# Solution of;
# Project Euler Problem 633: Square prime factors II
# https://projecteuler.net/problem=633
# 
# For an integer $n$, we define the square prime factors of $n$ to be the 
# primes whose square divides $n$. For example, the square prime factors of 
# $1500=2^2 \times 3 \times 5^3$ are $2$ and $5$. Let $C_k(N)$ be the number 
# of integers between $1$ and $N$ inclusive with exactly $k$ square prime 
# factors. It can be shown that with growing $N$ the ratio $\frac{C_k(N)}{N}$ 
# gets arbitrarily close to a constant $c_{k}^{\infty}$, as suggested by the 
# table below. \[\begin{array}{|c|c|c|c|c|c|}\hline& k = 0 & k = 1 & k = 2 & k 
# = 3 & k = 4 \\\hlineC_k(10) & 7 & 3 & 0 & 0 & 0 \\\hlineC_k(10^2) & 61 & 36 
# & 3 & 0 & 0 \\\hlineC_k(10^3) & 608 & 343 & 48 & 1 & 0 \\\hlineC_k(10^4) & 
# 6083 & 3363 & 533 & 21 & 0 \\\hlineC_k(10^5) & 60794 & 33562 & 5345 & 297 & 
# 2 \\\hlineC_k(10^6) & 607926 & 335438 & 53358 & 3218 & 60 \\\hlineC_k(10^7) 
# & 6079291 & 3353956 & 533140 & 32777 & 834 \\\hlineC_k(10^8) & 60792694 & 
# 33539196 & 5329747 & 329028 & 9257 \\\hlineC_k(10^9) & 607927124 & 335389706 
# & 53294365 & 3291791 & 95821 \\\hlinec_k^{\infty} & \frac{6}{\pi^2} & 3. 
# 3539\times 10^{-1} & 5. 3293\times 10^{-2} & 3. 2921\times 10^{-3} & 9. 
# 7046\times 10^{-5}\\\hline\end{array}\]Find $c_{7}^{\infty}$. Give the 
# result in scientific notation rounded to 5 significant digits, using a $e$ 
# to separate mantissa and exponent. E. g. if the answer were $0. 
# 000123456789$, then the answer format would be $1. 2346e\text{-}4$.
# 
# by lcsm29 http://github.com/lcsm29/project-euler
import timed


def dummy(n):
    pass


if __name__ == '__main__':
    n = 1000
    i = 10000
    prob_id = 633
    timed.caller(dummy, n, i, prob_id)
