Python程序检查二进制数中是否有K个连续的1？
首先我们将用户输入的字符串与1和0组合在一起，然后用1创建新的字符串，然后检查是否存在p个连续的1。如果存在，则显示找到，否则显示未找到。
示例
Binary number ::1111001111 Enter consecutive 1’s :3 Consecutive 1's is Found
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范例程式码
# 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)输出结果
Enter Binary number ::1111001111 Enter consecutive 1's ::3 Consecutive 1's is Found
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