fibonacci

问题 I write this code for show fibonacci series using recursion.But It not show correctly for n>43 (ex: for n=100 show:-980107325). #include<stdio.h> #include<conio.h> void fibonacciSeries(int); void fibonacciSeries(int n) { static long d = 0, e = 1; long c; if (n>1) { c = d + e; d = e; e = c; printf("%d \n", c); fibonacciSeries(n - 1); } } int main() { long a, n; long long i = 0, j = 1, f; printf("How many number you want to print in the fibonnaci series :\n"); scanf("%d", &n); printf("

问题 I got a question to solve but it is giving me errors: 2009-EE-182-Part2.c: In function ‘main’: 2009-EE-182-Part2.c:35:13: error: expected identifier or ‘(’ before ‘->’ token 2009-EE-182-Part2.c:40:22: error: expected identifier or ‘(’ before ‘->’ token 2009-EE-182-Part2.c:41:14: error: expected identifier or ‘(’ before ‘->’ token 2009-EE-182-Part2.c:42:22: error: expected expression before ‘shared_data’ 2009-EE-182-Part2.c:44:15: error: expected identifier or ‘(’ before ‘->’ token 2009-EE-182

问题 As we all know, the simplest algorithm to generate Fibonacci sequence is as follows: if(n<=0) return 0; else if(n==1) return 1; f(n) = f(n-1) + f(n-2); But this algorithm has some repetitive calculation. For example, if you calculate f(5), it will calculate f(4) and f(3). When you calculate f(4), it will again calculate both f(3) and f(2). Could someone give me a more time-efficient recursive algorithm? 回答1: One simple way is to calculate it iteratively instead of recursively. This will

问题 I have a method that takes n and returns nth Fibonacci number. Inside the method implementation I use BigDecimal to get the nth Fibonacci number then I use method toBigInteger() to get the number as a BigInteger object and that's surely because I am working with huge numbers in my application. I keep getting correct results until I pass 1475 as an argument for my method. I get NumberFormatException: Infinite or NaN in this case without any clear reason for me. Could you please explain me why

问题 var lookup = {}; function memoized(n) { if(n <= 1) { return 1; } if(lookup[n]) { return lookup[n]; } lookup[n] = n * memoized(n - 1); return lookup[n]; } vs. function fact(n) { if(n <= 1) { return 1; } return n * fact(n-1); } If we call fact(3) With the second method we get --> 3 * (2 * (1)) What is the efficiency gain of storing the result in a hash. Is it only for subsequent calls to the same function? I can't see how you would gain anything if you are only calling the function once. With

问题 I have been trying to come out with a solution regarding the problem of finding the last digit of the sum of large n Fibonacci series. I have been able to pass several test cases with large n. But I'm stuck at the following case where n = 832564823476. I know it can be solved using Pisano's period but I'm unable to come out with a efficient algo. Any help would be great. Thanks. My code that I have implemented is as follows- #include <iostream> using namespace std; int calc_fib(int n) { int

问题 In the current exercise assignment of the Functional Programming course I'm doing, we've got to make a memoized version of a given function. To explain memoization, the following example is given: fiblist = [ fibm x | x <- [0..]] fibm 0 = 0 fibm 1 = 1 fibm n = fiblist !! (n-1) + fiblist !! (n-2) But I don't fully understand how this works. Let's call fibm 3 . fibm 3 --> fiblist !! 2 + fibList 1 --> [fibm 0, fibm 1, fibm 2] !! 2 + [fibm 0, fibm 1] !! 1 --> fibm 2 + fibm 1 --> (fiblist !! 1 +

问题 In the current exercise assignment of the Functional Programming course I'm doing, we've got to make a memoized version of a given function. To explain memoization, the following example is given: fiblist = [ fibm x | x <- [0..]] fibm 0 = 0 fibm 1 = 1 fibm n = fiblist !! (n-1) + fiblist !! (n-2) But I don't fully understand how this works. Let's call fibm 3 . fibm 3 --> fiblist !! 2 + fibList 1 --> [fibm 0, fibm 1, fibm 2] !! 2 + [fibm 0, fibm 1] !! 1 --> fibm 2 + fibm 1 --> (fiblist !! 1 +

问题 I have this code for calculating fibonacci number in python . It works and gives the expected result. but when I translated the same to Java , it fails. Any idea of what is going wrong here? In python : def fib3(n): a,b=0,1 while n>0: a,b=b,a+b n-=1 return a fib3(12) --> 144 In Java : public static int fib2(int n){ int a = 0; int b =1; while(n-- >0){ a=b; b=a+b; } return a; } fib2(12) --> 2048 回答1: In this section: a=b; b=a+b; you're assigning b to a+b , but a is already b . So really you're

问题 I know that: int fib(int n) { if (n == 0 || n == 1) return 1; return fib(n − 1)+ fib(n − 2); } when n=5，fib(5) evaluates as: fib(5) fib(4) + fib(3) (fib(3) + fib(2)) + (fib(2) + fib(1)) ((fib(2) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1)) (((fib(1) + fib(0)) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1)) Notice that each base element being used for several times, is there a way to use map to store the previous value and simply do fib(n − 1) + fib（n − 2)? 回答1: