矩阵快速幂

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13372		Accepted: 9517

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

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

.

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

.

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <iostream>

using namespace std;

struct matrix{
	int arr[2][2];
};

matrix base,ans;

matrix matrix_multip(matrix a,matrix b){	
	matrix tmp;
	for(int i=0; i<2; i++){
		for(int j=0; j<2; j++){
			tmp.arr[i][j] = 0;
			for(int k=0; k<2; k++)
				tmp.arr[i][j] = (tmp.arr[i][j]+a.arr[i][k]*b.arr[k][j])%10000;
		}
	}
	return tmp;
}

void matrix_pow(matrix a,int n){
	while(n){
		if(n%2)
			ans = matrix_multip(ans,a);
		a = matrix_multip(a,a);
		n /= 2;
	}
}


int main(){
	base.arr[0][0] = base.arr[0][1] = base.arr[1][0] = 1;
	base.arr[1][1] = 0;

	int n;
	while(~scanf("%d",&n) && n!=-1){
		ans.arr[0][0] = ans.arr[1][1] = 1;
		ans.arr[0][1] = ans.arr[1][0] = 0;
		matrix_pow(base,n);
		printf("%d\n",ans.arr[0][1]);
	}
	
	return 0;
}