Problem :- Write A Program To Find Fibonacci Series Of Large Amount Of Integer
Logic :- For Print Fibonacci Series We Use Simple Method .As we know that Fibonacci Series is start with Zero (0) and next Element is One Then we add previous two element and print next element of Fibonacci Series .
Example :- Fibonacci Series is up-to 10 Element is below
Fibonacci Series :- 0 ,1 ,1 ,2 ,3 ,5 ,8 ,13 ,21 ,34 .
Explanation :- Fibonacci Series start with Zero and next element is one then first we print 0 and 1.
Now add two previous element and print then next element is 0+1=1.
repeat that process
1+1=2
1+2=3
2+3=5
3+5=8
5+8=13
8+13=21
13+21=34
Done.
Solution :-
#include<iostream>
using namespace std;
int main()
{
int n,a,b,i;
cout<<"How Many Term You Want To Check :\n";
cin>>n;
long long A[n];
a=0;
b=1;
cout<<"The Fibonacci Series Is :\n";
if(n<2)
{
A[0]=a; A[1]=b;
cout<<A[0]<<" , "<<A[i];
}
else
A[0]=a; A[1]=b;
cout<<A[0]<<" , "<<A[1]<<" , ";
for(i=2;i<n;i++)
{
A[i]=A[i-1]+A[i-2];;
cout<<A[i]<<" , ";
}
return 0;
}
Output:-
Logic :- For Print Fibonacci Series We Use Simple Method .As we know that Fibonacci Series is start with Zero (0) and next Element is One Then we add previous two element and print next element of Fibonacci Series .
Example :- Fibonacci Series is up-to 10 Element is below
Fibonacci Series :- 0 ,1 ,1 ,2 ,3 ,5 ,8 ,13 ,21 ,34 .
Explanation :- Fibonacci Series start with Zero and next element is one then first we print 0 and 1.
Now add two previous element and print then next element is 0+1=1.
repeat that process
1+1=2
1+2=3
2+3=5
3+5=8
5+8=13
8+13=21
13+21=34
Done.
Solution :-
#include<iostream>
using namespace std;
int main()
{
int n,a,b,i;
cout<<"How Many Term You Want To Check :\n";
cin>>n;
long long A[n];
a=0;
b=1;
cout<<"The Fibonacci Series Is :\n";
if(n<2)
{
A[0]=a; A[1]=b;
cout<<A[0]<<" , "<<A[i];
}
else
A[0]=a; A[1]=b;
cout<<A[0]<<" , "<<A[1]<<" , ";
for(i=2;i<n;i++)
{
A[i]=A[i-1]+A[i-2];;
cout<<A[i]<<" , ";
}
return 0;
}
Output:-
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Deletehow to find when the is out of long long int range? ( in c++)
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