이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma comment(linker, "/stack:200000000")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef int ll;
typedef long long lol;
typedef long double ld;
typedef pair <ll, ll> pll;
#ifdef SINA
#define dbg(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1> void __f(const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << std::endl; }
template <typename Arg1, typename... Args> void __f (const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...); }
#define dbg2(x, j, n) cout<< #x << " : "; output((j), (n), x, 1); cout.flush();
#else
#define dbg(...) 0
#define dbg2(x, j, n) 0
#endif
#define SZ(x) ((ll)((x).size()))
#define File(s, t) freopen(s ".txt", "r", stdin); freopen(t ".txt", "w", stdout);
#define input(j, n, a) for (int _i = (j); _i < (n)+(j); _i++) cin>> a[_i];
#define output(j, n, a, t) for (int _i = (j); _i < (n)+(j); _i++) cout<< a[_i] << (((t) && _i != (n)+(j)-1)? ' ' : '\n');
#define kill(x) return cout<< (x) << endl, 0
#define cl const ll
#define fr first
#define sc second
#define lc (v << 1)
#define rc (lc | 1)
#define mid ((l + r) >> 1)
#define All(x) (x).begin(), (x).end()
cl inf = sizeof(ll) == 4 ? (1e9 + 10) : (3e18), mod = 1e9 + 7, MOD = 998244353;
template <class A,class B> ostream& operator << (ostream& out,const pair<A,B>&a){return out<<'('<<a.first<<", "<<a.second<<')';}
template <class A> ostream& operator << (ostream& out, const vector<A> &a) {
out<< '['; for (int i = -1; ++i < int(a.size());) out<< a[i] << (i + 1 < int(a.size()) ? ", " : ""); return out<<']'; }
template <class T, typename _t = less <T> > using Tree = tree <T, null_type, _t, rb_tree_tag, tree_order_statistics_node_update>;
cl N = 1e5 + 3, K = 201;
ll ps [N], par [K][N], n, k, cur;
lol dp [2][N];
vector <ll> ans;
inline lol calc (cl &l, cl &r) { return lol(ps[r] - ps[l - 1]) * ps[l - 1]; }
void solve (cl &l = 1, cl &r = n, cl &optl = 1, cl &optr = n) {
ll m = mid, till = min(m, optl), t = (cur - 1) & 1;
ll bst = -1; lol mx = -1;
for (ll i = m; i >= till; i--) if (dp[t][i - 1] + calc(i, m) > mx) mx = dp[t][i - 1] + calc(i, m), bst = i;
dp[1 - t][m] = mx;
par[cur][m] = bst - 1;
if (l == r) return;
solve(l, m, optl, bst);
solve(m + 1, r, bst, optr);
}
int main ()
{
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
cin>> n >> k;
for (ll i = 1; i <= n; i++) cin>> ps[i], ps[i] += ps[i - 1];
for (cur = 1; cur <= k; cur++) solve();
cout<< dp[k & 1][n] << '\n';
for (; n; n = par[k][n], k--) ans.push_back(n);
for (ll i = SZ(ans) - 1; i; i--) cout<< ans[i] << ' ';
return 0;
}
/*
*/
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