proj_euler40.py

# This function solves the problem no. 40 of project-euler
# The link to the problem: http://projecteuler.net/problem=40
# We find a correspondence in the two below sequences s1 and s2:
# s1 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2, 1, 3, ...
# s2 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, ...
# the function nth_digit(n) finds the nth term of s1
# termcount is the term of s2 in which nth term of s1 lies at position numdigit
# from the left

import math

def digit(n, t):
    for i in range(t - 1):
        n /= 10
    return n % 10

def nth_digit(n):
    numlen = 1       
    termcount = 0    
    numdigit = 0

    while n > 0:
        temp = int(numlen * 9 * math.pow(10, numlen - 1))

        if n - temp >= 0:
            n -= temp
            termcount += int(9 * math.pow(10, numlen - 1))
            numlen += 1
        else:
            numnum = n / numlen
            numdigit = n % numlen
            termcount += numnum if numdigit == 0 else numnum + 1
            break

    if numdigit == 0:
        return termcount % 10
    else:
        return digit(termcount, numlen - numdigit + 1)