POJ 3070 Fibonacci(矩阵快速幂)

Fibonacci

http://poj.org/problem?id=3070

Time Limit:  1000MS

Memory Limit: 65536K

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

这题让我见识到了矩阵的巨大威力。。（稍微回忆了下矩阵在18~19世纪的发展）

完整代码：

/*0ms,164KB*/

#include <cstdio>
#include <cstring>
const int f0 = 0;
const int f1 = 1;
const int mod = 10000;

int n;

struct mat
{
	int v[2][2];
	mat()
	{
		memset(v, 0, sizeof(v));
	}
} matrix, tmp;

mat operator *(mat a, mat b)
{
	mat c;
	for (int i = 0; i < 2; i++)
		for (int j = 0; j < 2; j++)
			for (int k = 0; k < 2; k++)
				c.v[i][j] = (c.v[i][j] + a.v[i][k] * b.v[k][j]) % mod;
	return c;
}

mat matrix_qiuck_pow(mat a, int k)
{
	mat b;
	b.v[0][0] = 1;
	b.v[1][1] = 1;
	while (k)
	{
	    ///位运算求快速幂
	    ///由于十进制化二进制的过程是先算出低位的，刚好符合我们计算快速幂的方法！
		if (k & 1)
			b = b * a;
		k >>= 1;
		a = a * a;
	}
	return b;
}

int main()
{
    /*
    我是按
        F(n-1) Fn
        Fn     F(n+1)
    来排的
    */
	tmp.v[0][0] = 0;
	tmp.v[0][1] = 1;
	tmp.v[1][0] = 1;
	tmp.v[1][1] = 1;
	while (scanf("%d", &n), ~n)
	{
		matrix = tmp;
		if (n == 0)
			puts("0");
		else
			printf("%d\n", matrix_qiuck_pow(matrix, n - 1).v[1][1]);
	}
	return 0;
}