python中2的n次方怎么表达,python怎么算2的3次方

2401_84561313: def getNum(): ls=[] while True: num=input() if num: ls.append(eval(num)) else: break return ls def avg(ls): s=0 for i in ls: s+=i return s/len(ls) def dev(ls): a=avg(ls) s=0 for i in ls: s+=(i-a)**2 return s/len(ls) def median(ls): ls.sort() if len(ls)%2==0: return (ls[len(ls)//2]+ls[len(ls)//2]-1)/2 else: return ls[len(ls)//2] ls=getNum() print("数量",len(ls)) print("和",sum(ls)) print("平均值",avg(ls)) print("方差",dev(ls)) print("中位数",median(ls)) A = [ [1, 1, 1], [2, 2, 2], [3, 3, 3] ] B = [ [1, 0, 0], [0, 1, 0], [0, 0, 1] ] ls = [] for i in range(len(A)): ls.append([(A[i][j] + B[i][j]) for j in range(len(A[i]))]) print(ls) A = [ [1, 1, 1], [2, 2, 2], [3, 3, 3] ] B = [ [1, 0, 0], [0, 1, 0], [0, 0, 1] ] C = [] for i in range(len(A)): ls = [] for j in range(len(B)): s = 0 for k in range(len(A[i])): s += A[i][k]*B[k][j] ls.append(s) C.append(ls) for i in range(len(C)): for j in range(len(C[i])): print(C[i][j], end=' ') print() d = dict() for i in range(5): name = input() h = input() d[name] = h name = input() print('{} {}'.format(name, d[name])) ls = [] a = 0 b = 1 n = eval(input()) for i in range(n): ls.append(a) a, b = b, a+b print(ls) #输入指定的n n = eval(input()) #指定输入n个数，换行间隔 ls = [] for i in range(n): ls.append(eval(input())) ls.sort() #计算去掉一个最高分和最低分的平均值 ls = ls[1:-1] #输出最后得分 avg = sum(ls)/len(ls) print('该歌手最终成绩为{:.1f}'.format(avg))

云雾栀子青，: 那个辗转相除法那应该是b = t吧

2401_83275802: 代码中存在几个问题： 导入语句不正确。 最后的else块（在for循环外）实际上永远不会被执行，因为for循环内的break语句会中断循环。

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