# hdu 4549 M斐波那契数列 (矩阵快速幂+费马小定理+指数循环节)

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

111qqz_指数循环节学习笔记

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

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

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;
}
``````