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what is wrong in my code.Iam not get output
public int divisorSum(int n)
{
int sum =0;
int i=0;
while( i<=n)
{
if(n%i==0)
sum=sum+i;
i++;
}
return sum;
}
You're right, if we were talking prime numbers, sqrt(n) would make sense, but not for divisors.
[Edit]: Let me correct myself, if you add i and n / i at the same step, you're adding the other number that is also a divisor in same step, take for example 8, if i is 2, you'll end up adding sum += 2 + 8/2 which traduces to sum += 2 + 4, both 2 & 4 are divisors of 8, keep doing that and you'll soon notice that this way the i variable only needs to go up to sqrt(n).
Here is a tip, a number's divisor will always be 1 and the number itself.
So you can initialize the sum as n+1 itself.
And the start the loop from i=2 till i<=n/2 because the greatest divisor of the number will not be greater than its half. Hence by this you shorten your for loop.
Also place a condition at start for n=1 then just return 1. Otherwise that test case wont be successful
This one worked for me..
class Calculator implements AdvancedArithmetic{
public int divisorSum(int n){
int sum=n;
for(int i=1;i<=n/2;i++){
if(n%i == 0){
sum=sum+i;
}
}
return sum;
}
}
Day 19: Interfaces
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what is wrong in my code.Iam not get output public int divisorSum(int n) { int sum =0; int i=0; while( i<=n) { if(n%i==0) sum=sum+i; i++; } return sum; }
Your counter for the while loop starts at 0. It will compute the remainder modulo 0 in this first case, which does not make any sense.
change into int i = 1; insted of int i = 0;
Another point to note, is that After n/2, the only factor of n will be n itself. This reduces the running time of the whole algorithm
sqrt(n) will reduce it more.
just add i and n/i(if a pure int)
how is this correct for a number like 20? 10 is larger than sqrt(20), am confused here
You're right, if we were talking prime numbers,
sqrt(n)would make sense, but not for divisors.[Edit]: Let me correct myself, if you add
iandn / iat the same step, you're adding the other number that is also a divisor in same step, take for example8, ifiis2, you'll end up addingsum += 2 + 8/2which traduces tosum += 2 + 4, both2&4are divisors of8, keep doing that and you'll soon notice that this way theivariable only needs to go up tosqrt(n).yes i got it thank you very much
If n=1, then this would return 2, no? even though it should return 1.
Use proper format when u paste code.
Here is a tip, a number's divisor will always be 1 and the number itself. So you can initialize the sum as n+1 itself. And the start the loop from i=2 till i<=n/2 because the greatest divisor of the number will not be greater than its half. Hence by this you shorten your for loop. Also place a condition at start for n=1 then just return 1. Otherwise that test case wont be successful
i think this would fail for the case of 1, n+1 will be 2 where the correct answer is just 1
To account for a special case like 1, you one can leverage the ternary operator to assign value to sum like so:
This will mitigate logic errors in code later on.
thanks! this worked for me
This one worked for me.. class Calculator implements AdvancedArithmetic{ public int divisorSum(int n){ int sum=n; for(int i=1;i<=n/2;i++){ if(n%i == 0){ sum=sum+i; } } return sum; } }
put that i++ outside of if condition.
int sum=0; int i=1; while(i<=n) { if(n%i==0) { sum+=i; } i++; }