1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
|
#!env raku
# Helper Function
#
# For this task, you will most likely need a function f(a,b)
# which returns the count of different bits of binary representation of a and b.
#
# For example, f(1,3) = 1, since:
#
# Binary representation of 1 = 01
#
# Binary representation of 3 = 11
#
# There is only 1 different bit. Therefore the subroutine should return 1.
# Note that if one number is longer than the other in binary,
# the most significant bits of the smaller number are padded (i.e., they are assumed to be zeroes).
#
#
# Script Output
#
# You script should accept n positive numbers.
# Your script should sum the result of f(a,b) for every pair of numbers given:
#
# For example, given 2, 3, 4, the output would be 6, since f(2,3) + f(2,4) + f(3,4) = 1 + 2 + 3 = 6
sub f( Int:D $a, Int:D $b ) {
my $different-bits = 0;
my @a-bits = $a.base( 2 ).Str.comb.reverse;
my @b-bits = $b.base( 2 ).Str.comb.reverse;
# find the longest number
my $max-length = max( @a-bits.elems, @b-bits.elems );
# do the padding with zeros (to the end, the arra)
@a-bits.push: 0 for 0 .. ( $max-length - @a-bits.elems );
@b-bits.push: 0 for 0 .. ( $max-length - @b-bits.elems );
# compute the difference
for 0 ..^ @a-bits.elems {
$different-bits += 1 if ( @a-bits[ $_ ] != @b-bits[ $_ ] );
}
$different-bits;
}
################
my $sum = 0;
for 0 ..^ @*ARGS.elems -> $first {
for $first + 1 ..^ @*ARGS.elems -> $second {
$sum += f( @*ARGS[ $first ].Int, @*ARGS[ $second ].Int );
}
}
say "Sum is $sum";
|
