Is there any formula for calculating the count of prime factors of a number 'N'? I know the loose upper bound is log2(N) but is there a better upper bound? Thanks.
Is there any formula for calculating the count of prime factors of a number 'N'? I know the loose upper bound is log2(N) but is there a better upper bound? Thanks.
I'm trying this problem from some time but not even able to get one step towards the solution. Any help regarding the approach / concept required will be much helpful ThankYou
int a[100];
int sum = 0;
for(int i = 0; i < 100; ++i) {
cin >> a[i];
sum += a[i];
}
and
int sum = 0;
for(int i = 0; i < 100; ++i) {
int val;
cin >> val;
sum += val;
}
My doubt is whether both the code snippets consume same amount of memory ?
Thanks in advance!
Edit : I don't understand why it's getting downvoted despite being a valid doubt.