作者LimSinE (r=e^theta)
看板IMO_Taiwan
标题Re: [问题] IMO 2010 in Kazakhstan Day 1
时间Sat Jul 10 01:20:45 2010
1.
x=y=0: f(0)=f(0)[f(0)]
f(0)=0 or [f(0)]=1
If [f(0)]=1, then
x=0: f(0)=f(y)[f(0)]= f(y), f is constant C, with 1<=C<2
x=y=1: f(1)=f(1)[f(1)]
f(1)=0 or [f(1)]=1
If f(1)=0, then
x=1 f(y)=f(1)[f(y)] = 0, f=0
The remaining case is f(0)=0, [f(1)]=1
In this case,
y=1: f([x])=f(x)
But
x=2, y=0.5: 1 <= f(1) = f(2)[f(0.5)] = 0, contradiction.
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r=e^theta
即使有改变，我始终如一。
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