HDU 1568 Fibonacci (number theory)

Fibonacci
Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 3654 Accepted Submission (s): 1671


Problem Description arrived in 2007. After 2006 years of cultivation, mathematical prodigy Zouyu finally put 0 to 100000000 of Fibonacci series
The values (f[0]=0,f[1]=1;f[i] = f[i-1]+f[i-2] (i>=2) are all backed down.
Next, Codestar decided to test him, so each asked him a number, he will say the answer, but some numbers are too long. So the provision of more than 4 people as long as the first 4 can be said, but Codestar himself cannot remember. So he decided to write a program to test whether Zouyu said it was correct.
Input inputs several digits n (0 <= n <= 100000000), one row per digit. Read the end of the file.
Output output F[N] The first 4 digits (if less than 4 digits, all outputs).
Sample Input

0 1 2 3 4 5 35 36 37 38 39 40
Sample Output

0 1 1 2 3 5 9227 1493 2415 3908 6324 1023

Analysis: Number theory, IQ is not enough, the online great God of the arrangement of ideas briefly.

The formula for the Fibonacci sequence is used.

First look at the properties of logarithms, Loga (b^c) =c*loga (b), Loga (b*c) =loga (b) +loga (c);
Suppose given a number of 10234432, then log10 (10234432) =log10 (1.0234432*10^7) =log10 (1.0234432) +7;

LOG10 (1.0234432) is the small part of log10 (10234432).

LOG10 (1.0234432) =0.010063744
10^0.010063744=1.023443198
Then it's obvious to take a few.
First take the logarithm (to 10), and then get the result of the decimal part Bit,pow (10.0,bit) Later if the answer is still <1000 then always multiply by 10.
Note that I first processed the 0~20 item for ease of handling ~

This question takes the formula of the sequence: an= (1/√5) * [((1+√5)/2) ^n-((1-√5)/2) ^n] (n=1,2,3 ...)

The logarithm is taken out

LOG10 (AN) =-0.5*log10 (5.0) + ((double) n) *log (f)/log (10.0) +LOG10 (n ((1-√5)/(1+√5)) ^n) where f= (sqrt (5.0) +1.0)/2.0;
LOG10 ((1-√5)/(1+√5) ^n)->0
So it can be written as log10 (AN) =-0.5*log10 (5.0) + ((double) n) *log (f)/log (10.0);
Finally take its fractional part.

AC Code:

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
int f[21] = {0, 1, 1};

int main ()
{
//    freopen ("In.txt", "R", stdin);
	int n;
	for (int i = 2; i < ++i)
		f[i] = f[i-1] + f[i-2];
	while (scanf ("%d", &n)! = EOF)
	{
		if (n <=)
		{
			printf ("%d\n", F[n]);
			Continue;
		}
		else
		{
			Double temp = -0.5 * LOG (5.0)/log (10.0) + ((double) n) * Log ((sqrt (5.0) +1.0)/2.0)/log (10.0);
			Temp-= Floor (temp);
			temp = POW (10.0, temp);
			while (temp < *=)
				temp;
			printf ("%d\n", (int) temp);
		}
	}
	return 0;
}

Woo-woo ~ ~ ~ Number number problem, really bad math, flawed ah ...