Problem Statement
Recurrence relations are an important tool for the computer scientist. Many algorithms, particularly those that use divide and conquer, have time complexities best modeled by recurrence relations. A recurrence relation allows us to recursively define a sequence of values by defining the nth value in terms of certain of its predecessors.
Many natural functions, such as factorials and the Fibonacci sequence, can easily be expressed as recurrences. The function of interest for this problem is described below.
Let |An| denote the number of digits in the decimal representation of An. Given any numberA0, we define a sequence using the following recurrence:
Ai = |Ai-1| for i > 0
The goal of this problem is to determine the smallest positive i such that Ai = Ai-1.
Input consists of multiple lines, each terminated by an end-of-line character. Each line (except the last) contains a value for A0, where each value is non-negative and no more than a million digits. The last line of input contains the word END.
Output Format
For each value of A0 given in the input, the program should output one line containing the smallest positive i such that Ai = Ai-1.
Sample Input
9999
0
1
9999999999
END
Sample Output
3
2
1
4
Explanation
The first input value is A0 = 9999, resulting in A1 = |9999| = 4. Because 4 does not equal 9999, we find A2 = |A1| = |4| = 1. Since 1 is not equal to 4, we find A3 = |A2| = |1| = 1. A3 is equal to A2, making 3 the smallest positive i such that Ai = Ai-1.
The second input value is A0 = 0, resulting in A1 = |0| = 1. Because 0 does not equal 1, we findA2 = |A1| = |1| = 1. A2 is equal to A1, making 2 the smallest positive i such that Ai = Ai-1.
The third input value is A0 = 1, resulting in A1 = |1| = 1. A1 is equal to A0, making 1 the smallest positive i such that Ai = Ai-1.
The last input value is A0 = 9999999999, resulting in A1 = |9999999999| = 10. Because 10 does not equal 9999999999, we find A2 = |A1| = |10| = 2. Since 2 is not equal to 10, we find A3= |A2| = |2| = 1. Since 1 is not equal to 2, we find A4 = |A3| = |1| = 1. A4 is equal to A3, making 4 the smallest positive i such that Ai = Ai-1.
My Solution 🙂
__author__ = 'Rajind'
from sys import stdin
#Worked
string = raw_input()
count = 1
while string != "END":
num = int(string)
length = len(string)
while length != num:
count += 1
num = length
length = len(str(num))
print count
count = 1
string = raw_input()
~Rajind Ruparathna
Notebook 