Reputation: 175
hi i'm doing a java activity that will divide the two given numbers without using the "/" operator. I want to use a loop statement.
System.out.print("Enter Divident: ");
int ans1 = Integer.parseInt(in.readLine());
System.out.print("Enter Divisor: ");
int ans2 = Integer.parseInt(in.readLine());
The output is:
Enter Dividend: 25
Enter Divisor 5
5
How can solve this without using this "ans1/ans2"
Upvotes: 3
Views: 9725
Reputation: 4066
if you really want to use loop to divide two numbers, you can write it like code below
int c=0;
while(ans1 >= ans2){
ans1 -= ans2;
c++;
}
after loop c
equals quotient and ans1
equals reminder of division
if abs1
and abs2
are signed numbers, below code should be work for division
boolean n1 = (ans1 & (1<<31))!=0;
boolean n2 = (ans2 & (1<<31))!=0;
ans1 = Math.abs(ans1);
ans2 = Math.abs(ans2);
int c=0;
while(ans1 >= ans2){
ans1 -= ans2;
c++;
}
if(!n1 && n2) c = -c;
else if(n1 && !n2){
c = -c;
if(ans1 > 0){
ans1 = ans2 - ans1;
c--;
}
}else if(n1 && n2){
if(ans1 > 0){
ans1 = ans2 - ans1;
c++;
}
}
Upvotes: 6
Reputation: 3871
You can do it in the following way:
System.out.print("Enter Divident: ");
int ans1 = Integer.parseInt(in.readLine());
System.out.print("Enter Divisor: ");
int ans2 = Integer.parseInt(in.readLine());
int count=0;
while(ans1>=ans2)
{
ans1=ans1-ans2;
count++;
}
System.out.println(count);
Upvotes: 0
Reputation: 8101
Something like below ?
int i=1;
int mul;
while(true)
{
mul = i++;
if(mul*(ans2)==ans1)
{
System.out.println(mul);
break;
}
else if(mul*(ans2)>ans1)
{
System.out.println("Cannot be divided");
break;
}
}
Upvotes: -1
Reputation: 10789
BigInteger do the trick witout /
operator, literally.
(new BigInteger(ans1 + "")).divide(new BigInteger(ans2 + ""))
Upvotes: 0
Reputation: 2820
Using recursion:
// Calculate: a / b
public int divide (int a, int b) {
if ( a < b ) {
return 0;
} else {
return 1 + divide ( a - b, b );
}
}
Upvotes: 1
Reputation: 7905
If you really want to do it without the / operator and im guessing probably without shifting either the simplest loop method would be to count the number of times you need to subtract ans2 from ans1 without the remainder going below ans1.
Pseudo code:
numtimes init at 0
counter init at ans1
while counter is greater than ans2
subtract ans2 from counter
numtimes increase by 1
check numtimes
Upvotes: 0