Submission #598502

#	Time	Username	Problem	Language	Result	Execution time	Memory
598502		MohamedFaresNebili	The Big Prize (IOI17_prize)	C++14	100 / 100	52 ms	1960 KiB

prize

#include <bits/stdc++.h>
#include "prize.h"
/// #pragma GCC optimize ("Ofast")
/// #pragma GCC target ("avx2")
/// #pragma GCC optimize("unroll-loops")
 
                using namespace std;
 
                using ll = long long;
                using ld = long double;
                using ii = pair<ll, ll>;
                using vi = vector<int>;
 
                #define ff first
                #define ss second
                #define pb push_back
                #define all(x) (x).begin(), (x).end()
                #define lb lower_bound
 
                const int MOD = 1e9 + 7;
 
                int A[200005][2];
                int N, curr = 0, S = 0, res = -1;
 
                vector<int> query(int k) {
                    if(A[k][0] != -1)
                        return {A[k][0], A[k][1]};
                    vector<int> V = ask(k);
                    A[k][0] = V[0], A[k][1] = V[1];
                    return V;
                }
                void solve(int l, int r) {
                    if(res != -1) return;
                    vector<int> V = query(r);
                    if(V[0] + V[1] == curr && V[0] == S)
                        return;
                    if(l == r) {
                      	S++;
                        if(V[0] + V[1] == 0)
                            res = l;
                        return;
                    }
                    int md = (l + r) / 2;
                    solve(l, md); solve(md + 1, r);
                }
                void init() {
                    int calc = 0;
                    queue<pair<int, int>> Q;
                    Q.push({0, N - 1});
                    while(!Q.empty()) {
                        int l = Q.front().ff, r = Q.front().ss; Q.pop();
                        if(A[r][0] == -1) {
                            calc++; query(r);
                        }
                        curr = max(curr, A[r][0] + A[r][1]);
                        if(calc >= 500) return;
                        if(l == r) continue;
                        int md = (l + r) / 2;
                        Q.push({l, md});
                        Q.push({md + 1, r});
                    }
                }
                int find_best(int n) {
                    N = n;
                    memset(A, -1, sizeof A);
                    init(); S = 0; solve(0, N - 1);
                    return res;
                }

• Subtask 1: 20 / 20
• Subtask 2: 80 / 80