제출 #50050

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
50050		Benq	Languages (IOI10_languages)	C++14	90 / 100	4241 ms	8932 KiB

lang

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
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<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

#include "grader.h"
#include "lang.h"

#define SZ 100

//int language(int x) { cout << x << "\n"; return 0; }


struct hsh {
    size_t operator()(const pi& k) const {
        return ((ll)65536*k.f+k.s); // bad, but you get the point
    }
};

int m[56][65536];
unordered_map<pi,int,hsh> M[56];
map<int,int> tmp;
unordered_map<pi,int,hsh> TMP;
int tot[56];

double tri(int lang) {
    if (tot[lang] == 0) return MOD;
    
    double dif = 0, lef = 100;
    
    for (auto a: tmp) {
        lef -= (double)m[lang][a.f]/tot[lang];
        double z = a.s-(double)m[lang][a.f]/tot[lang];
        dif += abs(z);
    }
    
    dif += lef;
    return dif;
}

double get(int lang, pi t) {
    if (!M[lang].count(t)) return 0;
    return M[lang][t];
}

double TRI(int lang) {
    if (tot[lang] == 0) return MOD;
    
    double dif = 0, lef = 99;
    
    for (auto a: TMP) {
        double Z = get(lang,a.f);
        lef -= Z/tot[lang];
        double z = a.s-Z/tot[lang];
        dif += abs(z);
        //cout << a.f.f << " " << a.f.s << " " << Z << " " << a.s << " " << z << " " << lef << "\n";
    }
    
    dif += lef;
    //cout << dif << "\n";
    return dif;
}

void excerpt(int *E) {
    pair<double,int> bes = {MOD,0};
    tmp.clear(); TMP.clear();
    
    F0R(i,100) tmp[E[i]] ++;
    F0R(i,99) TMP[{E[i],E[i+1]}] ++;
    F0R(i,56) bes = min(bes,{tri(i)+2*TRI(i),i});
    
    int x = language(bes.s);
    tot[x] ++;
    F0R(i,100) m[x][E[i]] ++;
    F0R(i,99) M[x][{E[i],E[i+1]}] ++;
}

• Subtask 1: 30 / 30
• Subtask 2: 60 / 80