Submission #101677

#	Time	Username	Problem	Language	Result	Execution time	Memory
101677		arman_ferdous	Split the sequence (APIO14_sequence)	C++17	71 / 100	2045 ms	83784 KiB

sequence

#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
const int N = 100001;
const int K = 201;
const ll oo = 1e18;
 
int n, k;
ll S[N], dp[N][2]; int opt[N][K];

struct CHT{
	vector<ll> m, c, id; int last;
	void init() { last = 0; m.clear(); c.clear(); id.clear(); }
	bool bad(int i, int j, int k) {
		return 1.0 * (c[j] - c[i]) * (m[i] - m[k]) >= 1.0 * (c[k] - c[i]) * (m[i] - m[j]);
	}
	void insert(ll M, ll C, int id_) {
		m.push_back(M); c.push_back(C); id.push_back(id_);
		int sz = m.size();
		while(sz >= 3 && bad(sz-3, sz-2, sz-1)) {
			m.erase(m.end()-2); 
			c.erase(c.end()-2);
			id.erase(id.end()-2);
			sz--;
		}
	}
	ll f(int i, ll X) { return m[i] * X + c[i]; }
	pair<ll,int> get(ll X) {
		int sz = m.size();
		last = min(last, sz-1);
		while(last + 1 < sz && f(last,X) <= f(last+1,X))
			last++;
		return {f(last,X), id[last]};
	}
}ds;

int main() {
	scanf("%d %d", &n, &k);
	for(int i = 1; i <= n; i++) {
		scanf("%lld", &S[i]);
		S[i] += S[i-1];
	}
 
	for(int kk = 1; kk <= k; kk++) {
		int row = (kk&1);
		ds.init();
		ds.insert(0,0,0);
		for(int i = kk; i <= n; i++) {
			auto got = ds.get(S[i]);
			dp[i][row] = got.first;
			opt[i][kk] = got.second;
			ds.insert(S[i], dp[i][(row^1)] - S[i]*S[i], i);
		}
	}
	printf("%lld\n", dp[n][(k&1)]);
	int pos = n;
	for(int cnt = k; cnt > 0; cnt--) {
		printf("%d ", opt[pos][cnt]);
		pos = opt[pos][cnt];
	}
	return 0;
}

Compilation message (stderr)

sequence.cpp: In function 'int main()':
sequence.cpp:39:7: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
  scanf("%d %d", &n, &k);
  ~~~~~^~~~~~~~~~~~~~~~~
sequence.cpp:41:8: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
   scanf("%lld", &S[i]);
   ~~~~~^~~~~~~~~~~~~~~

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 17 / 17
• Subtask 5: 21 / 21
• Subtask 6: 0 / 29