设f(a,n)=floor{(ceil{sqrt{a}}+sqrt{a})^n} 已知f(5,2)=27,f(5,5)=3935 设G(n)=f(a,a^2). 已知G(1000) mod 999999937=163861845 求G(5000000) mod 999999937
import math
def f(a,n):
return math.floor((math.ceil(math.sqrt(a))+math.sqrt(a))**n)
>>> f(5,2)
27
>>> f(5,5)
3935
>>> def G(n):
... s=0
... for a in range(1,n+1) :s+=f(a,a*a)
... return s
...
>>> G(10)
31997447160056074891755545250568015145391367309910918290177259841659637836252530747180
https://www.cnblogs.com/wkfvawl/p/9125224.html