POJ_3070_Fibonacci

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15397		Accepted: 10803

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:

.

Source

Stanford Local 2006

• 普通的常係數線性齊次遞推
• 建立矩陣快速冪加速

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
typedef long long           LL ;
typedef unsigned long long ULL ;
const int    maxn = 1000 + 10  ;
const int    inf  = 0x3f3f3f3f ;
const int    npos = -1         ;
const int    mod  = 1e4        ;
const int    mxx  = 100 + 5    ;
const double eps  = 1e-6       ;

struct Matrix{
	int n;
	int a[mxx][mxx];
	void reset(int x){
		n=x;
		memset(a,0,sizeof(a));
	}
	Matrix operator * (const Matrix X){
		Matrix Y;
		Y.reset(n);
		for(int k=0;k<n;k++)
			for(int i=0;i<n;i++)
				if(a[i][k])
					for(int j=0;j<n;j++)
						if(X.a[k][j])
							Y.a[i][j]=
								((Y.a[i][j] + (a[i][k]*X.a[k][j])%mod)%mod + 2*mod)%mod;
		return Y;
	}
};
Matrix pow_mod(Matrix X, int k){
	int n=X.n;
	Matrix Y;
	Y.reset(X.n);
	for(int i=0;i<n;i++)
		Y.a[i][i]=1;
	while(k){
		if(1&k)
			Y=Y*X;
		X=X*X;
		k>>=1;
	}
	return Y;
}
int cal(int k){
	if(k<2){
		return k;
	}else{
		Matrix Y;
		Y.reset(2);
		Y.a[0][0]=1; Y.a[0][1]=1;
		Y.a[1][0]=1; Y.a[1][1]=0;
		Y=pow_mod(Y,k-1);
		return Y.a[0][0];
	}
}
int n, ans, f[2]={0,1};
int main(){
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	while(~scanf("%d",&n) && -1!=n){
		printf("%d\n",cal(n));
	}
	return 0;
}