이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("mmx,sse,sse2,sse3")
#include "bits/stdc++.h"
using namespace std;
#define INF 2000000100
#define INFLL 1000000000000000000ll
#define EPS 1e-9
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define pb push_back
#define fi first
#define sc second
using i64 = long long;
using u64 = unsigned long long;
using ld = long double;
using ii = pair<int, int>;
const int N = 2e3 + 100;
int dp[N], lo[N], hi[N];
ii dq[N][N];
void solve() {
int s, n;
cin >> s >> n;
fill(dp, dp + N, -INF);
dp[0] = 0;
for(int i = 0; i < n; ++i) {
int v, w, c;
cin >> v >> w >> c;
c = min(c, s / w);
for(int k = 0; k < w; ++k) lo[k] = hi[k] = 0;
for(int k = 0; k <= s; ++k) {
int X = dp[k], mod = k % w, j = k / w;
while(lo[mod] < hi[mod] && j - dq[mod][lo[mod]].sc > c) ++lo[mod];
if(lo[mod] != hi[mod]) dp[k] = max(dp[k], dq[mod][lo[mod]].fi + v * j);
if(X >= 0) {
while(lo[mod] < hi[mod] && dq[mod][hi[mod] - 1].fi < X - v * j) --hi[mod];
dq[mod][hi[mod]++] = {X - v * j, j};
}
}
}
cout << *max_element(dp, dp + N) << '\n';
}
int main() {
ios_base :: sync_with_stdio(false);
cin.tie(0);
int t = 1;
// cin >> t;
while(t--) solve();
return 0;
}
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