C PROGRAM TO CHECK WHETHER THE ENTERED NUMBER IS AUTOMORPHIC OR NOT.

An automorphic number is a number whose square contains the same number as last digits or digit. Example :, 5² = 25,6² = 36

Program:

#include <stdio.h>

main()

{

               int i,n,y,flag=0;

               printf("enter a number");

               scanf("%d",&n);

               y=n*n;

               while(n>0)

               {

                              if(n%10 == y%10)//if units place digits are same

                              {

                                             n=n/10;

                                             y=y/10;

                              }

                              else

                              {

                                             flag=1;

                                             break;

                              }

               }

               if(flag==0)

                              printf("given number is automorphic");

               else

                              printf("given number is not automorphic");

}

C program to print the numbers between 1 to n that do not appear in Fibonacci series

Problem:
Write a C program to print the numbers between 1 to n that do not appear in Fibonacci series.

Program:

#include<stdio.h>

void main()

{

int n,i,f0=0,f1=1,f2=f0+f1;

printf("enter the number of terms");

scanf("%d",&n);

while(f2<=n)

                {

                                for(i=f1+1;i<f2;i++) /*print all numbers between f1 and f2 */

                                 {

                                                printf("%d\t",i);

                                 }

                                 f0=f1;

                                 f1=f2;

                                 f2=f0+f1;

 }

}

Output:

(or)

#include <stdio.h>

main()

{

               int i,n,f0=0,f1=1,f2=f0+f1;

               printf("enter the number of terms");

               scanf("%d",&n);

               for(i=1;i<=n;i++)

               {

                              if(i != f2)

                                             printf("%d\t",i);

                              if(i==f2) /* we generate next fibonacci term if all the numbers till  f2 are printed */

                              {

                                             f0=f1;

                                             f1=f2;

                                             f2=f0+f1;

                              }

               }

}

Sherlock and Divisors hacker Rank solution in c

#include <stdio.h>

#include <math.h>

int main()

{

               int n,i,t,c;

               scanf("%d",&t);

               while(t--)

               {

                              c=0;

                              scanf("%d",&n);

                              for(i=1;i<=sqrt(n);i++)

                              {

                                             if(n%i == 0)    //i is divisor of n

                                             {

                                                            if(i%2 == 0)   //even divisor

                                                                           c++;

                                                            if((n/i)%2 == 0 && (n/i)!=i)   //if i is divisor of n then n/i is also divisor of n

                                                                           c++;

                                             }

                              }

                              printf("%d\n",c);

               }             

              

}

C program to find the smallest number divisible by 1 to 20.

Question: 

2520 is the smallest number divisible by 1 to 10. Write a C program to find the smallest number divisible by 1 to 20.

Program:

#include <stdio.h>

int main()

{

                int n,c,j;

                for(n=2520; ;n++)

                {

                                c=0;

                                for(j=1;j<=20;j++)

                                {

                                                if(n%j != 0)

                                                {

                                                                c=1;

                                                                break;

                                                }

                                }

                                if(c==0)

                                {

                                                printf("the smallest number divisible by 1 to 20 = %d",n);

                                                break;

                                }

                               

                }

               

}

Output:

Birthday Chocolate hacker rank solution in C

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int solve(int n, int s_size, int* s, int d, int m){
    int i,j,sum,c=0;
    for(i=0;i<=n-m;i++)
                {
                                sum=0;
                                for(j=i;j<m+i;j++)
                                {
                                                sum=sum+*(s+j);
                                }
                                if(sum == d)
                                                c++;
                }
    return c;
}

int main() {
    int n;
    scanf("%d", &n);
    int *s = malloc(sizeof(int) * n);
    for(int s_i = 0; s_i < n; s_i++){
       scanf("%d",&s[s_i]);
    }
    int d;
    int m;
    scanf("%d %d", &d, &m);
    int result = solve(n, n, s, d, m);
    printf("%d\n", result);
    return 0;
}

Write a C program to print prime pairs with distance <=3

Write  a program to print all possible pairs of prime numbers which are having distance 3 or less than 3  between them in a natural number line starting from 1 to given number N.

Example: prime pairs with distance <=3 – (2,3),  (2,5),  (3,5),  (5,7)

  (3,7), (5,11) are prime pairs with distance >3

Program:

#include<stdio.h>

main()

{

               int i,j,n,x=2,x1=3,x2=0,flag;

               printf("enter n value");

               scanf("%d",&n);

               printf("(%d,%d)\n",x,x1);

               for(i=4;i<=n;i++)

               {

                              flag=0;

                              for(j=2;j<i;j++)

                              {

                                             if(i%j == 0)

                                             {

                                                            flag=1;

                                                            break;

                                             }

                              }

                              if(flag == 0)

                              {

                                            

                                             x2=i;

                                             if(x2-x<=3)

                                                            printf("(%d,%d)\n",x,x2);

                                             if(x2-x1<=3)

                                                            printf("(%d,%d)\n",x1,x2);

                                             x=x1;

                                             x1=x2;

                              }

               }

}

              

Output: