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#! /opt/local/bin/perl
#
# count_set_match.pl
#
# TASK #1 › Count Set Bits
# Submitted by: Mohammad S Anwar
# You are given a positive number $N.
#
# Write a script to count the total number of set bits of the binary
# representations of all numbers from 1 to $N and return
# $total_count_set_bit % 1000000007.
#
# Example 1:
# Input: $N = 4
#
# Explanation: First find out the set bit counts of all numbers
# i.e. 1, 2, 3 and 4.
#
# Decimal: 1
# Binary: 001
# Set Bit Counts: 1
#
# Decimal: 2
# Binary: 010
# Set Bit Counts: 1
#
# Decimal: 3
# Binary: 011
# Set Bit Counts: 2
#
# Decimal: 4
# Binary: 100
# Set Bit Counts: 1
#
# Total set bit count: 1 + 1 + 2 + 1 = 5
#
# Output: Your script should print `5` as `5 % 1000000007 = 5`.
# Example 2:
# Input: $N = 3
#
# Explanation: First find out the set bit counts of all numbers
# i.e. 1, 2 and 3.
#
# Decimal: 1
# Binary: 01
# Set Bit Count: 1
#
# Decimal: 2
# Binary: 10
# Set Bit Count: 1
#
# Decimal: 3
# Binary: 11
# Set Bit Count: 2
#
# Total set bit count: 1 + 1 + 2 = 4
#
# Output: Your script should print `4` as `4 % 1000000007 = 4`.
#
# method:
# The way I see it, there are three parts to this task. Over a loop
# of values, we need to first create a binary representation of a
# number, then count and sum the 1s present in those values. For the
# third and seemingly, puzzlingly unrelated part, we then modulo the
# total by one billion and seven.
#
# In weeks past, we've found easy ways to convert decimal to binary.
# For the counting the 1s step, this is the same as summing every
# digit, as these, being binary, are only going to be either 1 or 0.
# Breaking the digit string into a list of elements and then summing
# these accomplishes this well; after this step the sum is added to
# a running total for the sequence of 1 up to the target value.
#
# As for the modulo phase, a little explanation is in order. The
# value 1000000007 is ultimately arbitrary, and is selected because
# it:
#
# 1. is large, but not too large, and
# 2. is prime
#
# Beyond these criteria, there is no meaning behind that specific
# choice of number, and 1,000,000,033 would do just as well, or
# 2,000,000,033. Speaking to the first point, what this selection
# does is produce a verifiable, reproducable result that fits into
# common integer data types without any risk of overflow.
#
# This can even be done at every step of a calculation involving
# very very large numbers, constructing the correct modulo result
# without ever exceeding the range of a 32-bit integer. But then
# again there is no requirement here either to process an unusually
# big range or for that matter work properly on 32-bit machines. So
# even should we wish to include values past 2^32 in our
# intermediary steps, the 64-bit norm these day gives us
# considerably more range.
#
# So we're not going to bother to do that. I do wonder if anyone
# will.
#
# 2020 colin crain
## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ##
use warnings;
use strict;
use feature ":5.26";
## ## ## ## ## MAIN:
my $input = shift // 100000;
my $tot;
for my $i (1..$input) {
my $bin = sprintf "%b", $i;
my $j = length $bin;
while (--$j >= 0 ) {
substr( $bin, $j, 1 ) and $tot++;
}
}
my $out = $tot % 1000000007;
say "out: $out";
|
