제출 #1017396

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1017396		ByeWorld	Bigger segments (IZhO19_segments)	C++14	27 / 100	94 ms	262144 KiB

segments

#include <bits/stdc++.h>
// #pragma GCC optimize("O3", "unroll-loops")
#define ll long long
#define int long long
#define pb push_back
#define fi first
#define se second
#define lf (id<<1)
#define rg ((id<<1)|1)
#define md ((l+r)>>1)
#define ld long double
using namespace std;
typedef pair<int,int> pii;
typedef pair<pii, int> ipii;
const int MAXN = 3010;
const int MAXA = 9e3+20;
const ll INF = 1e9+10;
const int LOG = 13;
const int SQRT = 450;
const vector<int> NOL = {};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const vector<int> dx = {1, -1, 0, 0};
const vector<int> dy = {0, 0, 1, -1};
vector <int> key = {29, 31};
vector <int> mod = {998244353, 1000000007};
void chmx(int &a, int b){ a = max(a, b); }

int n;
int dp[MAXN][MAXN], a[MAXN], pr[MAXN];
int sp[MAXN][MAXN][LOG+1];
int GET(int x, int y){
	return pr[y]-pr[x-1];
}
int SPR(int idx, int l, int r){
	int lg = log2(r-l+1);
	return max(sp[idx][l][lg], sp[idx][r-(1<<lg)+1][lg]);
}
signed main(){
	// ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin >> n;
	for(int i=1; i<=n; i++){ cin >> a[i]; pr[i] = pr[i-1]+a[i]; }
	for(int i=0; i<=n; i++){
		for(int j=1; j<=n; j++){
			dp[i][j] = -INF; sp[i][j][0] = -INF;
		}
		if(i!=0){
			dp[i][0] = 1; sp[i][0][0] = 1;
		}
	}
	dp[0][0] = 0; sp[0][0][0] = 0;
	int i = 0;
		for(int k=1; k<LOG; k++)
			for(int j=0; j+(1<<k)-1<=n; j++)
				sp[i][j][k] = max(sp[i][j][k-1], sp[i][j+(1<<(k-1))][k-1]);

	for(int i=1; i<=n; i++){
		for(int j=1; j<=i; j++){ // dp[i][j] --> terakhir take j-i
			int l=0, r=j-1, mid=0, cnt=-1;
			while(l<=r){
				mid = md;
				if(GET(mid, j-1) <= GET(j, i)){
					cnt = mid; r = mid-1;
				} else l = mid+1;
			}
			// chmx(dp[i][j], dp[j-1][j-1...cnt])
			if(cnt != -1) 
				chmx(dp[i][j], SPR(j-1, cnt, j-1)+1);
			sp[i][j][0] = dp[i][j];
			// for(int k=0; k<=j-1; k++){ // dp[j-1][k] --> terakhir take k-j
			// 	if(GET(k, j-1) <= GET(j, i)) 
			// 		chmx(dp[i][j], dp[j-1][k]+1);
			// }
		}
		// dp[i][...] kelar
		for(int k=1; k<LOG; k++)
			for(int j=0; j+(1<<k)-1<=n; j++)
				sp[i][j][k] = max(sp[i][j][k-1], sp[i][j+(1<<(k-1))][k-1]);
		
	}
	int MX = 0;
	for(int i=0; i<=n; i++) chmx(MX, dp[n][i]);
	cout << MX << '\n';
}

• Subtask 1: 13 / 13
• Subtask 2: 14 / 14
• Subtask 3: 0 / 10
• Subtask 4: 0 / 36
• Subtask 5: 0 / 27