이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <iomanip>
#include <stack>
#include <queue>
#include <deque>
using namespace std;
typedef long long ll;
typedef pair <int, int> pii;
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define pb push_back
#define ppb pop_back
#define mkp make_pair
#define F first
#define S second
#define show(a) cerr << #a <<" -> "<< a <<"\n"
#define fo(a, b, c, d) for(int (a) = (b); (a) <= (c); (a) += (d))
#define foo(a, b, c ,d) for(int (a) = (b); (a) >= (c); (a) -= (d))
//#define int ll
const int N = 2e5 + 5;
const int INF = 1e9 + 5;
int n, a[N], k[N], dp[N], p[N], posans;
int main () {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> n;
for(int i = 1; i <= n; ++i) {
cin >> a[i];
}
for(int i = 1; i <= n; ++i) {
cin >> k[i];
}
int ans = 0, posans = 0;
for(int i = 1; i <= n; ++i) {
int mx = 0;
for(int j = 1; j < i; ++j) {
if(__builtin_popcount((a[i] & a[j])) == k[i]) {
if(mx < dp[j]) {
mx = dp[j];
p[i] = j;
}
}
}
dp[i] = mx + 1;
if(ans < dp[i]) {
ans = dp[i];
posans = i;
}
}
vector <int> vec;
while(posans) {vec.pb(posans); posans = p[posans];}
reverse(all(vec));
cout << ans <<'\n';
for(int it : vec)
cout << it <<' ';
return 0;
}
/*
4
1 2 3 4
10 0 1 0
5
5 3 5 3 5
10 1 20 1 20
*/
/*
-------------------------------------------------------------------------------------------
If you only do what you can do,
You will never be more than you are now!
-------------------------------------------------------------------------------------------
We must all suffer from one of two pains: the pain of discipline or the pain of regret.
The difference is discipline weighs grams while regret weighs tons.
-------------------------------------------------------------------------------------------
*/
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