POJ - 3070 - Fibonacci （矩陣快速冪 + 斐波那契數列）

POJ - 3070 - Fibonacci （矩陣快速冪 + 斐波那契數列）

 

Fibonacci Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10096   Accepted: 7208

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

 

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通過這道題，練習了下矩陣快速冪的做法，還有學到了斐波那契數列的快速求法

 

AC代碼:

 

#include 
#include 
#include 
#include 
using namespace std;

const int MOD = 10000;

struct matrix {		//矩陣 
	int m[2][2];
}ans;

matrix base = {1, 1, 1, 0}; 

matrix multi(matrix a, matrix b) {	//矩陣相乘，返回一個矩陣 
	matrix tmp;
	for(int i = 0; i < 2; i++) {
		for(int j = 0; j < 2; j++) {
			tmp.m[i][j] = 0;
			for(int k = 0;  k < 2; k++)
				tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
		}
	}
	return tmp;
}

int matrix_pow(matrix a, int n) {	//矩陣快速冪，矩陣a的n次冪 
	ans.m[0][0] = ans.m[1][1] = 1;	//初始化為單位矩陣 
	ans.m[0][1] = ans.m[1][0] = 0;
	while(n) {
		if(n & 1) ans = multi(ans, a);
		a = multi(a, a);
		n >>= 1;
	}
	return ans.m[0][1];
}

int main() {
	int n;
	while(scanf("%d", &n), n != -1) {
		printf("%d\n", matrix_pow(base, n));
	}
	return 0;
}