Submission #201022

#	Time	Username	Problem	Language	Result	Execution time	Memory
201022		a1_N	Bigger segments (IZhO19_segments)	C++17	73 / 100	745 ms	262148 KiB

segments

#include <bits/stdc++.h>

#define F first
#define S second

using namespace std;

const int N = (int)5e5 + 5;
const long long M = (long long)5e14;
    
int n,a[N];

long long pref[N];

pair<int,long long> dp[N];

pair<int,int> ZeroPair = {0,0};

struct tree{
	tree *left,*right;
	pair<int,int> mx;
	tree(): left(NULL), right(NULL), mx(ZeroPair) {}
}; 

tree *root = new tree();

pair<int,int> get_mx(tree *t){
	return (t != NULL) ? (t->mx) : ZeroPair;
}

void upd(tree *t,long long tl,long long tr,long long pos,pair<int,int> val){
	if(tl == tr){
		t->mx = val;
		return;
	}
	long long tm = (tl + tr) / 2;
	if(pos <= tm){
		if(t->left == NULL) t->left = new tree();
		upd(t->left,tl,tm,pos,val);		
	}
	else{
		if(t->right == NULL) t->right = new tree();
		upd(t->right,tm + 1,tr,pos,val);	
	}
	t->mx = max(get_mx(t->left),get_mx(t->right));
}

pair<int,int> get(tree *t,long long tl,long long tr,long long l,long long r){
	if(l <= tl && tr <= r) return t->mx;
	if(l > tr || r < tl) return ZeroPair;
	long long tm = (tl + tr) / 2;
	pair<int,int> mx1 = (t->left != NULL) ? get(t->left,tl,tm,l,r) : ZeroPair;
	pair<int,int> mx2 = (t->right != NULL) ? get(t->right,tm + 1,tr,l,r) : ZeroPair;
	return max(mx1,mx2);
}

int 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];
		pref[i] = pref[i - 1] + a[i];
	}
	dp[1] = {1,a[1]};
	upd(root,1,M,pref[1] + dp[1].S,{dp[1].F,1});
	for(int i = 2; i <= n; i++){
		dp[i] = {dp[i - 1].F,dp[i - 1].S + a[i]};
		pair<int,int> cur = get(root,1,M,1,pref[i]);
		if(cur.F > 0) dp[i] = {cur.F + 1,pref[i] - pref[cur.S]};
		/* pref[i] - pref[j] >= dp[j].S 
			pref[i] >= pref[j] + dp[j].S

			1) we update value in position pref[j] + dp[j].S
			2) when we want to find answer satisfying condition for this position we use segment tree
		*/
		upd(root,1,M,pref[i] + dp[i].S,{dp[i].F,i});	
	}
	cout << dp[n].F;
	return 0;
}

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