Here is how to do it with a single call to rand and no looping.
int custRand(int n)
{
int a = rand(n*n+n);
int b = a / n;
a %= n;
return (b > a) ? n+a-b+1 : n-a+b;
}
On Aug 29, 10:48 am, Piyush Grover <[email protected]> wrote:
> Given a function rand(n) which returns a random value between 1...n assuming
> equal probability.
> Write a function CustRand(n) using rand(n) which returns a value between
> 1...n such that
> the probability of occurrence of each number is proportional to its value.
>
> I have a solution but wondering if I can get better than this or some other
> approaches:
>
> CustRand(n){
>
> sum = n*(n+1)/2;
>
> a = rand(sum);
> for(i = 1; i <= n; i++){
> h = i*(i+1)/2;
> l = i*(i-1)/2;
> if(a <= h && a > l)
> return i;
> }
>
> }
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