Lecture Note: n DIV d: integer part of dividing n by d n MOD d: the remainder n

Question

Lecture Note:

n DIV d: integer part of dividing n by d

n MOD d: the remainder

n = (n DIV d) * d + (n MOD d)

Example: 7=2*3+1, so

7 DIV 3 = 2

7 MOD 3 = 1

Recall the DIVMOD algorithm

This algorithm computed (n DIV d, n MOD d) for integers n and d, both greater than 0.

Using pseudocode in your programming language of choice, expand the algorithm to accept any integer n, both positive and negative (d is still assumed to be positive). Add comments when needed to make sure your logic is clear to the reader.

Explanation / Answer

Your program is as below :

//This program is to write function divMod algorithm. n = (n DIV d) *d + (n MOD d)

#include <stdio.h>

int divMod(int n,int d);    //Function Declaration

int main(void) {
  
   int n,d,result=0;

    //Input number and divisor

   printf("Please enter number :");
   scanf("%d",&n);
   label :
   printf("Please enter divisor : ");
   scanf("%d",&d);
   if(d<0){ //Checking for positive divisior
            printf("Please enter positive divisor");
            goto label;
   }
   result = divMod(n,d); //function call
   printf("Your result is : %d",result);
   return 0;
}
int divMod(int n,int d){
    int t = (n/d)*d+(n%d);
    return t;
}

Output :

(1)

Please enter number : 7

Please enter divisor : 3

Result is : 7

(2)

Plese enter number : 7

Please enter divisor : -3

Please enter positive integer

Please enter divisor : 3

Result is : 7

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