f(0)=0, f(x)+f(1-x)=1 ==> f(1/2)=1/2, f(1)=1
f(1)=1, f(x/5)=1/2f(x),f(x)+f(1-x)=1 ==> f(1/5)=1/2, f(4/5)=1/2
f(1/5)=1/2, f(1/2)=1/2, f(4/5)=1/2,f(x1)≤f(x2){0≤x1[1/5,4/5]区间f(x)=1/2
f(x/5)=1/2f(x) ==> [1/5^n, 4/5^n]区间上f(x)=1/2^n
1/2020 属于 [1/5^5, 4/5^5] ==> f(1/2020) = 1/2^5 =1/32