POJ3070Fibonacci(矩阵快速幂)(AC)_「poj3070」fibonacci-优快云博客

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Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16150		Accepted: 11354

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

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

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

int N = 0;//Int型够了
int binary[32]; //int最大32位，0~31

//矩阵快速幂 这种类型的题目没有做过
//矩阵的快速幂是用来高效地计算矩阵的高次方的
//简单的说就是把幂分成二级制 例如 A^19  =>  （A^16）*（A^2）*（A^1） A^4能通过(A^2)*(A^2)得到，A^8又能通过(A^4)*(A^4)得到
//网上找到的一段解释
//矩阵乘法的优越性究竟体现在哪里呢。其实，矩阵乘法只是体现了我们从之前求的数到现在要求的数的递推过程，就是说矩阵乘法可以完成多个元素的递推。
//不过这个我们用普通的递推就可以实现的啊~~认真想想我们就能发现，我们在矩阵乘法的过程中把上见面的A矩阵自己相乘了很多遍。就是说，我们可以求A矩阵的幂最后乘上B矩阵，既然要求幂，
//矩阵乘法满足结合律，那么我们就可以用快速幂啦~~矩阵乘法的优越性就体现在这里：在递推过程变成不断乘以一个矩阵，然后用快速幂快速求得从第一个到第n个的递推式，这样子就可以在短时间内完成递推了

//题目中已经告知公式
//因此不需要自己推导公式了

typedef struct matrix
{
	int m[2][2];
};

//最后算N-1次幂就可以得到Fn
matrix ma =  //题目中的矩阵，最后要算的就是矩阵的n次幂
{
	1, 1,
	1, 0
};


matrix azermi =  //0次幂
{
	1, 0,
	0, 1
};

matrix multi(matrix a, matrix b)
{
	matrix tmp;
	int i = 0;
	int j = 0;
	int k = 0;
	for (i = 0; i < 2;i++)
	{
		for (j = 0; j < 2; j++)
		{
			tmp.m[i][j] = 0;
			for (k = 0; k < 2; k++)
			{ 
				tmp.m[i][j] += (a.m[i][k] * b.m[k][j])% 10000; //只留最后四位
				tmp.m[i][j] %= 10000;
			}
		}
	}
	return tmp;
}

matrix getans()
{
	matrix tmp = azermi;
	matrix matmp = ma;

	while (N>0)
	{
		//如果N >0,则开始算，如果N是奇数，那必定最后一位是1
		if (N&1)
		{
			tmp = multi(tmp, matmp);
		}

		//举例10的9次方，= 10的8次方*10
		//1: tmp = 1*10 (9&1 = 1) 9/2 = 4 同时做(10*10）
		//2: (10*10）*(10*10) (4 &1 != 1) 4/2 = 2
		//3: (10*10*10*10）*(10*10*10*10) (2 &1 != 1) 2/2 = 1
		//4: 10* (10*10*10*10*10*10*10*10) (1 &1 != 1) 1/2 = 0
		matmp = multi(matmp, matmp); //这步是为什么 相当于幂相乘了，因为相当于自己又乘自己了//这个要好好体会下
		
		N >>= 1;//右移1位，相当于/2
	}

	return tmp;
}



int main()
{
	int i = 0;
	int tmpN = 0;
	matrix ans;

	freopen("input.txt", "r", stdin);
	while (1 == scanf("%d",&N) &&(-1 != N))
	{
		if (0 == N)
		{
			printf("0\n");
			continue;
		}

		/*
		//如果从N =1开始，那这个矩阵是
		tmpN = N;

		//转换成二进制用位运算
		for (i = 0; i < 32;i++)
		{
			binary[i] = tmpN & 1;
			tmpN = tmpN >> 1;
		}
		N = N;*/

		ans = getans();
		printf("%d\n", ans.m[0][1]);
	}
	
	return 0;
}