Computer Science, asked by loop55, 2 days ago

write a c program which will print odd numbers from 1 to 20.

Answers

Answered by ameyawadnerkar
0

#include <stdio.h>

   int main(){

   for(int i=1; i<=20; i+=2){

       printf("%d\n", i);

   }

   return 0;

}

Answered by eastcreature25
0

Answer:

Using For Loop :

#include <stdio.h>

#define EVEN 2

#define LIMIT 20

int main()

{

int Odd_Num[20];

int Even_Num[20];

int i , j , k ;

memset(Odd_Num, 0, sizeof(Odd_Num));

memset(Even_Num, 0, sizeof(Even_Num));

for(i = j = k = 0; i != LIMIT; i++)

{

if( i % EVEN )

{

Odd_Num[j++] = i ;

}

else

{

Even_Num[k++] = i ;

}

}

printf( "Even numbers from 0 to 20 are\n" ) ;

for (k = 0; k != LIMIT; k++)

{

printf("%d\t", Even_Num[k]) ;

}

printf("\n");

printf( "Odd numbers from 0 to 20 are\n" ) ;

for (j = 0; j != LIMIT; j++)

{

printf("%d\t", Odd_Num[j]) ;

}

}

Using While Loop :

#include <stdio.h>

#define EVEN 2

#define LIMIT 20

int main()

{

int Odd_Num[20];

int Even_Num[20];

int i = j = k = 0 ;

memset(Odd_Num, 0, sizeof(Odd_Num));

memset(Even_Num, 0, sizeof(Even_Num));

while( i != LIMIT )

{

if( i % EVEN )

{

Odd_Num[j++] = i++ ;

}

else

{

Even_Num[k++] = i++ ;

}

}

printf( "Even numbers from 0 to 20 are\n" ) ;

k = 0;

while( k != LIMIT )

{

printf("%d\t", Even_Num[k++]) ;

}

printf("\n");

printf( "Odd numbers from 0 to 20 are\n" ) ;

j = 0;

while( j != LIMIT )

{

printf("%d\t", Odd_Num[j++]) ;

}

}

Explanation:

hope you find this answer helpful

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