제출 #1066460

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1066460		beaconmc	수열 (APIO14_sequence)	C++14	0 / 100	35 ms	39928 KiB

sequence

#include <vector>
#include <bits/stdc++.h>
 
typedef long long ll;
#define FOR(i,x,y) for(ll i=x; i<y; i++)
#define FORNEG(i,x,y) for(ll i=x; i>y; i--)
 
using namespace std;

ll dp[10001][201];
ll prevs[1001][201];

deque<vector<ll>> sus;


bool sussy(ll a, ll b, ll mi, ll ci){
	return (sus[a][1] - ci) * (sus[a][0] - sus[b][0]) < (mi, - sus[a][0]) * (sus[b][1] - sus[a][1]);
}

int main(){
	

	ll n,k;
	cin >> n >> k;
	vector<ll> arr(n);
	vector<ll> pref(n);
	FOR(i,0,n) cin >> arr[i];
	pref[0] = arr[0];
	FOR(i,1,n) pref[i] = pref[i-1]+arr[i];

	FOR(i,0,10001)FOR(j,0,201) dp[i][j] = -1000000000;



	




	dp[0][0] = 0;
	FOR(j,1,k+1){
		sus.clear();
		sus.push_back({-pref[0]+arr[0], dp[0][j-1], 0});
		FOR(i,1,n+1){

			ll temp = (pref[n-1] - pref[i-1]);

			while (sus.size() >= 2 &&  (temp * sus[1][0] + sus[1][1]) > (temp * sus[0][0] + sus[0][1])){
				sus.pop_front();
			}

			ll ans = temp * sus[0][0] + sus[0][1] + (pref[n-1] - pref[i-1]) * pref[i-1];

			if (dp[i][j] < ans){
				dp[i][j] = ans;
				prevs[i][j] = sus[0][2];
			}
			ll newtemp = (pref[n-1] - pref[i]);
			while (sus.size()>=2 && sussy(sus.size()-2, sus.size()-1, (-pref[i]+arr[i]), dp[i][j-1])) {
				sus.pop_back();
			}
			sus.push_back({(-pref[i]+arr[i]), dp[i][j-1], i});
		}
	}




	ll maxi = -1;
	ll pos = -1;
	FORNEG(i,n-1,-1) if (dp[i][k] > maxi) maxi = dp[i][k], pos = i;

	cout << dp[pos][k] << endl;

	FOR(i,0,k){
		ll sus = prevs[pos][k-i];
		cout << pos << " ";
		pos  = sus;
		
	}

}

컴파일 시 표준 에러 (stderr) 메시지

sequence.cpp: In function 'bool sussy(ll, ll, ll, ll)':
sequence.cpp:17:55: warning: left operand of comma operator has no effect [-Wunused-value]
   17 |  return (sus[a][1] - ci) * (sus[a][0] - sus[b][0]) < (mi, - sus[a][0]) * (sus[b][1] - sus[a][1]);
      |                                                       ^~
sequence.cpp: In function 'int main()':
sequence.cpp:58:7: warning: unused variable 'newtemp' [-Wunused-variable]
   58 |    ll newtemp = (pref[n-1] - pref[i]);
      |       ^~~~~~~

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