먼저 1과 0의 조합으로 사용자 입력 문자열을 가져옵니다. 그런 다음 1로 새 문자열을 만든 다음 연속 1이 있는 p 번호가 있는지 확인합니다. 존재하는 경우 FOUND를 표시하고 그렇지 않으면 NOTFOUND를 표시합니다.
예
Binary number ::1111001111 Enter consecutive 1’s :3 Consecutive 1's is Found
알고리즘
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
예시 코드
# 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)
출력
Enter Binary number ::1111001111 Enter consecutive 1's ::3 Consecutive 1's is Found