Python：每日一题 29（答题领鱼币）

新手·ing

感谢冬雪的题目！！！

python有分数类型和相应的计算方法，要求不用分数模块（真是见鬼！）编写一个简单的分数加法和减法函数，实现分数与分数，分数与整数的加减法。如

3/5 - 1/5 = 2/5

1/2 - 1/3 = 1/6

1/2 + 1/2 = 1

1- 1/3 = 2/3

函数的用法

add(a, b, c, d)

a, b为第一个分数的分子与分母

c, d 为第二个分数的分子与分母

减法类似。

给大家不符合规则的：

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jerryxjr1220

#!/usr/bin/env python2
# -*- coding: utf-8 -*-
"""
Created on Sun Apr 16 09:45:31 2017

@author: jerry_xu
"""

def fadd(a,b,c,d):
    fenzi = a*d+b*c
    fenmu = b*d
    GCD = lambda x,y:x if y==0 else GCD(y, x%y)
    gcd = GCD(fenzi,fenmu)
    fenzi /= gcd
    fenmu /= gcd
    return fenzi, fenmu

print '1/6 + 1/3 = %d/%d' % fadd(1,6,1,3)

def fminus(a,b,c,d):
    fenzi = a*d-b*c
    fenmu = b*d
    GCD = lambda x,y:x if y==0 else GCD(y, x%y)
    gcd = GCD(fenzi,fenmu)
    fenzi /= gcd
    fenmu /= gcd
    return fenzi, fenmu

print '1/2 - 1/3 = %d/%d' % fminus(1,2,1,3)

输出：
1/6 + 1/3 = 1/2
1/2 - 1/3 = 1/6

ooxx7788

看看不符合规则的

新手·ing

ooxx7788 发表于 2017-4-16 10:27
看看不符合规则的

zhhmms

#!/usr/bin/env python2
# -*- coding: utf-8 -*-
"""
Created on Sun Apr 16 09:45:31 2017

@author: jerry_xu
"""

def fadd(a,b,c,d):
    fenzi = a*d+b*c
    fenmu = b*d
    GCD = lambda x,y:x if y==0 else GCD(y, x%y)
    gcd = GCD(fenzi,fenmu)
    fenzi /= gcd
    fenmu /= gcd
    return fenzi, fenmu

print '1/6 + 1/3 = %d/%d' % fadd(1,6,1,3)

def fminus(a,b,c,d):
    fenzi = a*d-b*c
    fenmu = b*d
    GCD = lambda x,y:x if y==0 else GCD(y, x%y)
    gcd = GCD(fenzi,fenmu)
    fenzi /= gcd
    fenmu /= gcd
    return fenzi, fenmu

print '1/2 - 1/3 = %d/%d' % fminus(1,2,1,3)

奔跑的小鱼

没看懂题目
def my_add(a,b,c,d):
    return a/b+c/d
感觉这个就能实现了

冬雪雪冬

奔跑的小鱼 发表于 2017-4-17 11:40
没看懂题目
def my_add(a,b,c,d):
    return a/b+c/d

返回的需要是分数

奔跑的小鱼

又看了下,看懂题目了
def Dicisor(a,b):
    #最大公约数
    dicisor_list = [i for i in range(1,a+1) if a%i==0 and b%i==0]
    return max(dicisor_list)

def my_add(a,b,c,d):
    molecular = a*d +c*b
    denominator =b*d
    result = molecular/denominator
    if result == int(result):
        print(int(result))
    else:
        dicisor = Dicisor(molecular,denominator)
        molecular_new = int(molecular/dicisor)
        denominator_new = int(denominator/dicisor)
        print("%s/%s"%(molecular_new,denominator_new))

奔跑的小鱼

zhhmms 发表于 2017-4-17 08:59
#!/usr/bin/env python2
# -*- coding: utf-8 -*-
"""

你这个递归用的牛

奔跑的小鱼

冬雪雪冬 发表于 2017-4-17 11:57
返回的需要是分数

看懂了,楼下是我的答案

余欲渔

ab=input('请输入需要计算的分式,如1/5-2/8:')
f=0
def yf(aa,bb):
    for i in range(aa,1,-1):
        if (aa/i)%1==0 and (bb/i)%1==0:
            aa=int(aa/i)
            bb=int(bb/i)
            break
    if bb>1:
        return str(aa)+r'/'+str(bb)
    else :return str(aa)
            
        
def jg(a1,a2,b1,b2):
    if f:
        aa=a1*b2+a2*b1
    else:
        aa=a1*b2-a2*b1
    bb=a2*b2
    return yf(aa,bb)
        
if '-'in ab:
    a,b=ab.split(sep='-')
elif '+'in ab:
    a,b=ab.split(sep='+')
    f=1
else :
    print('输入错误')
try:
    a1,a2=[int(i) for i in a.split(sep=r'/')]
except:
    a1=int(a)
    a2=1
try:
    b1,b2=[int(i) for i in b.split(sep=r'/')]
except:
    b1=int(b)
    b2=1
print(jg(a1,a2,b1,b2))

willLin

看看不符合规则的

solomonxian

应该要用到最大公约数来约分，还要判断一次是不是整数

def gcd(m,n):
    """辗转相除法，寻找n和m的最大公约数"""
    # 本来还想分大小，结果发现不分也可以，因为会换过来
    while n != 0:
        m,n = n, m%n
    return m

# 用公倍数处理，再除以最大公约数就可以相加了 (a*d)/(b*d) + (c*b)/(d*b)
def fractional_add(a,b,c,d):
    """分数加法，a/b + c/d """
    
    numerator = a*d + c*b   # 分子
    denominator = b*d      # 分母
    if numerator % denominator == 0:
        # 整除情况下打印整数
        print(numerator // denominator)
    else:
        divisor = gcd(numerator, denominator)
        print("%d / %d" % (numerator/divisor, denominator/divisor))

def fractional_minus(a,b,c,d):
    """分数减法, a/b - c/d"""
    
    numerator = a*d - c*b
    denominator = b*d
    if numerator % denominator == 0:
        # 整除情况下打印整数
        print(numerator // denominator)
    else:
        divisor = gcd(numerator, denominator)
        print("%d / %d" % (numerator/divisor, denominator/divisor))

keyi198707

看看

张大象

def yf(m,n):
        for i in range(2,m):
                if m%i == 0 and n%i == 0:
                        m,n = m//i,n//i
                        yf(m,n)
        return m,n

def add(a, b, c, d):
        x = b*d
        y = a*d
        z = c*b
        count = 0
        s = y + z
        s,x = yf(s,x)
        print('%s/%s'%(s,x))

renze

def add_f(a,b,c,d):
    e = a*d+b*c
    f = b*d
    return e ,f


def sub_f(a,b,c,d):
    e = a*d-b*c
    f = b*d
    return e ,f

gausser

from __future__ import division

# Get a and b 's the Maximum common divisor
def Max_Common_Divisor(a, b):
    while b != 0:
        c = a % b 
        a = b 
        b = c
    return a  # a is max common divisor

# Get a and b's the least common multiple
def Min_Common_Multiple(a, b):
    return a * b / a 

# Fractional Addition
# a is the first fraction's molecular, b is the first fraction's Denominator
# c is the second fraction's molecular, d is the second fraction's Denominator
# return the molecular and Denominator of Fraction_Add_Fraction's Add and Minus
def Add_Minus_Of_Fraction_Fraction(a, b, c, d):
    list = []
    New_Multiple = Min_Common_Multiple(b, d)
    b_Business = New_Multiple / b 
    d_Business = New_Multiple / d 
    
    a = a * b_Business
    c = c * d_Business
    
    # Get Add of Fractions
    Add_Molecular = a + c
    Add_Divisor = Max_Common_Divisor(New_Multiple, Add_Molecular)
    list.append(int(Add_Molecular / Add_Divisor))
    list.append(int(New_Multiple / Add_Divisor))
    
    # Get Minus of Fractions 
    Minus_Molecular = a - c 
    Minus_Divisor = Max_Common_Divisor(New_Multiple, Minus_Molecular)
    list.append(int(Minus_Molecular / Minus_Divisor))
    list.append(int(New_Multiple / Minus_Divisor))
    
    print "%d/%d + %d/%d = %d/%d" % (a, b, c, d, list[0], list[1])
    print "%d/%d - %d/%d = %d/%d" % (a, b, c, d, list[2], list[3])

石小龙

不会做

wyh2006

学习学习！

achencan

1