[POJ 3070] Fibonacci【矩乘优化】

  • 2018-01-07
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Problem:

Time Limit: 1000MS Memory Limit: 65536K

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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

Solution:

很经典的矩乘优化题,而且出题人很良心地给出了解题思路温馨提示~~

如果直接暴力递推 Fibonacci 数列的第 n 项,时间复杂度为 O(Tn),显然 TLE 炸上天。。

You're too young ~~ too simple ~~ too naive ~~

所以引入一种高大上的优化方法,也就是出题人告诉我们的公式 2

我们只要先模拟矩阵乘法(不知道矩阵乘法?点此速成 .(i_i).),再写一个快速幂就行了。

Code: O(Ts3logn), s为矩阵阶数(=2) [144K, 16MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cassert>
#include<iostream>
#include<algorithm>
using namespace std;
const int MOD = 10000;

int n;

struct Matrix{
	int ele[3][3], size;
	
	inline void setIdentity(int idensize){
		memset(ele, 0, sizeof(ele));
		size = idensize;
		for(register int i = 1; i <= size; i++)
			ele[i][i] = 1;
	}
	
	inline Matrix operator * (const Matrix &mat2) const {
		assert(size == mat2.size);  // Or the multiplication isn't defined
		Matrix res; res.size = size;
		for(register int i = 1; i <= size; i++)
			for(register int j = 1; j <= mat2.size; j++){
				res.ele[i][j] = 0;
				for(register int k = 1; k <= size; k++)
					res.ele[i][j] = (res.ele[i][j] + ele[i][k] * mat2.ele[k][j]) % MOD;
			}
		return res;
	}
	
	inline Matrix pow(int ex){
		Matrix bas = *this, res; res.setIdentity(size);
		while(ex){
			if(ex & 1) res = res * bas;
			ex >>= 1, bas = bas * bas;
		}
		return res;
	}
} Fib;

int main(){
	while(scanf("%d", &n) != EOF && n != -1){
		if(n < 2){
			if(n) puts("1"); else puts("0");
			continue;
		}
		Fib.size = 2;
		Fib.ele[1][1] = 1, Fib.ele[1][2] = 1;
		Fib.ele[2][1] = 1, Fib.ele[2][2] = 0;
		Fib = Fib.pow(n - 1);
		printf("%d\n", Fib.ele[1][1]);
	}
	return 0;
}

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OPEN AT 2017.12.10

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