Submission #468985

#	Time	Username	Problem	Language	Result	Execution time	Memory
468985		Millad	Liteh and Newfiteh (INOI20_litehfiteh)	C++14	0 / 100	2 ms	3660 KiB

Main

// In the name of god
#include <bits/stdc++.h>
     
#define F first
#define S second
#define pb push_back
#define debug(x) cerr << #x << " : " << x << '\n'
     
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
     
const ll maxn = 1e5 + 5;
const ll inf = 1e18;
const ll lg = 20;
ll n, a[maxn], dp[maxn][lg + 1][lg + 1][2], pw2[lg + 1], ANS[maxn];
int main(){
	pw2[0] = 1;
	for(ll i = 1; i < lg; i ++)pw2[i] = (pw2[i - 1] * 2ll);
    	cin >> n;
    	for(ll i = 0; i < n; i ++){
    		cin >> a[i];
    		for(ll j = 0; j < lg; j ++){
			for(ll k2 = 0; k2 < 2; k2 ++){
    				if(a[i] < j)dp[i][0][j][k2] = -1;
    				else if(a[i] == j)dp[i][0][j][k2] = 0;
    				else if(a[i] == j + 1)dp[i][0][j][k2] = 1;
    				else dp[i][0][j][k2] = -1;
			}
    		}
    		ll d;
    		ANS[i] = inf;
    		for(ll j = 1; pw2[j] < lg; j ++){
    			for(ll k = 0; k < lg; k ++){
				for(ll k2 = 0; k2 < 2; k2 ++){
    					if(pw2[j] > (i + 1ll)){
						dp[i][j][k][k2] = -1;
						continue;
					}
					d = inf;
					if((k < (lg - 1ll)) &&(dp[i][j - 1][k + 1][1] != -1) && (dp[i - pw2[j - 1]][j - 1][k + 1][1] != -1))
                                                d = min(d, dp[i - pw2[j - 1]][j - 1][k + 1][1] + dp[i][j - 1][k + 1][1] + 1ll);
					if(k2 == 0){
						if((dp[i][j - 1][k][0] != -1) && (dp[i - pw2[j - 1]][j - 1][k][0] != -1))
                                                d = min(d, dp[i - pw2[j - 1]][j - 1][k][0] + dp[i][j - 1][k][0]);
					}
					else{
                                                if((dp[i][j - 1][k][1] != -1) && (dp[i - pw2[j - 1]][j - 1][k][1] != -1))
                                                d = min(d, dp[i - pw2[j - 1]][j - 1][k][1] + dp[i][j - 1][k][1]);
					}
					if(d == inf)d = -1;
					dp[i][j][k][k2] = d;
				}
    			}
    			if(dp[i][j][0][0] != -1){
    				if(pw2[j] == i + 1ll)ANS[i] = min(ANS[i], dp[i][j][0][0]);
    				else if(ANS[i - pw2[j]] != -1)ANS[i] = min(ANS[i], dp[i][j][0][0] + ANS[i - pw2[j]]); 
    			}
    		}
    		if(ANS[i] == inf)ANS[i] = -1;
    	}
    	cout << ANS[n - 1] << endl;
    }

• Subtask 1: 0 / 30
• Subtask 2: 0 / 70