fibonacci principle

One: Recursive implementation Using the formula F[n]=f[n-1]+f[n-2], recursive calculation in turn, the recursive end condition is f[1]=1,f[2]=1. Two: Array implementation Spatial complexity and time complexity are all 0 (n), which is more efficient and faster than recursion. Three:vectorTime complexity is 0 (n), time complexity is 0 (1), is not aware of vector efficiency is not high, of course, Vector has its own properties will occupy resources. Four:queueOf course the queue is more suitable th

Approximate test instructions: Enter two nonnegative integers, a, B, and positive integer n. Calculates f (a^b)%n. Among them F[0]=f[1]=1, F[i+2]=f[i+1]+f[i]. That is to calculate the large Fibonacci number and then take the modulus.First saw the big Fibonacci number, think of the matrix fast power, output wait a few seconds before output, will definitely time out. Because all calculations are to be modulo,

Fibonaccitime limit: 1000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others) Total submission (s): 3569 accepted submission (s): 1627 Problem description is coming in 2007. After one year of practice in 2006, zouyu, a mathematical prodigy, finally ranked 0 to 100000000 In the Fibonacci series. (F [0] = 0, F [1] = 1; F [I] = f [I-1] + F [I-2] (I> = 2 )) all the values are backed up. Next, codestar decided to test him, so every time he a

The beauty of programming: 2.9 Fibonacci series, 2.9 Fibonacci The Fibonacci series is a classic example of recursion when we are learning the C language. Of course, there will also be an example of the tower of legends. This problem is defined as follows: 0 (x F (x) = 1 (x = 1) F (x-1) + f (x-2) (x> 1) After seeing this recursive formula, we can easily write th

10236-the Fibonacci Primes Time limit:3.000 seconds Http://uva.onlinejudge.org/index.php?option=com_onlinejudgeItemid=8category=24page=show_problem problem=1177 Note that the description of this topic is different from the description of Fibonacci prime on Wikipedia. The above-highlighted text shows that Fibonacci Prime can be screened with a similar number

10229-modular Fibonacci Time limit:3.000 seconds Http://uva.onlinejudge.org/index.php?option=com_onlinejudgeItemid=8category=115page=show_ problemproblem=1170 The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, ...) are defined by the recurrence: F0 = 0 F1 = 1 Fi = Fi-1 + Fi-2 for i>1 Write a program which calculates Mn = Fn mod 2m for given pair of N and M. 0 Input and Output Input consists of several lines

Luogp1962 Fibonacci series, P1962 Fibonacci SeriesBackground As we all know, the Fibonacci series is a series that meets the following characteristics: • F (1) = 1 • F (2) = 1 • F (n) = f (n-1) + f (n-2) (n ≥ 2 and n is an integer)Description Find the value of f (n) mod 1000000007.Input/Output Format Input Format: · Row 1st: an integer n Output Format: Row 3:

First: Using a For loopThe use of the For loop, does not involve functions, but this method for my small white is a good understanding of the function of a more abstract ...1 >>> fibs = [0,1]2 for in range (8):3 Fibs.append (Fibs[-2] + fibs[-1])45 >>> fibs6 [0, 1, 1, 2 , 3, 5, 8, 13, 21, 34]Or, enter a dynamic length:1fibs = [0,1]2num = input ('howmany Fibonacci numbers does you want? ' )3 for in range (Num-2):4 fibs.append (Fibs[-2] + Fi

Topic Transfer: UVA-10229Idea: The simple matrix quickly powers the Fibonacci sequence, then notice that the intermediate result can explode int, because 2^19 has more than 500,000AC Code:#include Uva-10229-modular Fibonacci (Matrix fast Power + Fibonacci)

The sum of the First n digits of Fibonacci and the n digits of Fibonacci Public class Fei_bo_na_qi {Public static void main (String [] args) {// main method, program entryInt I = 10; // declare an int type variable I, and assign a value of 10Int a = 0; // declare an int data type variableFor (int j = I; j> = 1; -- j) {// for Loop: j = I = 10, j> = 1A + = m1 (j); // call the m1 method and assign the value of

Question Link Give N and M, and find F (n) % m. f (x) is the Fibonacci sequence. Idea: Because N is big, it is very likely that it will use the formula to calculate it, so we can use the Matrix to quickly solve the power. | (1, 1), (1, 0) | * | f (n-1), F (n-2) | = | f (N), F (n-1) |, so f (N) equivalent to | F (1), F (0) | multiplied by N-1 | (1, 1), (1, 0) |. Code: #include Uva10299-modular Fibonacci

Logu P1306 Fibonacci common count, p1306 FibonacciDescription You should be familiar with the series of Fibonacci:, 13 ~~~ But now there is a very simple question: what is the maximum number of public appointments between the n and m items?Input/Output Format Input Format: Two positive integers n and m. (N, m Note: The data is large. Output Format: The maximum number of common Fn and Fm. Since reading b

Package C4;public class Fibtest {public static void main (String []args) {Long begin = System.currenttimemillis (); System.out.println (FIB); Long end = System.currenttimemillis (); System.out.println (End-begin);} public static long fib (int n) {if (nResult isS1:1 s2:2s1:2 s2:3s1:3 s2:5s1:5 s2:8s1:8 s2:13s1:13 s2:21s1:21 s2:34551　　　Fibonacci sequence Non-recursive algorithm (Fibonacci)

Js Fibonacci series implementation, js Fibonacci Series1. Recursion Function fib (n) {if (n = 1 | n = 2) {return 1;} return fbnq (n-1) + fbnq (n-2 );} fbnq (10); // 55 The time complexity is O (2 ^ n), and the space complexity is O (n) 2. Non-recursion Function fb (n) {var res = [1, 1]; if (n = 1 | n = 2) {return 1 ;}for (var I = 2; I The time complexity is O (n), and the space complexity is O (

Calculate the sum of the First n items of the Fibonacci fractional sequence (N is a constant, and the Fibonacci fractional sequence is 2/1, 3/2, 5/3, 8/5 ,...). # Include Stdio. h > # Include Conio. h > Void Main (){ Int I, N; Float F1 = 1 , F2 = 2 , F, Sum = 0 ;Scanf ( " % D " , N ); For (I = 0 ; I N; I ++ ){Sum + = F2 /

The meaning is to use matrix multiplication to find the last four digits of the nth of the Fibonacci series. If the last four digits are all 0, the output is 0. Otherwise Remove the leading 0 from the last four digits of the output... ... Yes... ... Output FN % 10000. The question is so clear .. I am still trying to find the last four digits of % and/and determine whether all the values are 0 and whether the output values are 0. Remove the leading 0.

Calculate the nth entry of a Fibonacci series. The upper limit of n is 10000. Solution: Because the upper limit of this question is 10000, indentation is required for high precision. # Include

Question: Fibonacci series, F [0] = 0, F [1] = 1, F [N] = f [N-2] + F [n-1], n> 1 Given N (0 Solution: N is too large, so we cannot use the cyclic recursive Calculation of O (n ). Conclusion: For matrices | 1 1 |, and matrices | FnFn-1 |, after the two are multiplied to obtain a new matrix | Fn + 1FN|, FN indicates the nth Number of fiber ACCI | 1 0 || Fn-1 Fn-2 || FnFn-1 Therefore, we can use the rapid power modulo + matrix multiplication of O (l

--Using SQL Server to calculate Fibonacci lawsDECLARE @number intDECLARE @A intDECLARE @B intDECLARE @c intSet @a=1Set @b=2Set @Number =3Select @[email Protected][email protected]while (@Number BeginSet @[email Protected][email protected]if (@ @ERROR Goto ErrorHandlerPrint N ' +convert (varchar), @number) +n ' number is ' +convert (varchar), @c)Set @[email protected]Set @[email protected]Set @[email protected]+1EndErrorHandler:Print N ' +convert (varc

clever Python programmer: fib=lambda n,x=0,y=1:x if not n else f(n-1,y,x+y) Note:Similar to the above example, the basic logic is tail recursion. The main difference is that the default parameters and ternary operators provided by Python are used to simplify the code to one line. As for the default parameters, students who have learned C ++ know this stuff. C #4.0 also introduces this stuff. 8) Python programmers who have just finished linear algebra: def fib(n): def m1(a,b): m=[[],[]] m[0].a