#!/usr/bin/python
# -*- coding: utf-8 -*-
def 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 t
def 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
return
def 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 s
def 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 mt
def 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+=1
def 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 s
def 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+b
class qi:
num0=1
st=''
def __init__(self, str1=None):
self.st=str1
self.num0=str1.find('')
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]!='':
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 = ""
s3 = ""
i = len(s3)-s1.find("")
s3=s3[i:]+s3[:i]+''
a=qi(s1)
dd=[s1]
ee={s1:''}
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!='':
qi(e).show()
print num
num+=1
e=ee[e]
return
if __name__ == '__main__':
s=""
for i in range(94,101):
s+='z'+str(i)+'()\n'
exec(s)