Optimization of Algorithm [closed]

Question

Here is the link to the problem.

The problem asks the number of solutions to the Diophantine equation of the form 1/x + 1/y = 1/z (where z = n!). Rearranging the given equation clearly tells that the answer is the number of factors of z².

So the problem boils down to finding the number of factors of n!² (n factorial squared).

My algorithm is as follows

1. Make a Boolean look up table for all primes <= n using Sieve of Eratosthenes algorithm.
2. Iterate over all primes P <= n and find its exponent in n!. I did this using step function formula. Let the exponent be K, then the exponent of P in n!² will be 2K.
3. Using step 2 calculate number of factors with the standard formula.

For the worst case input of n = 10⁶, my c code gives answer in 0.056s. I guess the complexity is no way greater than O(n lg n). But when I submitted the code on the site, I could pass only 3/15 test cases with the verdict on the rest as time limit exceeded.

I need some hints for optimizing this algorithm.

Code so far:

#include <stdio.h>
#include <string.h>

#define ULL unsigned long long int
#define MAXN 1000010
#define MOD 1000007

int isPrime[MAXN];

ULL solve(int N) {
    int i, j, f;
    ULL res = 1;
    memset(isPrime, 1, MAXN * sizeof(int));
    isPrime[0] = isPrime[1] = 0;
    for (i = 2; i * i <= N; ++i)
        if (isPrime[i])
            for (j = i * i; j <= N; j += i)
                isPrime[j] = 0;
    for (i = 2; i <= N; ++i) {
        if (isPrime[i]) {
            for (j = i, f = 0; j <= N; j *= i)
                f += N / j;
            f = ((f << 1) + 1) % MOD;
            res = (res * f) % MOD;
        }
    }
    return res % MOD;
}

int main() {
    int N;
    scanf("%d", &N);
    printf("%llu\n", solve(N));
    return 0;
}

Answer 1

You have an int overflow here:

for (j = i, f = 0; j <= N; j *= i)

If 46340 < i < 65536 and for many larger i, in the second iteration j will be negative if overflow wraps around (it is undefined behaviour, so anything could happen).

Replace it with e.g.

for(j = N / i, f = 0; j > 0; j /= i) {
    f += j;
}

It is, however, unlikely that the extra iterations due to the overflow would cause a "time limit exceeded", that will likely only cause wrong results.

So the generic advice is

• Sieve only odd numbers, perhaps also eliminate multiples of 3 from the sieve. Eliminating the odd numbers from the sieve roughly halves the time needed to sieve.
• Don't use an entire int for the flags, use bits or at least chars. That gives better cache locality and a further speedup.