事后回忆的解法,应该没错叭。(* ̄m ̄)。不过第二题题目啥意思呀,求个路过的好心人解惑一下。l=[2,2,3,4,5,6,7]avg=sum(l)/len(l)avgcount=0if avg-int(avg)==0: # l=[1,2,3,4,5] 3.0做差即可 for i in l: if avg-i>=1: count+=int(avg)-i print(count)else: down_=int(avg) up_=int(avg)+1 count_d=0 count_u=0 for i in l: if iup_: count_u+=i-up_ # (5,3)次选5次,因为[4,4,4,4,5,5,5]后还需要进行加1次(5-3)=2次,2次可以被容纳,最后变为 [4,4,5,5,5,5,5] print(max(count_d,count_u))