Python 程式檢查二進位制數中是否存在 K 個連續 1？

首先，我們輸入一個由 1 和 0 組合成的使用者字串。然後，建立一個由 1 組成的新字串，接下來檢查是否存在任意數量的連續 1。如果存在，則顯示 FOUND，否則顯示 NOTFOUND。

示例

Binary number ::1111001111
Enter consecutive 1’s :3
Consecutive 1's is Found

演算法

Step 1: input a string with the combination of 1’s, it’s stored in the variable X and 0’s and p is the consecutive 1’s in a binary number.
Step 2: form a new string of p 1’s.
   newstring=”1”*p
Step 3: check if there is p 1’s at any position.
   If newstring in X
      Display “FOUND”
   Else
      Display “NOT FOUND”
   End if

示例程式碼

# To check if there is k consecutive 1's in a binary number 
def binaryno_ones(n,p):
   # form a new string of k 1's 
   newstr = "1"*p

   # if there is k 1's at any position 
   if newstr in n:
      print ("Consecutive 1's is Found")
   else:
      print (" Consecutive 1's is Not Found")

# driver code
n =input("Enter Binary number ::")
p = int(input("Enter consecutive 1's ::"))
binaryno_ones(n, p)

輸出

Enter Binary number ::1111001111
Enter consecutive 1's ::3
Consecutive 1's is Found

karthikeya Boyini

更新於：30-Jul-2019

297 次瀏覽

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