0, 0, 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12, 6, 14, 7, 16, 8 This series is a mixture of 2 series all the odd terms in this series form even numbers in ascending order and every even terms is derived from the previous term using the formula (x/2)
Consider the below series :
0,0,2,1,4,2,6,3,8,4,10,5,12,6,14,7,16,8
- This series is a mixture of 2 series all the odd terms in this series form even numbers in ascending order
- Every even terms is derived from the previous term using the formula (x/2)
Write a program to find the nth term in this series.
- The value n in a positive integer that should be read from STDIN the nth term that is calculated by the program should be written to STDOUT.
- Other than the value of the nth term no other characters /strings or message should be written to STDOUT.
For example
if n=10,the 10 th term in the series is to be derived from the 9th term in the series. The 9th term is 8 so the 10th term is (8/2)=4. Only the value 4 should be printed to STDOUT.
You can assume that the n will not exceed 20,000.
C
C++
Java
Python
C
#include<stdio.h> int main() { int i, n, a=0, b=0; printf("enter number : "); scanf("%d",&n); for(i=1;i<=n;i++) { if(i%2!=0) { if(i>1) a = a + 2; } else { b = a/2; } } if(n%2!=0) { printf("%d",a); } else { printf("%d",b); } return 0; }
C++
#include<iostream>
using namespace std;
int main() { int i, n, a=0, b=0; cout << "enter number : "; cin >> n; for(i=1;i<=n;i++) { if(i%2!=0) { if(i>1) a = a + 2; } else { b = a/2; } } if(n%2!=0) { cout << a; } else { cout << b; } return 0; }
Java
//Java program to find nth element of the series
import java.util.Scanner;
class Main
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a = 0, b = 0;
if(n % 2 == 0)
{
for(int i = 1 ; i <= (n-2) ; i = i+2)
{
a = a + 2;
b = a / 2;
}
System.out.print(b);
}
else
{
for(int i = 1 ; i < (n-2) ; i = i+2)
{
a = a + 2;
b = a / 2;
}
a = a + 2;
System.out.print(a);
}
}
}
Python
n = int(input('enter the number:'))
a=0
b=0
for i in range(1,n+1):
if(i%2!=0):
a= a+2
else:
b= b+1
if(n%2!=0):
print('{}'.format(a-2))
else:
print('{}'.format(b-1))
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More efficient way in python:
num = int(input())
odd,even = 0,0
if(num%2==0):
even = (num-1)//2
print(even)
else:
odd = num – 1
print(odd)
n = int(input(‘enter the number:’))
if n%2!=0:
print(n-1)
else:
print((n-2)/2)
#include
int main(){
int i,k,c=0,v=0;
scanf(“%d”,&k);
int a[20000];
for(i=1;i<=k;i++){
a[i]=c;
c+=2;
a[++i]=v++;
}
printf("%d",a[k]);
return 0;
}
#include
int main()
{
int n;
scanf(“%d”,&n);
if(n%2!=0)
printf(“%d”,n-1);
else
printf(“%d”,(n/2)-1);
}
package tcs;
import java.util.Scanner;
public class Tcs4 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n=scan.nextInt();
int a=0;
int b=0;
System.out.println(a);
System.out.println(b);
for(int i=3;i<=n;i++)
{
if(i%2==0)
{
b=a/2;
System.out.println(b);
}
else
{
a=a+2;
System.out.println(a);
}
}
}
}
import java.util.Scanner;
public class series
{
public static void main(String[] args)
{
System.out.println(“enter n”);
Scanner obj = new Scanner(System.in);
int n = obj.nextInt();
int i=n/2;
System.out.println((n%2)!=0?i*2:i-1);
}
}
If n-1 is odd ,return (n-1)/2;
If n-1 is even, return (n-1);
Two line solution.
#PYTHON 3.7
#this program also gives the same result
i=int(input(‘N(positive integer only)=’))
if i%2==0: print(i//2-1,end=’,’)
else: print(i-1,end=’,’)
#include
int a[20];
void main()
{
int i,j,n;
scanf(“%d”,&n);
for(i=1;i<=n;i=i+2)
{
a[i]=(i-1)/2;
}
for(j=0;j<=n;j=j+2)
{
a[j]=j;
}
printf("%d",a[n-1]);
}
n=int(input())
if n%2==0:
n=int(n/2)-1
else:
n=n-1
print(n)
Recursive code in C
main()
{
int n;
scanf(“%d”,&n);
printf(“%d”,term(n));
}
int term(int n){
int a;
if(n%2!=0)
return (n-1);
else
return(term(n-1)/2);
}
n=int(input())
if n>20000:
print(0)
else:
if n%2==1:
print(2*(n//2))
else:
print((n//2)-1)