Python Exercise 7

#!/usr/bin/python#-*-coding:utf-8-*-defgcd (x, y):#Max male Factor    ifx>y:x%=y whileX:x,y=y%x,xreturnydefLCM (x, y):#least common multiple    returnx*y/gcd (x, y)defJhua (x): T=GCD (x[0],x[1])    return[x[0]/t,x[1]/T]defz42 (): x, y= (120,36)    PrintX, Y,"The least common multiple is", LCM (x, y)PrintX, Y,"The maximum common factor is", gcd (x, y)defz43 ():#Compare the size of a two score    defBijiao (x, y): x, y=Jhua (x), Jhua (y) m,n=x[0]*y[1],x[1]*Y[0]ifM>N:return '>'        elifm<N:return '<'        Else:            return '='x, y=[1235,2356],[1357,2468]    PrintX,bijiao (x and y), ydefz44 ():#Find out the solution of equation 1/x+1/y+1/z+1/a=1    defaddf (x, y): x, y=Jhua (x), Jhua (y)returnJhua ([x[0]*y[1]+x[1]*y[0],y[1]*x[1]]) AI=range (2,13) aII=range (2,43) Num=1 forIinchAI: forJinchAI: forMinchAI: forKinchaII:ifi<=j<=m<=K:r=[I,j,m,k] t=reduce (ADDF, Map (LambdaX:[1, x],r))ift==[1,1]:                            Printnum,r Num+=1defz45 ():defFena (x, y):#the fraction of the molecule 1 is called the Egyptian score. decomposes a fraction A/b into n Egyptian fractions andsh,yu=divmod (x, y) re=[]        ifYu:t=sh+1Re+=[T] y,x=y*t-x,x*T re+=Fena (x, y)Else: Re+=[x/y]returnRePrintFena (99,19)defz46 ():#find the simplest fraction of a molecule that is less than 40 and the denominator equals 40, meaning that the numerator and denominator have no common factor fractions.Num=1 forIinchRange (1,40): Y=40ifGCD (i,y) ==1:            PrintNum,[i,y] Num+=1defz47 ():defdivf (x, y): Yu=[] SH=[] sh1,yu1=divmod (x, y) sh+=[str (SH1),'.']        ifYu1==0:return "'. Join (Sh[:-1])         whileyu1:sh1,yu1=divmod (yu1*10, y) sh+=[Str (SH1)]ifYu1 not inchYu:yu+=[YU1]Else: T=Yu.index (YU1) Sh.insert (t+3,'(') Sh.append (')')                 Break        return "'. Join (SH)PrintDIVF (3737,27000)    PrintDIVF (12,100)    PrintDIVF (120,12)if __name__=='__main__': S=""     forIinchRange (42,48): S+='Z'+str (i) +'() \ n'    exec(s)

Python Exercise 7