Online Judge | Problem Set | Authors | Online Contests | User | ||||||
---|---|---|---|---|---|---|---|---|---|---|
Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest |
Re:我的推导过程In Reply To:我的推导过程 Posted by:xuguangshengqq at 2007-07-19 22:05:39 > 1/a = (1/b+1/c) / (1 - (1/b) * (1/c)) > 1/a = (b+c)/bc / (1-1/bc) > a = (bc-1) / (b+c) > 令 m = b+c > 则 c = m - b; > m = (b^2+1)/(b-a) > 令 i = b-a > 则 m = i + 2a + (a^2+1)/i //最后表达式 > i+ (a^2+1)/i >= 2*sqrt(a^2+1) > 所以: i >= sqrt(a^2+1) 或 i<= sqrt(a^2+1) > 则 m >= 2*sqrt(a^2+1) + 2a > 所以可以令 i 从 a 开始循环 i--; > 直到 (a^2+1)%i == 0 再求出m即可 > 请问为什么一定是for(i=a;;i--)呢 难道i++不行? Followed by: Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator