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>
using namespace std;
typedef long long ll;
#define vt vector
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
#define R_EP(i, a, b, s) for (int i = (a); (s) > 0 ? i < (b) : i > (b); i += (s))
#define R_EP1(e) R_EP(i, 0, e, 1)
#define R_EP2(i, e) R_EP(i, 0, e, 1)
#define R_EP3(i, b, e) R_EP(i, b, e, 1)
#define R_EP4(i, b, e, s) R_EP(i, b, e, s)
#define GET5(a, b, c, d, e, ...) e
#define R_EPC(...) GET5(__VA_ARGS__, R_EP4, R_EP3, R_EP2, R_EP1)
#define rep(...) R_EPC(__VA_ARGS__)(__VA_ARGS__)
#define trav(x, a) for (auto x : a)
template<typename T> inline bool umax(T&x, const T& y) { if (x >= y) return false; x = y; return true; }
template<typename T> inline bool umin(T&x, const T& y) { if (x <= y) return false; x = y; return true; }
const int MOD = int(1e9) + 7, naxm = 200000, INF = int(1e9);
int N, M, a[naxm + 5], dp[naxm + 5];
vt<int> f;
int main() {
ios_base::sync_with_stdio(false), cin.tie(nullptr);
cin >> N >> M;
rep(i, 1, N + 1) cin >> a[i];
// maximize number of a[i] do not change
// a[i] <= M * i => M * i - a[i] >= 0
// a[i] <= a[j] + M * (i - j) (j < i) <=> M * j - a[j] <= M * i - a[i]
// so we define M * x - a[x] = f[x] => f[0] >= 0 && f[x] <= f[x + 1] => longest non-decreasing subsequence
rep(i, 1, N + 1) {
if (M * i - a[i] >= 0) f.pb(M * i - a[i]);
}
vt<int> dp;
trav(v, f) {
int l = lower_bound(all(dp), v + 1) - dp.begin();
if (l < sz(dp)) dp[l] = v;
else dp.pb(v);
}
cout << N - sz(dp) << "\n";
return 0;
}
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |