A1918. 最大划分乘积

欧拉工程183题 有改动

Let N be a positive integer and let N be split into k equal parts, r = N/k, so that N = r + r + ... + r.

Let P be the product of these parts, P = r ×r × ... × r = rk.


将正整数N分为k个相等的部分r,有r=N/k

另P=r×r×...×r=rk


For example, if 11 is split into five equal parts, 11 = 2.2 + 2.2 + 2.2 + 2.2 + 2.2, then P = 2.2^5 = 51.53632.


例如，将11分为五等份 则P=2.2^5



Let M(N) = Pmax for a given value of N.

另M(N)为满足要求的对应N的最大的P


It turns out that the maximum for N = 11 is found by splitting eleven into four equal parts which leads to Pmax = (11/4)^4; that is, M(11) = 14641/256 = 57.19140625, which is a terminating decimal.


容易知道，N=11时，Pmax=57.19140625，是有穷小数


However, for N = 8 the maximum is achieved by splitting it into three equal parts, so M(8) = 512/27, which is a non-terminating decimal.


而N=8时，Pmax=512/27，是无尽小数


Let D(N) = N if M(N) is a non-terminating decimal and D(N) = -N if M(N) is a terminating decimal.


设D(n)=n(若M(n)无尽),D(n)=-n(若M(n)有尽)

输入文件仅一行 为正整数a(5≤a≤32767).

输出文件仅一行 为D(5)+D(6)+...+D(a)的值.