一道Codility上的面试题

A positive integer N is given. Write a function:            function solution(N)； that returns the sum of the values of the digits in a decimal representation of N! (fractorial of N). For example, give N = 14, the function should return 45, because N! = 87178291200 and the sum of the values of its digits equals: 8+7+1+7+8+2+9+1+2+0+0 = 45 The function should return -1 if the result exceeds 100,000,000. Assume that： N is an integer within the range [1...2,000] Complexity: expected worst-case time complexity is O(N^2) expected worst-case space complexity is O(N)

Assuming the space complexity asked is O(N) word complexity (i.e. O(N) 32-bit integers or something), and arithmetic of words is O(1) time.

1. Precompute everything, hardcode them in a table, do a table look up. O(1) time, if 100,000,000 fits in one word this is O(N) space.

2. Use linked list to implement large number multiplication in a big enough base, say base N. As N!<N^N, which in base-N representation only needs N numbers, so space complexity is O(N) words. Computing N! is O(N^2) time (actually no, see edit note below), once you have N! in base-N, computing the digits in base-10 should be easy.

Edit: In the 2nd approach computing N! in base N should be O(N^2 log N) using FFT, not sure how to get it down to O(N^2) for now, maybe do (N / log N) based arithmetic instead.

LZ能不能share下这题的题目名称是什么？谢谢了~

nemoczech 发表于 2014-12-22 01:10
LZ能不能share下这题的题目名称是什么？谢谢了~

不知道叫什么啊 online 做的 不是demo题 没有名字的

- = 我第一个想到的是斯特灵公式, 然后感觉那么复杂不会出..然后就不知道了

- = 确实想不到不算出来n!就解决此题的方法...要是让算出来能有好多办法..

readman 发表于 2014-12-22 15:39
- = 确实想不到不算出来n!就解决此题的方法...要是让算出来能有好多办法..

算出来肯定overflow了

有没有想出什么好的解法？我在网上搜了一圈似乎也没有什么满意的答案...顺便问问LZ做的是哪家公司的OA啊？

xutopia 发表于 2014-12-24 02:49
Assuming the space complexity asked is O(N) word complexity (i.e. O(N) 32-bit integers or something) ...

大神 你认识徐硕 长起么？ 我好像知道你

题目的名字是：FactorialDigitsSum