Fibonacci

最新推荐文章于 2019-03-26 00:15:50 发布

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最新推荐文章于 2019-03-26 00:15:50 发布

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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

Sample Output

#include<cstdio>
#include<string.h>
#define MOD 10000
using namespace std;
struct Mat{
	long long a[5][5];
}A,B;
void init()
{
	A.a[1][1] = A.a[1][2] = A.a[2][1] = 1;A.a[2][2] = 0;
	B.a[1][1] = B.a[1][2] = B.a[2][1] = 1;B.a[2][2] = 0;
}
Mat Ma(Mat x,Mat y)
{
	Mat t;
	memset(t.a,0,sizeof(t.a));
	for(int i = 1; i <= 2; i++)
	for(int j = 1; j <= 2; j++){
		if(x.a[i][j] == 0) continue; 
		for (int q = 1 ; q <= 2 ; q++)  
             t.a[i][q] = (t.a[i][q] + x.a[i][j] * y.a[j][q] % MOD) % MOD;
        t.a[i][j] = t.a[i][j] % MOD;
	}
	return t;
}
void MatQuck(int k)
{
	while(k){
		if(k&1) 
		B = Ma(A,B);
		A = Ma(A,A);
		k >>= 1;
	}
}




int main()
{
	int f[5]={0,1,1,2,3};
	long long n;
	while(~scanf("%lld",&n)&&(n!=-1)){
		if(n<5) printf("%d\n",f[n]);
		else {
			init();
			MatQuck(n-4);
			long long sum = 0;
			for(int i = 1; i <= 2; i++)
			for(int j = 1; j <= 2; j++)
			sum = (sum + B.a[i][j]) % MOD;
			printf("%lld\n",sum);
		}
	}
 }