fibonacci

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 7 years ago . Given a series Fib(1) * Fib (n+2) + Fib(2) * Fib(n+1) + Fib(3) * Fib(n) + ...... + Fib(n-1) * Fib(4) or Summation Fib(x) * Fib (n-x+3) where x varies from 1 to n-1 where Fib(n) is nth number of Fibonacci series to evaluate this series Fib(n) can be calculated using matrix exponentiation . But the complexity for

问题 I am working on a python program to calculate numbers in the Fibonacci sequence . Here is my code: import math def F(n): return ((1+math.sqrt(5))**n-(1-math.sqrt(5))**n)/(2**n*math.sqrt(5)) def fib(n): for i in range(n): print F(i) My code uses this formula for finding the Nth number in the Fibonacci sequence: This can calculate many of the the numbers in the Fibonacci sequence but I do get overflow errors. How can I improve this code and prevent overflow errors? Note: I am using python 2.7.

问题 I'm a newbie in writing assembly code and I need some help. My task is to write a program in NASM (on Linux), which computes n-th Fibonacci number, where n is read from STDIN with read syscall and converted to int/long with C atoi/atol. The calculated number will be written to STDOUT (I can use C printf). I managed to write the working 32-bit code and I'm stuck at converting it to 64-bit (using 64-bit registers, 64-bit long int). I tried to do it naively (changing eax -> rax, esp -> rsp and

问题 I'm struggling a bit with dynamic programming. To be more specific, implementing an algorithm for finding Fibonacci numbers of n. I have a naive algorithm that works: int fib(int n) { if(n <= 1) return n; return fib(n-1) + fib(n-2); } But when i try to do it with memoization the function always returns 0: int fib_mem(int n) { if(lookup_table[n] == NIL) { if(n <= 1) lookup_table[n] = n; else lookup_table[n] = fib_mem(n-1) + fib_mem(n-2); } return lookup_table[n]; } I've defined the lookup

问题 This is my first question here so sorry for any mistakes :S. I have recently picked up python, and I have made some very simple text based application. Now I tried to make one with a proper GUI. I have the code bellow. I have made the GUI, and it's working fine. Apart of a little mistake. The idea is that the user enters a number and the app will return a Fibonacci number that lies at the same position in the sequence as specified by the user. But when I try it out, all that is shown is the

问题 I'm trying to filter a list that contains the Fibonacci numbers. What I need is only odd numbers, and less or equal than N . This is what I have so far: fib n | n == 0 = 0 | n == 1 = 1 | otherwise = fib (n-1) + fib (n-2) fibs n = [a | a <- [fib x | x <- [1..]], odd a, a < n] That will give me what I want, but at the same time that solution won't work because I don't know how to stop retrieving elements from fib function. Of course, that is because of x <- [1..] . I have thought about two

问题 I am new to Prolog and was tasked with a Fibonnaci predicate fib( N, F) where N is the number in sequence, and F is the value. What I came up with does not work, but the solution I found seems identical to me... I cannot understand the difference. My version: /* MY VERSION, DOES NOT WORK */ fib( 0, 0). fib( 1, 1). fib(N,F) :- N > 1, fib(N-1,F1), fib(N-2,F2), plus(F1,F2,F). The working version: /* FOUND SOLUTION, DOES WORK */ fib( 0, 0). fib( 1, 1). fib(N,F) :- N > 1, N1 is N-1, N2 is N-2, fib

问题 The code below is able to determine the correct sequence up to a point namely 70 using the data type unsigned long long . I know the sequence can become large thus I mod 10,000 the results. I want to determine the nth term for 10,000 using the best data type or improve the algo to calculate the nth term. #define MOD %10000 unsigned long long calc(long nth) { return (pow( 1 + sqrt(5), nth ) - pow( 1 - sqrt(5), nth )) / (pow(2.0, nth)*(sqrt(5))); } int main() { long t, nth; for (std::cin>>t; t-

问题 Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... I made the program but my answer doesnt match. #include<stdio.h> int main() { long unsigned int i,sum=0,x=1,y=2,num; for(i=0;i<4000000;i++) { num=x+y; if(i%2==0) sum+=num; x=y; y=num; } printf("%lu\n",sum); getchar(); return 0; } 回答1: Three problems I can see: You should start with x = 1, y = 1 , since otherwise you

问题 I have tried to find examples of tail recursion and I really don't see the difference between regular and tail. If this isn't tail recursion, can someone tell me why its not? public static long fib(long index) { // assume index >= 0 if (index == 0) // Base case return 0; else if (index == 1) // Base case return 1; else // Reduction and recursive calls return fib(index - 1) + fib(index - 2); } // end of method fib(long index) 回答1: No, the method in the question does not use a tail recursion. A