# -*- coding: cp936 -*-#第十二章----其他趣味程序#:《c趣味编程》94-100题#21:51 2006-11-4def z94(): #斐波那契数列 def filie(x): a,b,t=1,1,0 if x==1 or x==2:return 1 while t!=x-2: a,b,t=b,a+b,t+1 return b for i in range(1,30): print filie(i)def z95(): #把三位数字转化成罗马数字 def lm(x): fy=[["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"], ["","X","XX","XXX","XL","L","LX","LXX","LXXX","XCC"], ["","I","II","III","IV","V","VI","VII","VIII","IX"]] e=reduce(lambda x,y:x+y, map(lambda x,y:x[y],fy,map(lambda x:int(x),list(str(x))))) print e lm(863)def z96(): #7个选手的得分分别是{5,3,4,7,3,5,6}序号是1~7，规则是得分越高，名次越低 #而相同的得分名次一样,输出名次是{3,1,2,5,1,3,4} z=[5, 3, 4, 7, 3, 5, 6] k=[0]*7 max1=0 minc=0 b=map(lambda x:list(x),zip(range(1,8),z,k)) b=sorted(b,cmp=lambda x,y: cmp(x[1], y[1])) for i in b: if i[1]>max1: max1=i[1] minc+=1 i[2]=minc b=sorted(b) t=map(lambda x:x[2],b) print tdef z97(): #满足一定条件的序列：如3，2，2，1四数,他们的和是8,并且3>=2>=2>=1;找出所有这样的序列 n=18 r=[1]*4 num=1 for r[1] in range(r[0],n-sum(r[:0])): for r[2] in range(r[1],n-sum(r[:1])): for r[3] in range(r[2],n-sum(r[:2])): t=n-sum(r) if t>=r[3]: tt=r[1:]+[t] print num,tt num+=1 returndef z97_b(): exec(zz97(23,6)) z97_a()def zz97(n,k): s='def z97_a():\n' tb=' ' s+=tb+'n='+str(n+1)+'\n' s+=tb+'r=[1]*'+str(k)+'\n' g=lambda x:tb*(x)+'for r['+str(x)+'] in range(r['+str(x-1)+\ '],n-sum(r[:'+str(x-1)+'])):\n' for i in range(1,k): s+=g(i) e=tb*(k+1) s+=e+'t=n-sum(r)\n'+e+'if t>=r['+str(k-1)+']:\n' e+=tb s+=e+'tt=r[1:]+[t]\n'+e+'print tt' print s return sdef remo(x1,y): mt=[] for now in range(1,y+1): t=True for x in range(len(x1)): m=x-len(x1) if now==x1[x] or now==x1[x]+m or now==x1[x]-m: t=False if t: mt+=[now] return mtdef z98_4(): # 四皇后问题 ,就是把四个国际象棋里的皇后放到棋盘里问怎么放不互相吃 n=4 k=[1]*n num=1 for k[0] in remo(k[:0],n): for k[1] in remo(k[:1],n): for k[2] in remo(k[:2],n): for k[3] in remo(k[:3],n): print num,k num+=1def zz98(n): s='def z98_a():\n' tb=' ' s+=tb+'n='+str(n)+'\n' s+=tb+'k=[1]*n\n' s+=tb+'num=1\n' g=lambda x:tb*(x+1)+'for k['+str(x)+'] in remo(k[:'+str(x)+'],n):\n' for i in range(0,n): s+=g(i) e=tb*(n+1) s+=e+'print num,k\n' s+=e+'num+=1\n' print s return sdef z98(): # 八皇后问题 ,就是把八个国际象棋里的皇后放到棋盘里问怎么放不互相吃 exec(zz98(8)) z98_a()def z99(): #超长正整数的加法 a=122414354367l b=23157465721578l print a,"+",b,"=",a+bclass qi: num0=1 st='1' def __init__(self, str1=None): self.st=str1 self.num0=str1.find('0') return def show(self): st=list(self.st) e=st[:] e[7],e[8],e[5],e[4],e[3]=st[5],st[4],st[3],st[8],st[7] m=0 for i in e: m+=1 print i, if m%3==0:print return def move(self,x): st=list(self.st) if st[x]!='1': st[x],st[self.num0]=st[self.num0],st[x] return ''.join(st) def z100(): '''数字移动 ::在图上的九个点上，空出中间的点，其余的 点上任意填上1~8 8个数字，然后移动除了一之外其余的 数字，使1到8顺时针从小到大排列，规则是，只能将数字 沿线移动到空白的位置。''' def nexts(d,e): tt=[] def add (a,b,t=tt,ee=e): if a not in ee: ee[a]=b t+=[a] return for i in d: x=qi(i) if (x.num0==8): for j in range(0,8): add(x.move(j),i) else: t=x.num0 add(x.move((t+1)%8),i) add(x.move((t+7)%8),i) add(x.move(8),i) return tt s1 = "856321740" s3 = "12345678" i = len(s3)-s1.find("1") s3=s3[i:]+s3[:i]+'0' a=qi(s1) dd=[s1] ee={s1:'0'} while s3 not in ee: dd=nexts (dd,ee) e=ee[s3] num=1 qi(s3).show() num=1 print num num+=1 while e!='0': qi(e).show() print num num+=1 e=ee[e] returnif __name__ == '__main__': s="" for i in range(94,101): s+='z'+str(i)+'()\n' exec(s)
