제출 #211929

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
211929		mode149256	Cluedo (IOI10_cluedo)	C++14	100 / 100	18 ms	384 KiB

cluedo

#include<bits/stdc++.h>
#include "grader.h"
#include "cluedo.h"
using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}


mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

void Solve() {

	vi sk[4];

	for (int i = 1; i <= 6; ++i)
	{
		sk[1].emplace_back(i);
		sk[3].emplace_back(i);
	}
	for (int i = 1; i <= 10; ++i)
		sk[2].emplace_back(i);

	for (int i = 1; i <= 3; ++i)
		shuffle(sk[i].begin(), sk[i].end(), rng);

	int ans;

	do {
		ans = Theory(sk[1].back(), sk[2].back(), sk[3].back());

		if (ans) sk[ans].pop_back();
		else return;
	} while (ans);
}

• Subtask 1: 50 / 50
• Subtask 2: 50 / 50