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瞬秒爆加速 发表于 2018-3-11 13:27:05

def gcd(x,y):
while y:
   x,y = y,x%y
return x
def add(a,b,c,d):
if b==d:
   print("%d/%d" % ((a+c),b))
else:
   z = gcd((a*d + c*b),b*d)
   print("%d/%d" % ((a*d + c*b)//z,b*d//z))

大头目 发表于 2018-4-4 10:50:11

#学习
def Dicisor(a,b):
#最大公约数
dicisor_list =
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))

my_add(1,1,3,4)

小王溜溜 发表于 2018-4-4 22:57:39

666

kwm 发表于 2018-4-5 16:22:52

123

tsembrace 发表于 2018-4-11 22:42:49

'''
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 = 11- 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。
'''
def foo(n,m):
for i in range(2,min(n,m)+1):
   if n%i==0 and m%i==0:
         while n%i==0 and m%i==0:
            n=n//i
            m=m//i
return str(n)+'/'+str(m)

def add(a,b,c,d):
stri='%d/%d+%d/%d=' %(a,b,c,d)
n=a*d+c*b
m=b*d
if n%m==0:
   return stri+str(n//m)
else:
   return stri+foo(n,m)

def sub(a,b,c,d):
stri='%d/%d-%d/%d=' %(a,b,c,d)
n=a*d-c*b
m=b*d
if n%m==0:
   return n//m
else:
   strx=foo(abs(n),m)
if n<0:
   return stri+'-'+strx
else:
   return stri+strx

print(sub(1,2,2,6))
print(add(1,2,2,6))

小庆子 发表于 2018-5-15 09:24:43

def add(a,b,c,d):
print('%s／%s + %s／%s = '%(a,b,c,d),end='')
return (a/b)+(c/d)

萧丹夜 发表于 2018-5-24 15:46:52

本帖最后由萧丹夜于 2018-5-24 16:02 编辑

def add(a,b,c,d):
y1 = a * d + b * c
y2 = b * d
return simp(y1,y2)

def sub(a,b,c,d):
y1 = a * d - b * c
y2 = b * d
return simp(y1,y2)

def simp(y1,y2):
i = 2
while i <= y2:
   if y1 % i == 0 and y2 % i == 0:
         y1,y2 = y1 // i,y2 // i
         i = 1
   i += 1
return y1,y2

def jug(x):
if ' ' not in x:
   return int(x),1
else:
   return map(int,x.split())

x1 = input('请输入第一个数分子和分母：')
x2 = input('请输入第二个数分子和分母：')
a,b = jug(x1)
c,d = jug(x2)
k = input('请输入你想要做的运算（+/-）：')
if k == '+':
y1,y2 = add(a,b,c,d)
elif k == '-':
y1,y2 = sub(a,b,c,d)
else:
print('输入有误！')

if y1 % y2 != 0:
if b == 1:
   print('%d %c %d/%d = %d/%d ' %(a,k,c,d,y1,y2))
elif d == 1:
   print('%d/%d %c %d = %d/%d ' %(a,b,k,c,y1,y2))
else:
   print('%d/%d %c %d/%d = %d/%d ' %(a,b,k,c,d,y1,y2))
else:
if b == 1:
   print('%d %c %d/%d = %d ' %(a,k,c,d,y1//y2))
elif d == 1:
   print('%d/%d %c %d = %d ' %(a,b,k,c,y1//y2))
else:
   print('%d/%d %c %d/%d = %d ' %(a,b,k,c,d,y1//y2))

jrro452 发表于 2018-5-30 17:26:00

#求最大公约数：
def gcd(a, b):
if b == 0:return a
return gcd(b, a % b)
#最小公倍数
def lcm(a,b):
return a*b//gcd(a,b)
a,b,c,d =input('请输入4个数，空格隔开：').split()
e = input('请输入‘+’或‘-’：')
def caculating(a,b,c,d,e):
lcmNum = lcm(int(b),int(d))
S = a+'/'+b+e+c+'/'+d+'='
if e =='-':
x =int(int(a)*lcmNum/int(b)-int(c)*lcmNum/int(d))
else:
x =int(int(a)*lcmNum/int(b)+int(c)*lcmNum/int(d))
y=gcd(lcmNum,x)
x=int(x/y)
lcmNum = int(lcmNum/y)
return S+str(x)+'/'+str(lcmNum)
print(caculating(a,b,c,d,e))

天下独我意 发表于 2018-6-14 19:31:20

{:10_275:}

可乐必须加冰 发表于 2018-8-2 21:03:15

def add(a, b, c, d):
if b == d:
   if a + c == b:
         print("1")
   else:
         print((a + c), "/", b)
else:
   if a * d + b * c == b * d:
         print("1")
   else:
         print((a * d + b * c), "/", b * d)

小木子c 发表于 2018-8-3 21:37:00

from fractions import gcd

def add(a,b,c,d):
#m、n表示加数和被加数的最简形式，下面的减法同理
m = a//b if a % b == 0 else str(a//gcd(a,b)) + '/' + str(b//gcd(a,b))
n = c//d if c % d == 0 else str(c//gcd(c,d)) + '/' + str(d//gcd(c,d))
k = a*d + b*c
fm = b*d
if k % fm == 0:
   print(m,'+',n,'= %d' % (k//fm))
else:
   result = str(k//gcd(k,fm)) + '/' + str(fm//gcd(k,fm))
   print(m,'+',n,'= %s' % result)

def sub(a,b,c,d):
m = a//b if a % b == 0 else str(a//gcd(a,b)) + '/' + str(b//gcd(a,b))
n = c//d if c % d == 0 else str(c//gcd(c,d)) + '/' + str(d//gcd(c,d))
k = a*d - b*c
fm = b*d
if k % fm == 0:
   print(m,'-',n,'= %d' % (k//fm))
else:
   result = str(k//gcd(k,fm)) + '/' + str(fm//gcd(k,fm))
   print(m,'-',n,'= %s' % result)

测试结果
>>> add(8,2,1,4)
4 + 1/4 = 17/4
>>> add(6,10,2,7)
3/5 + 2/7 = 31/35
>>> add(3,3,4,6)
1 + 2/3 = 5/3
>>> sub(8,2,1,4)
4 - 1/4 = 15/4
>>> sub(1,4,1,5)
1/4 - 1/5 = 1/20
>>> sub(1,4,5,4)
1/4 - 5/4 = -1

uilUVUBWP 发表于 2018-8-18 08:43:42

我做的也是不符合规定的{:9_240:}
import fractions
def 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
x = fractions.Fraction(a, b)
y = fractions.Fraction(c, d)
print(x + y)
def subtraction(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
x = fractions.Fraction(a, b)
y = fractions.Fraction(c, d)
print(x - y)

z = add(1, 5, 3, 5)
print(z)
q = subtraction(3, 5, 1, 5)
print(q)

lyslyslys 发表于 2018-9-23 09:37:26

感谢

double-lee 发表于 2018-12-9 16:18:58

def add(a,b,c,d):
str1 = str(a)+'/'+str(b)+' + '+str(c)+'/'+str(d)+' = '
fz = a*d + c*b
fm = b*d
if fz == 0:
   return str1+'0'
elif fz == fm:
   return str1+'1'
elif fz > fm:
   if fz % fm == 0:
         return str1+str(int(fz/fm))
   else:
         n = fz % fm
         m = int(fz // fm)
         for i in range(n,0,-1):
            if fm % i == 0 and n % i == 0:
               return str1+str(m)+'('+str(int(n/i))+'/'+str(int(fm/i))+')'
else:
   for i in range(fz,0,-1):
         if fm % i == 0 and fz % i == 0:
            return str1+str(int(fz/i))+'/'+str(int(fm/i))

def sub(a,b,c,d):
str1 = str(a)+'/'+str(b)+' - '+str(c)+'/'+str(d)+' = '
fz = a*d - c*b
fm = b*d
if fz == 0:
   return str1+'0'
elif fz == fm:
   return str1+'1'

elif fz > fm:
   if fz % fm == 0:
         return str1+str(int(fz/fm))
   else:
         n = fz % fm
         m = int(fz // fm)
         for i in range(n,0,-1):
            if fm % i == 0 and n % i == 0:
               return str1+str(m)+'('+str(int(n/i))+'/'+str(int(fm/i))+')'
else:
   for i in range(fz,0,-1):
         if fm % i == 0 and fz % i == 0:
            return str1+str(int(fz/i))+'/'+str(int(fm/i))

1/2 + 1/2 = 1
6/2 + 9/3 = 6
6/2 + 8/3 = 5(2/3)
1/2 + 1/3 = 5/6
1/2 - 1/3 = 1/6
1/2 - 3/6 = 0
4/2 - 2/3 = 1(1/3)
9/5 - 4/5 = 1

sunyuanency 发表于 2019-1-3 14:20:26

持之以恒

咕咕鸡鸽鸽 发表于 2019-1-5 10:40:36

看看

北城以北灬 发表于 2019-3-15 14:39:57

def simplify(zi, mu):
x = 2
while x <= mu:
   if zi % x == 0 and mu % x == 0:
         zi = zi / x
         mu = mu / x
         return simplify(zi, mu)
   else:
         x += 1
if mu == 1:
   print(zi)
else:
   print(zi, "/", mu)

def add(a, b, c, d):
zi = a * d + b * c
mu = b * d
simplify(zi, mu)

def sub(a, b, c, d):
zi = a * d - b * c
mu = b * d
simplify(zi, mu)

add(1, 3, 1, 2)

5 / 6

Process finished with exit code 0

捞鬼发表于 2019-4-8 21:59:53

捞

永恒的蓝色梦想 发表于 2019-8-20 12:17:07

本帖最后由永恒的蓝色梦想于 2019-8-20 12:18 编辑

from math import gcd
def add(a,b,c,d):a=a*d+c*b;b=b*d;k=gcd(a,b);return f'{a//k}/{b//k}'
def sub(a,b,c,d):return add(a,b,-c,d)

永恒的蓝色梦想 发表于 2019-8-20 12:20:44

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

老哥上来就抄的么？

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