题目

For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference, product and quotient.

Input Specification:

Each input file contains one test case, which gives in one line the two rational numbers in the format a1/b1 a2/b2. The numerators and the denominators are all in the range of long int. If there is a negative sign, it must appear only in front of the numerator. The denominators are guaranteed to be non-zero numbers.

Output Specification:

For each test case, print in 4 lines the sum, difference, product and quotient of the two rational numbers, respectively. The format of each line is number1 operator number2 = result. Notice that all the rational numbers must be in their simplest form k a/b, where k is the integer part, and a/b is the simplest fraction part. If the number is negative, it must be included in a pair of parentheses. If the denominator in the division is zero, output Inf as the result. It is guaranteed that all the output integers are in the range of long int.

Sample Input 1:

1

2/3 -4/2

Sample Output 1:

1
2
3
4

2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

Sample Input 2:

1

5/3 0/6

Sample Output 2:

1
2
3
4

1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf

题解

思路

• 为了方便我使用了Fraction类，该类可以直接读取“-1/4”的字符串并当作数来存储以及加减乘除，它可以自动转换为最简形式，但是不能转为分离整数的形式。
• 如果不记得这个类的话，就硬怼，最后用最大公因数化简分子分母。
• 读入就按照Fraction类来读入
• 简单使用加减乘除便可得到结果
• 要注意除法的话，要区别除数是否为0,为0的话除法就是Inf
• 复杂的是把这个结果转为我们想要的格式
• 我这里写了一个函数方便转换。
• 首先是如果是负数的话，要将其反转
• 并且记录一下输出的时候要再外面包裹(-和)。
• 我们把分数分为整数部分和小数部分，两者０与非０共组成４种组合。需要对这四种组合分别讨论。
• 看整数部分，由ans.numerator // ans.denominator决定。即分子除以分母
  • 当它不为０的时候，要将返回的字符串添加上它。
    • 这时候再看小数部分，即分子与分母求余
    • 当小数部分不为０的时候，字符串添加“空格　＋　分子　＋　除号　＋　分母”
    • 为０的时候什么都不添加
  • 当它为０的时候，什么都不添加
    • 这时候再看小数部分，即分子与分母求余
    • 看小数部分，如果不为０，添加“分子　＋　除号　＋　分母”
    • 当它也是０的时候，添加一个“０”
• 如果是负数的话，别忘了输出的时候要再外面包裹(-和)
• 返回这个新的字符串即可

数据结构

+　a,b是原来的两个数
+　 _sum, diff, prod, quo　是和，差，积，商
+　在fraction函数中
+　ans 是原Fraction
+　string 是要返回的新字符串
+　negative 记录是否是负数

算法

• 照着思路里的算就可以了

代码

• 因为使用Python 可以AC，因此只放了Python的解

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from fractions import Fraction


def fraction(ans):
    string = ""
    negative = False
    if ans < 0:
        ans = -ans
        negative = True
    if ans.numerator // ans.denominator != 0:
        string += str(ans.numerator // ans.denominator)
        if ans.numerator % ans.denominator != 0:
            string += (" " + str(ans.numerator % ans.denominator) + "/" + str(ans.denominator))
    else:
        if ans.numerator % ans.denominator != 0:
            string += (str(ans.numerator % ans.denominator) + "/" + str(ans.denominator))
        else:
            string += "0"
    if negative:
        string = "(-" + string + ")"
    return string


a, b = list(map(Fraction, input().split()))
if b != 0:
    _sum, diff, prod, quo = fraction(a + b), fraction(a - b), fraction(a * b), fraction(a / b)
else:
    _sum, diff, prod, quo = fraction(a + b), fraction(a - b), fraction(a * b), "Inf"
a, b = fraction(a), fraction(b)
# 输出
print(a, "+", b, "=", _sum)
print(a, "-", b, "=", diff)
print(a, "*", b, "=", prod)
print(a, "/", b, "=", quo)