# -*- coding: cp936 -*-#第六章----分数趣题#《c趣味编程》42-47题#21:46 2006-11-4def gcd(x,y): #最大公因子 if x>y:x%=y while x: x,y=y%x,x return ydef lcm(x,y): #最小公倍数 return x*y/gcd(x,y)def jhua(x): t=gcd(x[0],x[1]) return [x[0]/t,x[1]/t]def z42(): x,y=(120,36) print x,y,"的最小公倍数是",lcm(x,y) print x,y,"的最大公因子是",gcd(x,y)def z43(): #比较两个分数的大小 def bijiao(x,y): x,y=jhua(x),jhua(y) m,n=x[0]*y[1],x[1]*y[0] if m>n: return '>' elif m<n: return '<' else: return '=' x,y=[1235,2356],[1357,2468] print x,bijiao(x,y),ydef z44(): #找出方程1/x+1/y+1/z+1/a=1的解 def addf(x,y): x,y=jhua(x),jhua(y) return jhua([x[0]*y[1]+x[1]*y[0],y[1]*x[1]]) ai=range(2,13) aii=range(2,43) num=1 for i in ai: for j in ai: for m in ai: for k in aii: if i<=j<=m<=k: r=[i,j,m,k] t=reduce(addf, map(lambda x:[1,x],r)) if t==[1,1]: print num,r num+=1 def z45(): def fena(x,y): #分子为1的分数叫做埃及分数把一个分数a/b分解成n个埃及分数的和 sh,yu=divmod(x,y) re=[] if yu: t=sh+1 re+=[t] y,x=y*t-x,x*t re+=fena(x,y) else: re+=[x/y] return re print fena(99,19)def z46(): #找出分子小于40，分母等于40的最简真分数,就是说分子和分母没有公因子的分数 num=1 for i in range(1,40): y=40 if gcd(i,y)==1: print num,[i,y] num+=1def z47(): def divf(x,y): yu=[] sh=[] sh1,yu1=divmod(x,y) sh+=[str(sh1),'.'] if yu1==0:return ''.join(sh[:-1]) while yu1 : sh1,yu1=divmod(yu1*10,y) sh+=[str(sh1)] if yu1 not in yu: yu+=[yu1] else: t=yu.index(yu1) sh.insert(t+3,'(') sh.append(')') break return ''.join(sh) print divf(3737,27000) print divf(12,100) print divf(120,12)if __name__ == '__main__': s="" for i in range(42,48): s+='z'+str(i)+'()\n' exec(s)
