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>
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#define fi first
#define se second
#define p_b push_back
#define pll pair<ll,ll>
#define pii pair<int,int>
#define m_p make_pair
#define all(x) x.begin(),x.end()
#define sset ordered_set
#define sqr(x) (x)*(x)
#define pw(x) (1ll << x)
using namespace std;
typedef long long ll;
typedef long double ld;
const int MAXN = 1123456;
const int N = 1e6 + 2;
const int inf = 3e18;
int buben = 2000;
mt19937_64 rnd(chrono::system_clock::now().time_since_epoch().count());
template <typename T> void vout(T s){cout << s << endl;exit(0);}
int f[1025][1025][21], f1[1025][1025][21];
int main(){
ios_base :: sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
vector <int> a(n + 1), k(n + 1);
for(int i = 1; i <= n; i++)cin >> a[i];
for(int i = 1; i <= n; i++)cin >> k[i];
vector <int> dp(n + 1), pr(n + 1);
for(int i = 1; i <= n; i++){
dp[i] = 1;
int F = (pw(10) - 1) & a[i], S = a[i] >> 10;
for(int j = 0; j < pw(10); j++){
int K = k[i] - __builtin_popcount(j & S);
if(K >= 0){
if(dp[i] < f[F][j][K] + 1){
dp[i] = f[F][j][K] + 1;
pr[i] = f1[F][j][K];
}
}
}
for(int j = 0; j < pw(10); j++){
if(dp[i] > f[j][S][__builtin_popcount(F & j)]){
f[j][S][__builtin_popcount(F & j)] = dp[i];
f1[j][S][__builtin_popcount(F & j)] = i;
}
}
}
int Mx = *max_element(dp.begin() + 1, dp.end());
cout << Mx << "\n";
for(int i = 1; i <= n; i++)if(dp[i] == Mx){
int x = i;
vector <int> ans;
while(x){
ans.p_b(x);
x = pr[x];
}
reverse(all(ans));
for(auto i : ans)cout << i << " ";
cout << "\n";
return 0;
}
return 0;
}
컴파일 시 표준 에러 (stderr) 메시지
subsequence.cpp:23:17: warning: overflow in implicit constant conversion [-Woverflow]
const int inf = 3e18;
^~~~
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