Difference between revisions of "KGS math club/solution 11 2"
(added credit, C code) |
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| Line 7: | Line 7: | ||
He also supplied this C code to compute it: | He also supplied this C code to compute it: | ||
| − | main(int a, char **v) { | + | 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); | |
| − | } | + | } |
Revision as of 10:45, 14 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 also 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);
}
