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 ll long long
#define ld long double
#define ull unsigned long long
#define ii pair <int, int>
#define ill pair <ll, ll>
#define ild pair <ld, ld>
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define file "test"
using namespace std;
const int N = 1e5 + 2;
const ll MOD = 998244353;
const ll INF = 1e18;
const ll base = 311;
const int BLOCK_SIZE = 400;
int n;
int a[N], b[N];
int cnt[1 << 10 + 2][1 << 10 + 2];
ii f[1 << 10 + 2][1 << 10 + 2][12];
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
for (int i = 0; i < (1 << 10); i ++)
for (int j = 0; j < (1 << 10); j ++)
cnt[i][j] = __builtin_popcount(i & j);
cin >> n;
for (int i = 1; i <= n; i ++) cin >> a[i];
for (int i = 1; i <= n; i ++) cin >> b[i];
vector <int> trace(n + 1);
int n1 = n / 2, n2 = n - n1;
int ans = 1, k = 1;
for (int i = 1; i <= n; i ++) {
int l = a[i] >> n2;
int r = a[i] & ((1 << n2) - 1);
int cur = 1;
for (int s = 0; s < (1 << n1); s ++) {
int w = b[i] - cnt[s][l];
if (0 > w || w > n2) continue;
if (f[s][r][w].fi + 1 > cur)
cur = f[s][r][w].fi + 1, trace[i] = f[s][r][w].se;
}
if (cur > ans)
ans = cur, k = i;
for (int s = 0; s < (1 << n2); s ++)
if (f[l][s][cnt[s][r]].fi < cur)
f[l][s][cnt[s][r]] = {cur, i};
}
cout << ans << '\n';
vector <int> res;
while (k > 0) {
res.push_back(k);
k = trace[k];
}
reverse(all(res));
for (int u: res) cout << u << ' ';
}
/*
/\_/\ zzZ
(= -_-)
/ >u >u
*/
컴파일 시 표준 에러 (stderr) 메시지
subsequence.cpp:21:17: warning: suggest parentheses around '+' inside '<<' [-Wparentheses]
21 | int cnt[1 << 10 + 2][1 << 10 + 2];
| ~~~^~~
subsequence.cpp:21:30: warning: suggest parentheses around '+' inside '<<' [-Wparentheses]
21 | int cnt[1 << 10 + 2][1 << 10 + 2];
| ~~~^~~
subsequence.cpp:22:14: warning: suggest parentheses around '+' inside '<<' [-Wparentheses]
22 | ii f[1 << 10 + 2][1 << 10 + 2][12];
| ~~~^~~
subsequence.cpp:22:27: warning: suggest parentheses around '+' inside '<<' [-Wparentheses]
22 | ii f[1 << 10 + 2][1 << 10 + 2][12];
| ~~~^~~
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