Case 3: Fibonacci numbers is famous formula in mathematics that generates a sequence of numbers in which the next number is computed based on the previous two numbers (1, 1, 2, 3, 5, 8, 13, 21, 34,...).
Case 3: Fibonacci numbers is famous formula in mathematics that generates a sequence of numbers in which the next number is computed based on the previous two numbers (1, 1, 2, 3, 5, 8, 13, 21, 34,...).
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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![Case 3: Fibonacci numbers is famous formula in mathematics that
generates a sequence of numbers in which the next number is computed
based on the previous two numbers (1, 1, 2, 3, 5, 8, 13, 21, 34,..).
Program 1 (recursive method)
Program 2 (Dynamic Programming method)
#include <iostream>
#include <iostream>
using namespace std;
using namespace std;
void fib(double A[], int n)
{
int fib(int n)
{
A[0]=0;A[1]=1;
for (int i=2;i<n;i++)
if (n<=1) return n;
else
A[i]=A[i-1]+A[i-2];
return fib(n-1)+fib(n-2); main()
{
}
main()
{
int n;
cout<<"input n:\n";
cin>>n;
double A[n];
int n;
cout<<"input n:\n";
cin>>n;
for (int i=0;i<n;i++)
cout<<fib(i)<<" ";
fib(A,n);
for (int i=0;i<n;i++)
cout<<A[i]<<endl;
}
}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F959cfe71-4ad4-4936-a810-ead187049518%2F2acff6f4-6906-473c-9a1c-a64f46f306c8%2Fxerryv7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Case 3: Fibonacci numbers is famous formula in mathematics that
generates a sequence of numbers in which the next number is computed
based on the previous two numbers (1, 1, 2, 3, 5, 8, 13, 21, 34,..).
Program 1 (recursive method)
Program 2 (Dynamic Programming method)
#include <iostream>
#include <iostream>
using namespace std;
using namespace std;
void fib(double A[], int n)
{
int fib(int n)
{
A[0]=0;A[1]=1;
for (int i=2;i<n;i++)
if (n<=1) return n;
else
A[i]=A[i-1]+A[i-2];
return fib(n-1)+fib(n-2); main()
{
}
main()
{
int n;
cout<<"input n:\n";
cin>>n;
double A[n];
int n;
cout<<"input n:\n";
cin>>n;
for (int i=0;i<n;i++)
cout<<fib(i)<<" ";
fib(A,n);
for (int i=0;i<n;i++)
cout<<A[i]<<endl;
}
}
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