,&b,&n))
    {
        if (n==0)
        {
        printf("%lld\n",a);
        continue;
        }
        if (n==1)
        {
        printf("%lld\n",b);
        continue;
        }
        init();
        Mat ans;
        ans.clear();
        ans = (M ^(n-1))*M1;
        LL x = ans.mat[0][0];
        LL y = ans.mat[1][0];
        LL ret =(ksm(a,x)*ksm(b,y))%mod;
        printf("%lld\n",ret);
    }
  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

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