POJ 3070 Fibonacci

最新推荐文章于 2019-03-26 00:15:50 发布

涅槃phy

最新推荐文章于 2019-03-26 00:15:50 发布

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分类专栏：快速幂

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本文介绍了一种高效计算斐波那契数列第n项最后四位数字的方法，利用矩阵快速幂技巧，解决了大规模数值计算的问题，并提供了一个完整的C语言实现示例。

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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 thelast 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 thenumber −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

999999999

1000000000

-1

Sample Output

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 0thpower gives the identity matrix:

题目简介：求Fibonacci的n项的最后四位上的数字。

方法：快速幂。注意特殊的n=0。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define P 10000

struct multiply
{
	int f[2][2];
};

struct multiply M(struct multiply a,struct multiply b)
{    
	struct multiply c; 
	int i, j, k;
	memset(c.f,0,sizeof(c.f)); 
	for(i = 0;i<2;i++)
	{
		for(j = 0;j<2;j++)
		{
			for(k = 0;k<2;k++)
			{
				c.f[i][j]=(c.f[i][j] + (a.f[i][k] * b.f[k][j]) % P ) % P;
			}
		}
	}
	return c;
};

struct multiply solve(struct multiply a,int x)
{
	if(x==1)
	{
		return a;
	}
	struct multiply b = solve(a,x/2);
	if(x&1)
	{
		return M(M(b, b), a);
	}
	else
		return M(b, b);
};

int main()
{
	long long int n;
	struct multiply F, ans;
	F.f[0][0] = 0;
	F.f[0][1] = 1;
	F.f[1][0] = 1;
	F.f[1][1] = 1;
	while(scanf("%lld",&n)&&n!=-1)
	{
		if(n==0)
		{
			printf("0\n");
			continue;
		}

		ans = solve(F,n);
		printf("%d\n",ans.f[1][0]);
	}
	return 0;
}