发现值域很小,而且怎么异或都不会超过1023……然后可以使用类似基数排序的思想,每次扫一遍就行了。

复杂度O(k*1024)。

#include<cstdio>
#include<cstring>
using namespace std;
int n,k,x,cnts[1110],tmpcnts[1110];
int main()
{
// freopen("c.in","r",stdin);
int X;
scanf("%d%d%d",&n,&k,&x);
for(int i=1;i<=n;++i)
{
scanf("%d",&X);
++cnts[X];
}
for(int i=1;i<=k;++i)
{
memset(tmpcnts,0,sizeof(tmpcnts));
int now=0;
for(int j=0;j<=1023;++j)
{
int t=cnts[j];
if(now%2==0)
{
cnts[j]/=2;
tmpcnts[j^x]+=(t-cnts[j]);
}
else
{
cnts[j]=(cnts[j]+1)/2;
tmpcnts[j^x]+=(t-cnts[j]);
}
now+=t;
}
for(int j=0;j<=1023;++j)
cnts[j]+=tmpcnts[j];
}
for(int i=1023;i>=0;--i)
if(cnts[i])
{
printf("%d ",i);
break;
}
for(int i=0;i<=1023;++i)
if(cnts[i])
{
printf("%d\n",i);
break;
}
return 0;
}

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