Question

C Programming Language

Jo Number Jojo likes Jo numbers. A Jo numbers are positive integers whose decimal representation contains only 7 and 8. For example, 8 and 778 are Jo while 18 and 286 are not. Jojo calls a number semi-Jo if it could be evenly divided by some Jo number. Given integer N, help him find out if it's an semi-Jo number Format Input The input begins with an integer T indicating the number of test cases. In each test case, there is an integer N, the number that needs to be checked Format Output For each test case, output YES if number N is semi-Jo. Otherwise, output NO Constraints 1N <- 1000000000 Sample Input Sample Output Case #1: YES Case #2 : YES Case #3: NO 28 5605

The code below matches the sample input/output but is marked wrong. Are there other ways to do this problem? The focus is on recursion and functions.

#include<stdio.h>

int joCheck(unsigned long long int number)

{

if(number % 7 == 0 || number % 8 == 0)

{

return 1;

}

else return 0;

}

int main()

{

int cases = 0;

scanf("%d", &cases);

for(int i = 1; i<=cases; i++)

{

unsigned long long int number;

scanf("%d", &number);

if(joCheck(number) == 1)

{

printf("Case #%d: YES ", i);

}

else printf("Case #%d: NO ", i);

}

}

Answer 1

#include<stdio.h>

int joCheck(unsigned long long int number)

{

if (number % 7 == 0 || number % 8 == 0)

{

return 1;

}

else

return 0;

}

int main()

{

int cases = 0;

unsigned long long int number;

printf("enter the number of cases \n");

scanf("%d", &cases);

for(int i = 1; i<=cases; i++)

{

printf("enter number \n");

scanf("%d", &number);

if (joCheck(number) == 1)

{

printf("Case #%d: YES\n", i);

}

else printf("Case #%d: NO \n", i);

}
return 0;
}

The second method is

#include<stdio.h>

int isDivisibleBy7( int num )
{

// If number is negative, make it positive

if ( num < 0 )

return isDivisibleBy7( -num );

  

// Base cases

if ( num == 0 || num == 7 )

return 1;

if ( num < 10 )

return 0;

  

// Recur for ( num / 10 - 2 * num % 10 )  

return isDivisibleBy7( num / 10 - 2 * ( num - num / 10 * 10 ) );
}

int isDivisibleBy8( int num )
{

// If number is negative, make it positive

if ( num < 0 )

return isDivisibleBy8( -num );

  

// Base cases

if ( num == 0 || num == 8 )

return 1;

if ( num < 10 )

return 0;

  

// Recur for ( num / 10 - 2 * num % 10 )  

return isDivisibleBy8( num / 10 - 2 * ( num - num / 10 * 10 ) );
}

int main()

{

int cases = 0,se,ei;

unsigned long long int number;

printf("enter the number of cases \n");

scanf("%d", &cases);

for(int i = 1; i<=cases; i++)

{

printf("enter number \n");

scanf("%d", &number);

se=isDivisibleBy7(number);

ei=isDivisibleBy8(number);

if (se==1|ei==1)

{

printf("Case #%d: YES\n", i);

}

else printf("Case #%d: NO \n", i);

}
return 0;
}

6:10 PM 1.95K/s TAB enter the number of cases enter number Case #1: YES enter number 28 Case #2: YES enter number 5605 Case #3: No Program finished]