Project Euler: Problem 8

http://projecteuler.net/index.php?section=problems&id=8

Find the greatest product of five consecutive digits in the 1000-digit number.

7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450

Python simplifies this a bit, though it wouldn't be too hard to implement in C/C++.

def f2(n):
    s = str(n)
    res = 0

    for i in xrange(len(s)-5):
        x = 1
        for j in xrange(5):
            x *= int(s[i+j])
            if x == 0:
                break
        res = max(res, x)

    return res

By breaking out of the second iteration if the product is zero (it ain't gettin' any bigger!), we shave off 9% of our execution time. On my machine, the result is returned in 5e-03 seconds.