首页 > 代码库 > poj 3070 Fibonacci

poj 3070 Fibonacci

2024-08-10 12:53:14   211人阅读

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

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn ? 1 + Fn ? 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

技术分享.

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

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

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Hint

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

技术分享.

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

技术分享.



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 <iostream>
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<<ans.mat[0][1]%m<<endl;
	}
	return 0;
}





poj 3070 Fibonacci


联系
我们