Tuesday, April 2, 2013


SPOJ Problem Set (Classical) 5970. Finding Primes

Fun problem! First it wants you to run a sieve from 2 to N, and then do the similar operations starting from N to 2, except, this time, you have to mark the factors of N and then find the next largest N1 not yet marked, then mark factors of N1 and so on... Finally, you have to tell how many numbers will be unmarked by both algorithms up to N.

First observation is, you only need to consider those numbers which are primes, so, we can run a sieve up to 107, which will allow us to test whether a number is prime or not in O(1) time.

Now, for each N, we can pre-calculate the result. Just note that, say if you know ans[n-1] and want to determine ans[n] from it, this inequality will hold |ans[n] - ans[n-1]| <= 1. Because the current N will be added as an unmarked, or there may be a number which you added in the past, will be removed by one of the factors of this N. Or, if N is not a prime, then it will be already removed by the first algorithm i.e. sieve.

If N is a prime number, then no N1 has removed this N already, so in this case the ans[n] = ans[n-1] + 1.

If N is an even number, and if N/2 was a prime, you have added that with your result, but this should be removed at this stage. So in such a case, ans[n] = ans[n-1].

Why don't you need to consider other factors like N/3, N/5 ... ? 2 is the only even prime factor and it can produce even numbers by multiplying other primes with it. For other primes, p = 3, 5 and so on, if N/p is prime, then N cannot be even. So the basic idea is as follows:

for( i = 2; i <= N; i++ ) {
    ans[i] = ans[i - 1];
    if( i is even and i/2 is prime ) ans[i] = ans[i] - 1;
    if( i is prime ) ans[i] = ans[i] + 1;

I had absolutely no clue at the first glance. Really nice problem.