Project Euler problem 479
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1/x = (k/x)2(k+x2) - kx => k*x^3 - k^2*x^2 + x - k^3 = 0
由一元三次方程的韦达定理可知:
ak*bk*ck = k^2
=> ak+bk+ck = k
ak*bk+bk*ck+ak*ck = 1/k
S(n) = Σ (ak+bk)p(bk+ck)p(ck+ak)p
= Σ (k-ak)^p*(k-bk)^p*(k-ck)^p
= Σ [k^3-(ak+bk+ck)*k^2+(ak*bk+ak*ck+bk*ck)*k-ak*bk*ck]^p
= Σ [k^3 - k^3 + 1 - k^2]^p
= Σ (1-k^2)^p 1<= k,p <=n
0 0
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