HDU

Description

The Fibonacci sequence is the sequence of numbers such that every element is equal to the sum of the two previous elements, except for the first two elements f0 and f1 which are respectively zero and one.

What is the numerical value of the nth Fibonacci number?

 

Input

For each test case, a line will contain an integer i between 0 and 10 ⁸ inclusively, for which you must compute the ith Fibonacci number fi. Fibonacci numbers get large pretty quickly, so whenever the answer has more than 8 digits, output only the first and last 4 digits of the answer, separating the two parts with an ellipsis (“...”).

There is no special way to denote the end of the of the input, simply stop when the standard input terminates (after the EOF).

 

Sample Input

 0
1
2
3
4
5
35
36
37
38
39
40
64
65 

 

Sample Output

 0
1
1
2
3
5
9227465
14930352
24157817
39088169
63245986
1023...4155
1061...7723
1716...7565 

题意：求第n个斐波那契数的前4个和后4个

思路：对于前四个我们能够採用科学计数发的方式得到，

斐波那契数的通项公式是：f(n)=1/sqrt(5)(((1+sqrt(5))/2)^n+((1-sqrt(5))/2)^n)，对于40个后((1-sqrt(5))/2)^n能够忽略不计了，

后4个我们採用矩阵高速幂的方法获得，构造的矩阵是：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn = 10;
const int mod = 10000;

int cnt;
struct Matrix {
	ll v[maxn][maxn];
	Matrix() {}
	Matrix(int x) {
		init();
		for (int i = 0; i < maxn; i++) 
			v[i][i] = x;
	}
	void init() {
		memset(v, 0, sizeof(v));
	}
	Matrix operator *(Matrix const &b) const {
		Matrix c;
		c.init();
		for (int i = 0; i < cnt; i++)
			for (int j = 0; j < cnt; j++)
				for (int k = 0; k < cnt; k++)
					c.v[i][j] = (c.v[i][j] + (ll)v[i][k]*b.v[k][j]) % mod;
		return c;
	}
	Matrix operator ^(int b) {
		Matrix a = *this, res(1);
		while (b) {
			if (b & 1)
				res = res * a;
			a = a * a;
			b >>= 1;
		}
		return res;
	}
} a, b, tmp;

int main() {
	int f[40], n;
	f[0] = 0, f[1] = 1;
	for (int i = 2; i < 40; i++) 
		f[i] = f[i-1] + f[i-2];
	while (scanf("%d", &n) != EOF) {
		if (n < 40) {
			printf("%d
", f[n]);
			continue;
		}
		double k = log10(1.0/sqrt(5.0)) + (double)n * log10((1.0 + sqrt(5.0))/2.0);
	 	double num = k;
		num = k - (int)num;
		a.init();
		a.v[0][0] = a.v[0][1] = a.v[1][0] = 1;
		cnt = 2;
		tmp = a^n;
		printf("%d...%0.4d
", (int)(1000.0*pow(10.0, num)), tmp.v[0][1]%mod);
	}
}