ProjectEuler 362 Squarefree factors - 拼圖
By Audriana
at 2011-12-19T19:31
at 2011-12-19T19:31
Table of Contents
※ [本文轉錄自 chyrliin 信箱]
作者: babufong (嗶嗶) 看板: puzzle
標題: [中譯] ProjectEuler 362 Squarefree factors
時間: Sun Dec 11 15:30:32 2011
362. Squarefree factors
http://projecteuler.net/problem=362
來看看 54 這個數字
54 可以拆成 7 種全然不同的一個或多個因數相乘的組成方法,而這些因數都大於 1
如:54 , 2 X 27 , 3 X 18 , 6 X 9 , 3 X 3 X 6 , 2 X 3 X 9 和 2 X 3 X 3 X 3
如果我們要求這些因數都要是「無平方數因數」的數所組成,那就只剩下 2 種組法:
3 X 3 X 6 和 2 X 3 X 3 X 3
Fsf(n) 為 n 有幾種上述條件的方法去組成,所以 Fsf(54) = 2
S(n) 為 ΣFsf(k) , k = 2 到 n
已知 S(100) = 193
請求出 S(10000000000)
--
作者: babufong (嗶嗶) 看板: puzzle
標題: [中譯] ProjectEuler 362 Squarefree factors
時間: Sun Dec 11 15:30:32 2011
362. Squarefree factors
http://projecteuler.net/problem=362
來看看 54 這個數字
54 可以拆成 7 種全然不同的一個或多個因數相乘的組成方法,而這些因數都大於 1
如:54 , 2 X 27 , 3 X 18 , 6 X 9 , 3 X 3 X 6 , 2 X 3 X 9 和 2 X 3 X 3 X 3
如果我們要求這些因數都要是「無平方數因數」的數所組成,那就只剩下 2 種組法:
3 X 3 X 6 和 2 X 3 X 3 X 3
Fsf(n) 為 n 有幾種上述條件的方法去組成,所以 Fsf(54) = 2
S(n) 為 ΣFsf(k) , k = 2 到 n
已知 S(100) = 193
請求出 S(10000000000)
--
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