choose k numbers k1,k2...km such that sigma(cos((2*ki+1)*pi/n))==0&&sigma(sin((2*ki+1)*pi/n))==0

that is  sigma(cos((2*ki+1)*pi/n)+sin((2*ki+1)*pi/n)*i)==0
because cos(k*theta)+i*sin(k*theta)==(cos(theta)+i*sin(theta)^k

so it is sigma(cos(pi/n)+i*sin(pi/n))^(2*ki+1)==0

I think this problem must involved this...

is there O(polynomial)algo ? I only came up with a brute_force search idea..maybe it can pass..

brute_force search compnent part which is central symmetry

Edited by author 22.11.2017 08:15

seems can turn to minimum cut problem because there are only two prime factor,I didn't see it...