[Python](PAT)1060 Are They Equal（25 分）

最新推荐文章于 2020-08-21 16:07:42 发布

Amos H

最新推荐文章于 2020-08-21 16:07:42 发布

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分类专栏： PAT Python 文章标签： PAT1060 python PAT 字符串操作

本文链接：https://blog.youkuaiyun.com/qq_35499060/article/details/82184826

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该博客讨论了PAT1060题目的解决方案，涉及Python中浮点数的比较，特别是在机器只能保存有限位数的显著数字时。文章详细解释了输入输出规格，提供样本输入和输出，并分析了如何处理不同类型的数字，以判断它们是否在机器中被视为相等。最后，给出了Python代码实现。

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If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123×105 with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.

Input Specification:

Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10100, and that its total digit number is less than 100.

Output Specification:

For each test case, print in a line YES if the two numbers are treated equal, and then the number in the standard form 0.d[1]...d[N]*10^k (d[1]>0 unless the number is 0); or NO if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.

Note: Simple chopping is assumed without rounding.

Sample Input 1:

3 12300 12358.9

Sample Output 1:

YES 0.123*10^5

Sample Input 2:

3 120 128

Sample Output 2:

NO 0.120*10^3 0.128*10^3

题目大意

给定N，A，B三个数字，将A和B转换成科学计数法，要求保留N位小数，如果转换出来的结果相同，则输出YES和转换结果，否知输出NO和A、B的转换结果。

分析

本来想用format()函数转换成科学计数法，然后进行格式化输出，但是format()的缺陷是超过6位就不精确了，所以还是老实来吧。

把数字分为四类，进行不同的处理：

1>等于0. 最简单的，直接格式化输出字符串

2>小于1但是大于0.获取科学计数法的小数和指数部分

3>大于1但是包含小数点的。先去除掉小数点前一个0之前的所有0。（考虑00000.123这个例子）去除小数点，并在最前面加上'0.'，并在最后加上N个0

4>大于1且没有小数点的，去除掉前面所有的0（考虑0005这个例子）。在前面加上'0.',在后面加上N个'0'。

最后输出的格式是统一的，w是指数，而q因为最后都加了N个0，所以输出q[:n+2]。

Python实现

def change(num, n):
    if float(num) < 1:
        if  float(num) == 0:
            q = '0.'+'0'*n
            w = 0
        else:
            a = str(int(num[2:]))
            w = -(len(num) - 2 - len(a))
            q = '0.' + a + '0'*n
    else:
        start = 0
        if '.' in num:
            for x in range(len(num)):
                if num[x+1] == '.' or int(num[x]) >0:
                    break
            num = num[x:]
            w = num.find('.')
            q = '0.'+num.replace('.' ,'') + '0'*n
        else:
            for x in range(len(num)):
                if int(num[x]) > 0:
                    break
            num = num[x:]
            w = len(num)
            q = '0.'+num+'0'*n
    return '{}*10^{}'.format(q[:n+2], w)
            
if __name__ == "__main__":
    line = input().split(" ")
    n = int(line[0])
    a, b = line[1], line[2]
    first, second  = change(a, n), change(b, n)
    if first == second:
        print("YES", first)
    else:
        print("NO", first, second)

python写PAT甲级，答案都在这了