【每日一题20220525】整数次方

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1

:mage:‍ 实现函数 solution(base:float, x:int)，求base的x次方。
注意：
1.base和x不同时为0。

【示例】
输入：base=2.00000, x=3
输出：8.00000
解释：因为2.00000的三次方运算后是8.00000

题目难度：中等
题目来源：牛客网-整数次方

def solution(base:float, x:int)-> float:
    # your code here

assert solution(2.00000,3) == 8.00000
assert solution(2.10000,3) == 9.26100
2 
def solution(base: float, x: int) -> float:
    # your code here
    if base == 0 and x == 0:
        return float(-1)
    else:
        return round(base ** x, 5)


assert solution(2.00000, 3) == 8.00000
assert solution(2.00000, -2) == 0.25000
assert solution(2.10000, 3) == 9.26100
assert solution(0, 0) == -1.0
3 
def solution(base:float, x:int)-> float:
        if x <0:
            base=1/base
            x=-x
        sum=1
        for i in range(x):
            sum=sum*base
        return round(sum,5)

assert solution(2.00000,3) == 8.00000
assert solution(2.10000,3) == 9.26100
4 

def solution(base:float, x:int)-> float:
    # your code here 这个题有点一言难尽了啊
    return round(pow(base,x),5)
assert solution(2.00000,3) == 8.00000
assert solution(2.10000,3) == 9.26100
6 
def solution(base:float, x:int)-> float:
    if base == 0 and x <= 0:
        return -1
    else:
        return ("%.5f" %(base ** x))
7 
def solution(base:float, x:int)-> float:
    multiplication=1
    count = 0
    while count< abs(x):
        multiplication = multiplication*base
        count += 1
    if x>=0:
        return '%.5f' %multiplication
    else:
        return '%.5f' %(1/multiplication)
8 
def solution(base: float, x: int) -> float:
    # your code here
    # return round(base ** x, 5)
    return round(pow(base, x), 5)


assert solution(2.00000, 3) == 8.00000
assert solution(2.10000, 3) == 9.26100
9 
def solution(base:float, x:int)-> float:
    return round((base**x),5)

assert solution(2.00000,3) == 8.00000
assert solution(2.10000,3) == 9.26100