이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define FORD(i,a,b) for(int i=a;i>=b;i--)
#define FORE(it,v) for(__typeof((v).begin() it=(v).begin();it!=(v).end();i++)
#define IN(v) cerr<<v<<endl
#define SET(a,x) memset((a),(x),sizeof(a))
#define mod 1000000000
#define fi first
#define se second
#define mp make_pair
using namespace std;
typedef long long ll;
typedef pair<ll,ll> ii;
typedef pair<ii,ll> iii;
typedef pair<ll,int> lii;
template <typename T> inline void read(T &x)
{
int am=0;char k;
while(!(isdigit(k=getchar()))&&k!='-');
if (k=='-') k=getchar(),am=1;
x=k-48;
while(isdigit(k=getchar())) x=x*10+k-48;
if (am) x=-x;
}
template <typename T> inline void writep(T x)
{
if (x>=10) writep(x/10);
putchar(x%10+48);
}
template <typename T> inline void write(T x)
{
if(x<0) putchar ('-'),x=-x;
writep(x);
}
template <typename T> inline void writeln(T x)
{
write(x);
putchar('\n');
}
template <typename T> inline void writesp(T x)
{
write(x);
putchar(' ');
}
const int N=1000005;
int n, a[N], k[N], f[N],trace[N];
ii bit[10004];
int bitcount(int so)
{
int dem=0;
while(so) dem+=so%2,so=so/2;
return dem;
}
void Solve()
{
cin >> n;
FOR(i,1,n) cin >> a[i];
FOR(i,1,n) cin >> k[i];
int mx=1,vt=1;
FOR(i,1,n) f[i]=1;
FOR(i,2,n)
FOR(j,1,i-1) if (bitcount(a[i]&a[j])==k[i])
if(f[i]<f[j]+1)
{
f[i]=f[j]+1;
trace[i]=j;
if(mx<f[i]) mx=f[i],vt=i;
}
cout << mx <<endl;
deque<int> g;
while(mx) g.push_front(vt),mx--,vt=trace[vt];
FOR(i,0,(int)g.size()-1) cout << g[i] << " ";
}
void sub3()
{
cin >> n;
FOR(i,1,n) cin >> a[i];
FOR(i,1,n) cin >> k[i];
int mx=1,vt=1;
FOR(i,1,n) f[i]=1;
bit[a[1]]=mp(1,1);
FOR(i,2,n)
FOR(j,0,1000) if (bitcount(a[i]&j)==k[i])
if(f[i]<bit[j].fi+1)
{
f[i]=bit[j].fi+1;
trace[i]=bit[j].se;
bit[a[i]]=mp(f[i],i);
if(mx<f[i]) mx=f[i],vt=i;
}
cout << mx <<endl;
deque<int> g;
while(mx) g.push_front(vt),mx--,vt=trace[vt];
FOR(i,0,(int)g.size()-1) cout << g[i] << " ";
}
int main()
{
ios_base::sync_with_stdio(false);
sub3();
//if(n>=5000) sub3();else Solve();
}
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