Python：每日一题 174,Python交流,技术交流,鱼C论坛

冬雪雪冬 发表于 2018-4-5 17:55:27

Python：每日一题 174

本帖最后由冬雪雪冬于 2018-4-7 20:21 编辑

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题目：找出斐波那契数列中包含0~9的连续10个数字。
斐波那契数列大家都很熟悉，1, 1, 2, 3, 5, 8, 13, 21, 34, 55......
如果把这些数量组成一个字符串就成为：'11235813213455'，这里面'2134'是包含1234这四个数字的，继续下去可以找到第一个含有0~9这10个数字的地方（10个数字的次序随意），可能处于一个斐波那契数，有可能跨越2个或多个斐波那契数，现在要求你找到它，并打印出这十个数字，例如：9034621587

shigure_takimi 发表于 2018-4-5 20:01:52

本帖最后由 shigure_takimi 于 2018-4-5 21:05 编辑

def fib(n):
s = ''
a = 0
b = 1
for i in range(n):
a,b = b,a+b
s += str(a)
return s

n = 400
s = fib(n)
length = len(s)
isFind = False
while not isFind:
for i in range(length):
if len(set(s)) == 10:
print(s)
isFind = True
break
else:
n += 200
s = fib(n)
length = len(s)
#第一个数为9034621587。
#如果去掉break，前400个斐波那契数组成的字符串包含如下符合条件的数。
9034621587
5037496128
4538679120
5386791204
0685291437
1307952468
6823514097
0625379814
3495718260
4872635019
8094635172

晓屁屁 发表于 2018-4-5 20:26:47

真的没明白题的意思

天圆突破 发表于 2018-4-5 20:34:40

本帖最后由天圆突破于 2018-4-7 14:54 编辑

def feb():
a = b = 1
while True:
   yield str(a)
   a,b = b,a+b

def st(string):
n = len(string)
for i in range(n-10):
   if len(set(string)) == 10:
         return True,string
else:
   return False,None

def count():
f, string = feb(), ''
while True:
   while len(string) <= 10:
         string += next(f)
   p = st(string)
   if p == True:
         print(p)
         input()
   string = string[-10:]

if __name__ == '__main__':
count()

溯影发表于 2018-4-5 21:15:51

#!usr/bin/env python
#_*_ coding:utf-8 _*_

def fib(number):
lst = [];
lst.append(1);
lst.append(1);

string = '';
string = string + str(lst);

for i in range(1,number):
   lst.append(lst+lst);
   string = string + str(lst);
return string;

def test(string):
result = '';
for i in range(len(string)):
   temp = string;
   if len(temp) == 10:
         set1 = set(temp)
         if len(set1) == 10:
            result = result + temp;
            return result;

return result;



if __name__ == '__main__':
number = int(input("请输入你想写多少项:"));
if len(test(fib(number))):
   print(test(fib(number)));
else:
   print("Nothing");
运行结果:
请输入你想写多少项:200
9034621587

塔利班 发表于 2018-4-5 21:18:41

本帖最后由塔利班于 2018-4-7 21:27 编辑

a,b,c=0,1,1
s=''
while c:
a,b=b,a+b
s+=str(a)
try:
   for i in range(len(s)-10):
         if len(set(s))==10:
            print(s)
            c=0
            break
except:
   pass

yunying12321 发表于 2018-4-5 21:29:18

def fib():
a = 0
b = 1
while True:
   a, b = b, a+b
   yield a

def check_num(ss):
a = []
for each in ss:
   if each not in a:
         a.append(each)
   else:
         break

if len(a) == 10:
   return True
else:
   return False

def main():
ss = ''
num_list = []
NUM = 50
f = fib()
running = True
m = n = 0

while True:
   for i in range(NUM):
         ss += str(next(f))

   m = len(ss)

   for i in range(n, len(ss)):
         if i <= len(ss)-10:
            if check_num(ss):
               print('\n找到这个数了：',ss)
               print('m=%d, n=%d, NUM=%d' %(m,n,NUM))
               running = False
               break
         else:
            break

   if running == False:
         break

   print(ss[(NUM-50):NUM],end='')
   n = len(ss) - 10
   NUM += 50

if __name__ == '__main__':
main()

达锅发表于 2018-4-5 21:38:47

def findDigits():
m,n,txt=1,1,"11"#mn分别是数列前后项
while len(set(txt[:10]))!=10:#每次判断前10位，因为10个数字不一样，所以转换成集合后长度为10的符合条件
   while len(txt)<11:#文本长度不足时生成数列下一个值
         m,n=n,m+n#生成下一个值
         txt +=str(n)#新增字符
   txt=txt #去掉第一位
return txt[:10]
print(findDigits())

总觉得有点啰嗦，但是又精炼不了

凌九霄 发表于 2018-4-5 22:51:57

本帖最后由凌九霄于 2018-4-5 23:00 编辑

def Fibonacci(n):
fibo =
if n == 1:
   return fibo
elif n == 2:
   return fibo
else:
   for i in range(3, n + 1):
         fibo.append(fibo[-2] + fibo[-1])
return fibo

#解决的方法比较笨拙，抛砖引玉看高手们的简便解法

fibstr = ''

for i in Fibonacci(500):
fibstr += str(i)

for j in range(0, len(fibstr)):
numstr = fibstr
if '0' in numstr and '1' in numstr and '2' in numstr and '3' in numstr and '4' in numstr and '5' in numstr and '6' in numstr and '7' in numstr and '8' in numstr and '9' in numstr:
   print(numstr)

吴有才 发表于 2018-4-5 23:28:08

shu=1000
a=0
b=10
def feibo(shu):
"生成一个斐波那契数列的list并转换为字符串"
a=0
b=1
list_1=[]
for i in range(shu):
   a,b=b,a+b
   list_1.append(str(a))
list_3 = ''.join(list_1)
return list_3

print(feibo(shu))
while b<len(feibo(shu)):
if '1'in feibo(shu):
   if '2' in feibo(shu):
         if '3' in feibo(shu):
            if '4' in feibo(shu):
               if '5' in feibo(shu):
                     if '6' in feibo(shu):
                        if '7' in feibo(shu):
                           if '8' in feibo(shu):
                                 if '9' in feibo(shu):
                                    if '0' in feibo(shu):
                                       print(feibo(shu))
                                       break
a=a+1
b=b+1

阿bang 发表于 2018-4-5 23:58:14

import time

def fibo():
a = 1
b = 1
while True:
   a, b = b, a+b
   yield b

def findallnum(n):
allstr = '11'
fibonum = fibo()
beginindex = 0
count = 0
while True:
   for i in range(beginindex, len(allstr) - 9):
         checknum = allstr
         numberlength = len(set(list(checknum)))
         if numberlength == 10:
            print(allstr)
            count += 1
            break
   beginindex = len(allstr) - 9
   allstr += str(next(fibonum))
   if count == n:
         break

if __name__ == '__main__':
start = time.time()
findallnum(10)
print(time.time() - start)

风凌渡 发表于 2018-4-6 01:20:16

125zl 发表于 2018-4-6 08:59:21

def fb():
a = 1
b = 1
c = a + b
s = True
str1 = ''
while s:
   str1 += str(a)
   a = b
   b = c
   c = a + b
   for i in range(len(str1) - 10):
         str2 = str1
         if fd(str2):
            print(str2)
            s = False
            break
def fd(str2):
list1 = ['0','1','2','3','4','5','6','7','8','9']
for i in str2:
   if i in list1:
         list1.remove(i)
   else:
         return False
return True
fb()

lan358088 发表于 2018-4-6 09:37:31

本帖最后由 lan358088 于 2018-4-6 10:10 编辑

def fei_bo(n):
a = 0
b = 1
i = 0
fei = []
while i < n:
   fei.append(str(b))
   a,b = b,a+b
   i+=1
qq = ''.join(fei)
return qq

n=10
flag=1
while flag:
qq=fei_bo(n)
num = len(qq)
for h in range(num-10):
   aa=list(qq)
   bb=aa.copy()
   bb.sort()
   if bb==['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']:
         print('成功了')
         print(aa)
         flag=0
         break
   else:
         n+=1

坑得飞起 发表于 2018-4-6 15:13:24

本帖最后由坑得飞起于 2018-4-10 00:41 编辑

def findd(ss):
for i in range(10):
   if ss.find(str(i))==-1:
         return False
print('Answer: ',ss)
return True
def fun(n):
a,b,i,jilu=0,1,0,''
while i<n :
   a,b,i,jilu=b,a+b,i+1,jilu[-10:]+str(b)
   for j in range(1,len(str(b))+1):
         if findd(jilu):
            pass
n=int(input())
fun(n)
input()

南城顾她 发表于 2018-4-6 15:43:51

def log(*args, **kwargs):
print(*args, **kwargs)

def fun(number):
if number == 1 or number == 2:
   return 1
elif number > 2:
   return fun(number-1) + fun(number-2)
else:
   return None

list1 = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
num = 11
a = 1
while True:
b = 0
sum1 = ''
sum3 = ''
c = False
dict1 = {}
for x in range(a, num):
   result = fun(x)
   sum1 += str(result)
for lis in list1:
   if lis in sum1:
         b += 1
   if b == 10:
         log(sum1)
         for sum2 in sum1:
            dict1 = 1
         for key in dict1.keys():
            sum3 += key
         log(sum3)
         c = True
if c == True:
   break
else:
   a += 10
   num += 10

BlueSand 发表于 2018-4-6 16:34:14

'''
题目：找出斐波那契数列中包含0~9的连续10个数字。
斐波那契数列大家都很熟悉，1, 1, 2, 3, 5, 8, 13, 21, 34, 55......
如果把这些数量组成一个字符串就成为：'11235813213455'，
这里面'2134'是包含1234这四个数字的，
继续下去可以找到第一个含有0~9这10个数字的地方（10个数字的次序随意），
可能处于一个斐波那契数，有可能跨越2个或多个斐波那契数，
现在要求你找到它，并打印出这十个数字，例如：9034621587
'''

#检测给定字符串是否合格
def KO(s):
for i in range(10):
if str(i) not in s:
return False
return True

n = 10
b = True
#为了提高效率，这里不用FB产生数列的项，而是用前面的产生后面的项
a1 = 1
a2 = 1

strfb = "11"
for i in range(1,5):
a1, a2 = a2, a1 + a2
strfb = strfb + str(a2)

print(strfb)
while b:
for i in range(len(strfb)-9):
if KO(strfb):
print(strfb)
print("在", a2, "里找到的，是第", n, "项")
b = False
break
if b:
print(n, "接着找......")
n += 1
a1, a2 = a2, a1 + a2
strfb = strfb[-9:] + str(a2)

BlueSand 发表于 2018-4-6 16:34:44

kerwin_lui 发表于 2018-4-6 16:41:53

啊啊

Chase_Kas 发表于 2018-4-6 22:54:14

def fibs():
a, b = 0, 1
while True:
   a, b = b, a+b
   yield a

string, f = '', fibs()
while True:
if len(set(string)) != 10:
   string += str(next(f))
   string = string[-11:-1]
else:
   print(string)
   break

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