Difference between revisions of "KGS math club/solution 11 2"
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<nowiki>$_=shift;print $_;while(s/[A-Z][a-z]//){}while(s/^[a-z](.*)[A-Z]$/\1/){}print "^$_\n";</nowiki> | <nowiki>$_=shift;print $_;while(s/[A-Z][a-z]//){}while(s/^[a-z](.*)[A-Z]$/\1/){}print "^$_\n";</nowiki> | ||
| − | which requires presenting the subset as a word in mixed upper (members) and lower (non- | + | which requires presenting the subset as a word in mixed upper (members) and lower (non-member) case. |
Latest revision as of 17:08, 21 February 2011
Given n, k, and a k-member subset K of the n-member set N, we form the (n-k)-member subset it is paired with as follows.
Assign integers modulo n to the members of N, and cyclically order its members using them. Remove from N a member of K with a non-member of K that immediately follows it; and repeat until there are no members of K left. Append what we have left to K.
Solution by iceweasel.
He supplied this C code to compute it:
main(int a, char **v) {
int n=atoi(v[1]),
ss=atoi(v[2]),
x0=0,
x1=0,
m;
for ( m=1<<n ; m>>=1 ; (m&ss?(x1|=m):x1?(x1&=(x1-1)):(x0|=m)) );
for ( m=1<<n ; m>>=1 ; m&x0&&x1&&(x1&=(x0^=m,x1-1)) );
printf("%d\n",ss^x0^x1);
}
The arguments to the program are two integers: n, and one whose set bits correspond to the elements of K.
He also supplied this Perl code:
$_=shift;print $_;while(s/[A-Z][a-z]//){}while(s/^[a-z](.*)[A-Z]$/\1/){}print "^$_\n";
which requires presenting the subset as a word in mixed upper (members) and lower (non-member) case.
