poj 3070 Fibonacci-白红宇

poj 3070 Fibonacci

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发布时间：2019-05-10

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Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9769		Accepted: 6959

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_n_{− 1} + F_n_{− 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

099999999991000000000-1

Sample Output

0346266875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

解题思路：单纯的快速矩阵幂

快速矩阵幂模板：

Matrix add(Matrix a,Matrix b){	Matrix ans;	for (int i=0;i<2;i++){		for (int j=0;j<2;j++){			ans.mat[i][j]=a.mat[i][j]+b.mat[i][j];			if (ans.mat[i][j]>=m){				ans.mat[i][j]%=m;			}		}	}	return ans;}Matrix mul(Matrix a,Matrix b){	Matrix ans;	for (int i=0;i<2;i++){		for (int j=0;j<2;j++){			ans.mat[i][j]=0;			for (int k=0;k<2;k++){				ans.mat[i][j]+=a.mat[i][k]*b.mat[k][j];				if (ans.mat[i][j]>=m){					ans.mat[i][j]%=m;				}			}		}	}	return ans;}Matrix Init(){	Matrix ans;	for (int i=0;i<2;i++){		for (int j=0;j<2;j++){			if (i==j)				ans.mat[i][j]=1;			else				ans.mat[i][j]=0;		}	}	return ans;}Matrix exp(Matrix a,int k){	Matrix ans=Init();	while (k){		if (k&1)			ans=mul(ans,a);		a=mul(a,a);		k>>=1;	}	return ans;}

参考代码：

#include 
   
    using namespace std;int m=10000;struct Matrix{	long long mat[2][2];};Matrix add(Matrix a,Matrix b){	Matrix ans;	for (int i=0;i<2;i++){		for (int j=0;j<2;j++){			ans.mat[i][j]=a.mat[i][j]+b.mat[i][j];			if (ans.mat[i][j]>=m){				ans.mat[i][j]%=m;			}		}	}	return ans;}Matrix mul(Matrix a,Matrix b){	Matrix ans;	for (int i=0;i<2;i++){		for (int j=0;j<2;j++){			ans.mat[i][j]=0;			for (int k=0;k<2;k++){				ans.mat[i][j]+=a.mat[i][k]*b.mat[k][j];				if (ans.mat[i][j]>=m){					ans.mat[i][j]%=m;				}			}		}	}	return ans;}Matrix Init(){	Matrix ans;	for (int i=0;i<2;i++){		for (int j=0;j<2;j++){			if (i==j)				ans.mat[i][j]=1;			else				ans.mat[i][j]=0;		}	}	return ans;}Matrix exp(Matrix a,int k){	Matrix ans=Init();	while (k){		if (k&1)			ans=mul(ans,a);		a=mul(a,a);		k>>=1;	}	return ans;}int main(){	Matrix a;	a.mat[0][0]=a.mat[0][1]=a.mat[1][0]=1;	a.mat[1][1]=0;	int n;	while (cin>>n&&n!=-1){		Matrix ans=exp(a,n);		cout<
    
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