Submission #1001102

#	Time	Username	Problem	Language	Result	Execution time	Memory
1001102		c2zi6	The Big Prize (IOI17_prize)	C++14	20 / 100	56 ms	5480 KiB

prize

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define replr(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define reprl(i, a, b) for (int i = int(a); i >= int(b); --i)
#define rep(i, n) for (int i = 0; i < int(n); ++i)
#define mkp(a, b) make_pair(a, b)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VPI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<VPI> VVPI;
typedef pair<ll, ll> PLL;
typedef vector<ll> VL;
typedef vector<PLL> VPL;
typedef vector<VL> VVL;
typedef vector<VVL> VVVL;
typedef vector<VPL> VVPL;
template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;}
template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;}
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<class T>
using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#include "prize.h"

VVI qr;
VI myask(int ind) {
    if (qr[ind].size()) return qr[ind];
    return qr[ind] = ask(ind);
}
int sum(int ind) {
    return qr[ind][0] + qr[ind][1];
}

int dfs(int l, int r) {
    /*cout << "NOW PROCESSING " << l+1 << " " << r+1 << endl;*/
    myask(l);
    if (sum(l) == 0) return l;
    myask(r);
    if (sum(r) == 0) return r;
    if (sum(l) < sum(r)) return dfs(l, r-1);
    if (sum(l) > sum(r)) return dfs(l+1, r);
    if (qr[l][0] == qr[r][0]) return -1;
    int L = l, R = r;
    while (L + 1 < R) {
        int M = (L + R) / 2;
        myask(M);
        if (sum(M) == 0) return M;
        if (sum(M) != sum(l)) {
            int ret;
            ret = dfs(l, M);
            if (ret != -1) return ret;
            ret = dfs(M, r);
            if (ret != -1) return ret;
        } else {
            if (qr[M][0] - qr[l][0] == 0) L = M;
            else R = M;
        }
    }
    return -1;
}

int find_best(int n) {
    qr = VVI(n);
    return dfs(0, n-1);
}

• Subtask 1: 20 / 20
• Subtask 2: 0 / 80