Submission #743932

#	Time	Username	Problem	Language	Result	Execution time	Memory
743932		maomao90	Izvanzemaljci (COI21_izvanzemaljci)	C++17	0 / 100	1 ms	340 KiB

izvanzemaljci

// Hallelujah, praise the one who set me free
// Hallelujah, death has lost its grip on me
// You have broken every chain, There's salvation in your name
// Jesus Christ, my living hope
#include <bits/stdc++.h> 
using namespace std;

#define REP(i, s, e) for (int i = (s); i < (e); i++)
#define RREP(i, s, e) for (int i = (s); i >= (e); i--)
template <class T>
inline bool mnto(T& a, T b) {return a > b ? a = b, 1 : 0;}
template <class T>
inline bool mxto(T& a, T b) {return a < b ? a = b, 1: 0;}
typedef long long ll;
typedef long double ld;
#define FI first
#define SE second
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> iii;
#define ALL(_a) _a.begin(), _a.end()
#define SZ(_a) (int) _a.size()
#define pb push_back
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ii> vii;
typedef vector<iii> viii;

#ifndef DEBUG
#define cerr if (0) cerr
#endif

const int INF = 1000000005;
const ll LINF = 1000000000000000005ll;
const int MAXN = 200005;

int n, k;
ii xy[MAXN];

namespace st1 {
    int main() {
        int mnx = INF, mxx = -INF, mny = INF, mxy = -INF;
        REP (i, 0, n) {
            mnto(mnx, xy[i].FI);
            mxto(mxx, xy[i].FI);
            mnto(mny, xy[i].SE);
            mxto(mxy, xy[i].SE);
        }
        cout << mnx << ' ' << mny << ' ' << max(mxx - mnx, mxy - mny) << '\n';
        return 0;
    }
}
namespace st2 {
    int main() {
        int mnx = INF, mxx = -INF, mny = INF, mxy = -INF;
        REP (i, 0, n) {
            mnto(mnx, xy[i].FI);
            mxto(mxx, xy[i].FI);
            mnto(mny, xy[i].SE);
            mxto(mxy, xy[i].SE);
        }
        cout << mnx << ' ' << mxx << ' ' << max(mxx - mnx, mxy - mny) << '\n';
        return 0;
    }
}

int main() {
#ifndef DEBUG
    ios::sync_with_stdio(0), cin.tie(0);
#endif
    cin >> n >> k;
    REP (i, 0, n) {
        cin >> xy[i].FI >> xy[i].SE;
    }
    if (k == 1) {
        return st1::main();
    } else if (k == 2) {
        return st2::main();
    }
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 21
• Subtask 3: 0 / 12
• Subtask 4: 0 / 30
• Subtask 5: 0 / 32