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2的幂次方

[编程题]2的幂次方

热度指数：7972 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 64M，其他语言128M
算法知识视频讲解

Every positive number can be presented by the exponential form.For example, 137 = 2^7 + 2^3 + 2^0。 Let's present a^b by the form a(b).Then 137 is presented by 2(7)+2(3)+2(0). Since 7 = 2^2 + 2 + 2^0 and 3 = 2 + 2^0 , 137 is finally presented by 2(2(2)+2 +2(0))+2(2+2(0))+2(0). Given a positive number n,your task is to present n with the exponential form which only contains the digits 0 and 2.

输入描述:

    For each case, the input file contains a positive integer n (n<=20000).

输出描述:

    For each case, you should output the exponential form of n an a single line.Note that,there should not be any additional white spaces in the line.

示例1

输入

输出

2(2(2+2(0))+2)+2(2(2+2(0)))+2(2(2)+2(0))+2+2(0)

算法知识视频讲解

查看答案及解析

Python

TimeMac

简单易懂，最简洁实现。一个转换函数，递归实现

def convenBin(num):
    numBin = bin(num).replace('0b','')        #转换成2进制
    numPower = len(numBin)-1                  #对应长度的2进制次幂
    result = []                           
    while numBin:
        if numBin[0] == '1':                  #如果该二进制位为1我们才加入结果
            if numPower >= 2:                 #次幂大于等于2我们要递归它
                result.append('2(%s)'% convenBin(numPower))
            elif numPower == 1:               #1和0不能通用解决，就单独列出来
                result.append('2')
            else:
                result.append('2(0)')
        numPower -= 1
        numBin = numBin[1:]
    return '+'.join(result)                   #返回用+相接的结果
while True:
    try:
        print(convenBin(int(input)))
    except Exception:
        break

编辑于 2018-10-14 10:19:46 回复(0)