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#include
int prime(int);
int fibo(int);
int main(){
int n;
scanf(“%d”,&n);
if(n%2==0)
printf(“%d”,prime(n/2));
else
printf(“%d”,fibo((n/2)+1));
return 0;
}
int prime(int num){
int i,j,flag=0,count;
for(i=2;i<=100;i++){
count=0;
for(j=2;j<=i;j++){
if(i%j==0){
count++;
}
}
if(count==1){
flag++;
}
if(flag==num){
return i;
}
}
}
int fibo(int n){
if(n==1){
return 1;
}else if(n==0){
return 0;
}
return fibo(n-1)+fibo(n-2);
}
def prime(n):
i=1
j=1
count=0
flag=0
for i in range(2,int(n+1)):
flag = 0
for j in range(2,i/2 +1):
if(i%j==0):
flag+=1
if(flag==0):
count+=1
if(count==n):
print(i)
break
def fib(m):
a = 0
b = 1
for i in range(1,int(m+1)):
c = a+b
a=b
b = c
if(i==m):
print(a)
if __name__==’__main__’:
n = int(input())
e = 0
if(n%2==0):
e = float(n/2)
prime(e)
else:
e = float((n/2) +1)
fib(e)
def fib(n):
frst=1
scnd=1
list1=[frst,scnd]
for i in range(n-2):
res=frst+scnd
list1.append(res)
frst,scnd=scnd,res
return list1
def prime(n):
count=n-1
list2=[2]
for i in range(3,100):
flag=0
if count==0:break
for j in range(2,i):
if i%j==0:
flag=1
break
if flag==0:
count=count-1
list2.append(i)
return list2
n=int(input())
e=n//2
if n%2==0:
p=prime(e)
f=fib(e)
else :
p=prime(e)
f=fib(e+1)
if(n%2==0):
print(p[-1])
else:
print(f[-1])
#Python code 3.6
def prime(num):
count = 0
for i in range(2, 100):
flag = 0 # after loop finish flag =0 bcz next iteration will be check
for j in range(2, i):
if i%j==0:
flag = 1
if (flag == 0):
count += 1
if count == num:
return i # return i value
def fib(num):
if num==1 or num==2:
return 1
else:
return fib(num-1) + fib(num-2) #1,1,2,3 == fib(n-1)+fib(n-2) == 3+2=5
num = int(input())
if num%2==0:
n = num//2 # 3/2==1.5 or 3//2==1
print(prime(n))
else:
n =(num+1)//2 # 2//2==1 or 2+1//2==1
print(fib(n))
I found some mistake in the c code
If(i%j==0)
{flag=1;
break; }
If(flag==0)
If(++count==n)
{printf(%d, I);
break; }
Hey guys,
This is the python code,hope it is helpful :
def Prime(n):
result=[]
x=2
while len(result) < n:
flag=0
for i in range(2,x):
if x%i == 0:
flag+=1
break
if flag == 0:
result.append(x)
x+=1
return result
def Fib(n):
if n<=1:
return n
else:
return(Fib(n-1)+Fib(n-2))
n=int(input())
c=n
n=int(n/2)
p=Prime(n)
fib_list=[]
for i in range(n+1):
f=Fib(i+1)
fib_list.append(f)
actual=[]
p1=0
p2=0
for i in range(1,c+1):
if(i%2==0):
actual.append(p[p2])
p2+=1
else:
actual.append(fib_list[p1])
p1+=1
print(actual[c-1])
import java.util.Scanner;
public class NewClass4 {
static void fibonacci(int n)
{
int i, t1 = 0, t2 = 1, nextTerm;
for (i = 1; i<=n; i++)
{
nextTerm = t1 + t2;
t1 = t2;
t2 = nextTerm;
}
System.out.print(t1);
}
static void prime(int n)
{
int MAX = 1000;
int i, j, flag, count =0;
for (i=2; i<=MAX; i++)
{
flag = 0;
for (j=2; j<i; j++)
{
if(i%j == 0)
{
flag = 1;
break;
}
}
if (flag == 0)
if(++count == n)
{
System.out.print(i);
break;
}
}
}
public static void main(String[] args)
{
int n;
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
if(n%2 == 1)
fibonacci (n/2 + 1);
else
prime(n/2);
}
}
//the FLAG values were by mistakely given wrong values by me in one of the explanation. This is the correct one
/* the “i” acts as a variable that will eventually contain the prime no. and will be printed. “j” loop is used to check if “i”(which is our prime no. ) has any factors if “i” has a factor, the control breaks out of “j” loop and flag is set to 1. And if “i” has no factor then the count is increased by 1 and checked if (count == n) and if turns out to be true then “i” value is printed. For e.g. if argument passed to function prime is 1 i.e. void prime(1) then for(i=2;i<MAX;i++) is executed and flag is set to 0, then for(j=2;j<i;j+) is executed which results in "FALSE" and hence we do not enter the loop and thus flag remains 0 and then the condition(++count==n) gets "TRUE"( as ++count changes count to 1 and the value of n is also 1). Therefore, we print "i" which is 2 (and the 1st prime is 2)… */
#include
#define MAX 1000
void fibonacci(int n)
{
int i, t1 = 0, t2 = 1, nextTerm;
for (i = 1; i<=n; i++)
{
nextTerm = t1 + t2;
t1 = t2;
t2 = nextTerm;
}
printf("%d", t1);
}
void prime(int n)
{
int i, j, flag, count =0;
for (i=2; i<=MAX; i++)
{
flag = 0;
for (j=2; j<i; j++)
{
if(i%j == 0)
{
flag = 1;
break;
}
}
if (flag == 0)
{
if(++count == n)
{
printf("%d", i);
printf("%d",count);
break;
}
}
}
}
int main()
{
int n;
scanf("%d", &n);
if(n%2 == 1)
{
fibonacci (n/2 + 1);
}
else
{
prime(n/2);
}
return 0;
}
/* the “i” acts as a variable that will eventually contain the prime no. and will be printed. “j” loop is used to check if “i”(which is our prime no. ) has any factors if “i” has a factor, the control breaks out of “j” loop and flag is set to 1. And if “i” has no factor then the count is increased by 1 and checked if (count == n) and if turns out to be true then “i” value is printed. For e.g. if argument passed to function prime is 1 i.e. void prime(1) then for(i=2;i<MAX;i++) is executed and flag is set to 1, then for(j=2;j<i;j+) is executed which results in "FALSE" and hence we do not enter the loop and thus flag remains 1 and then the condition(++count==n) gets "TRUE"( as ++count changes count to 1 and the value of n is also 1). Therefore, we print "i" which is 2 (and the 1st prime is 2)… 😛
#include
#define MAX 1000
void fibonacci(int n)
{
int i, t1 = 0, t2 = 1, nextTerm;
for (i = 1; i<=n; i++)
{
nextTerm = t1 + t2;
t1 = t2;
t2 = nextTerm;
}
printf("%d", t1);
}
void prime(int n)
{
int i, j, flag, count =0;
for (i=2; i<=MAX; i++)
{
flag = 0;
for (j=2; j<i; j++)
{
if(i%j == 0)
{
flag = 1;
break;
}
}
if (flag == 0)
{
if(++count == n)
{
printf("%d", i);
printf("%d",count);
break;
}
}
}
}
int main()
{
int n;
scanf("%d", &n);
if(n%2 == 1)
{
fibonacci (n/2 + 1);
}
else
{
prime(n/2);
}
return 0;
}
the java code of the above question is—
/*package whatever //do not write package name here */
import java.io.*;
import java.util.Scanner;
class Main {
public static void main (String[] args) {
int n,e;
Scanner sc=new Scanner(System.in);
n=sc.nextInt();
e=n/2;
Main obj=new Main();
if(n%2==0)
obj.prime(e);
else
obj.fibo(e+1);
}
public void prime(int e){
int i,j,no,flag=0,count=0;
for(i=1;i<=100;i++){
flag=0;
for(j=2;j<=i/2;j++){
if(i%j==0)
flag=0;
else
flag=1;
}
if(flag==1)
count++;
if(count==e){
System.out.println(i);
break;
}
}
}
public void fibo(int e){
int n0=0,n1=1,n2=0;
for(int i=3;i<=e;i++)
{
n2=n0+n1;
n0=n1;
n1=n2;
}
System.out.println(n2);
}
}
I have a better solution…
gjg
#include
#define MAX 1000
void fibonacci(int n)
{
int i, t1 = 0, t2 = 1, nextTerm;
for (i = 1; i<=n; i++)
{
nextTerm = t1 + t2;
t1 = t2;
t2 = nextTerm;
}
printf("%d", t1);
}
void prime(int n)
{
int i, j, flag, count =0;
for (i=2; i<=MAX; i++)
{
flag = 0;
for (j=2; j<i; j++)
{
if(i%j == 0)
{
flag = 1;
break;
}
}
if (flag == 0)
if(++count == n)
{
printf("%d", i);
break;
}
}
}
int main()
{
int n;
scanf("%d", &n);
if(n%2 == 1)
fibonacci (n/2 + 1);
else
prime(n/2);
return 0;
}
this code seems to be giving the wrong output …. i have a better solution…
#include
int main ()
{
int n, flag;
int i, t = 1, j;
int a, b, c, r, g;
a = 0;
b = 1;
c = 0;
r = 1;
scanf (“%d”, &n);
if (n % 2 != 0)
{
while (r <= (n / 2 + 1))
{
g = b;
c = a;
a = b;
b = a + c;
r++;
}
printf ("%d ", g);
}
else
{ //for prime no
for (i = 2; t <= (n / 2); i++)
{
flag = 0;
for (j = 2; j < i; j++)
{
if (i % j == 0)
{
flag = 1;
break;
}
}
if (flag == 0)
{
t++;
}
}
printf ("%d ", i – 1);
}
return 0;
}