Submission #490414

#	Time	Username	Problem	Language	Result	Execution time	Memory
490414		mathking1021	Languages (IOI10_languages)	C++17	97 / 100	5705 ms	32572 KiB

lang

//#include <stdlib.h>
//#include <stdio.h>
//#include <cmath>
//#include <unordered_map>
//
//#include "grader.h"
//#include "lang.h"
//
//#define SZ 100
//
//using namespace std;
//
//typedef long long ll;
//
//const int M = 100;
//const int N = 2;
//const int K = 15;
//int a[65555][65];
//int y[65];
//unordered_map<ll, int> ma;
//int now = 1;
////int cnt = 0;
//
//void excerpt(int *E)
//{
////    int mx2 = -1, mi2 = 0;
////    for(int i = 0; i < 56; i++) y[i] = 0;
////    for(int i = 0; i < 100; i++)
////    {
////        int mx = -1, mi = 0, mx2 = -1;
////        for(int j = 0; j < 56; j++)
////        {
////            if(x[E[i]][j] > mx) mx = x[E[i]][j], mi = j;
////            else if(x[E[i]][j] > mx2) mx2 = x[E[i]][j];
////        }
////        y[mi] += M + N * sqrt(mx - mx2) + K * sqrt(sqrt(mx - mx2));
////    }
////    mx2 = -1;
////    for(int i = 0; i < 56; i++)
////    {
////        if(y[i] > mx2) mx2 = y[i], mi2 = i;
////    }
//    int mx = -1, mi = -1;
////    for(int i = 0; i < 56; i++)
////    {
////        int cnt = 0;
////        for(int j = 0; j < 1; j++)
////        {
////            ll t = E[j] * 65536LL * 65536LL + E[j + 1] * 65536LL + E[j + 2];
////            ll t2 = E[j + 1] * 65536LL + E[j + 2];
////            cnt += 5 * x.count(t * 100 + i);
////        }
////        for(int j = 5; j < 10; j++)
////        {
////            ll t = E[j] * 65536LL * 65536LL + E[j + 1] * 65536LL + E[j + 2];
////            ll t2 = E[j + 1] * 65536LL + E[j + 2];
////            cnt += 1 * x.count(t * 100 + i);
////        }
////        for(int j = 95; j < 100; j++)
////        {
////            ll t = E[j] * 65536LL * 65536LL + E[j + 1] * 65536LL + E[j + 2];
////            ll t2 = E[j + 1] * 65536LL + E[j + 2];
////            cnt += 1 * x.count(t * 100 + i);
////        }
////        if(mx < cnt) mx = cnt, mi = i;
////    }
//    ll cnt[59];
//    for(int i = 0; i < 56; i++) cnt[i] = 0;
//    for(int i = 0; i < 100; i++)
//    {
//        int mx1 = -1, mx2 = -1, mi = -1;
//        int mx3 = 1;
//        for(int j = 0; j < 56; j++)
//        {
//            mx3 += a[E[i]][j];
//            if(mx1 < a[E[i]][j]) mx2 = mx1, mx1 = a[E[i]][j], mi = j;
//            else if(mx2 < a[E[i]][j]) mx2 = a[E[i]][j];
//        }
//        cnt[mi] += (mx1 * 50 / (mx3)) * (mx1 * 50 / (mx3));
//        if(i == 99) break;
//        for(ll j = 0; j < 56; j++) cnt[j] += 70000 * (a[E[i]][j]) / now + 700000 * (ma[E[i] * 6553600 + E[i + 1] * 100 + j]) / now;
//    }
//    int mxx = -1, mii = 0;
//    for(int i = 0; i < 56; i++)
//    {
//        if(mxx < cnt[i]) mxx = cnt[i], mii = i;
//    }
//    int k = language(mii);
//    for(int i = 0; i < 100; i++)
//    {
////        ll t = E[i] * 65536ULL * 65536ULL + E[i + 1] * 65536ULL + E[i + 2];
////        x.insert(t * 100 + k);
////        ll t2 = E[i] * 65536LL + E[i + 1];
////        xxx.insert(t2 * 100 + k);
//        a[E[i]][k]++;
//        if(i < 99) ma[E[i] * 6553600 + E[i + 1] * 100 + k] += 1, now++;
//    }
////    cnt++;
//}
#include <stdlib.h>
#include <stdio.h>
#include <cmath>
#include <set>
#include <map>

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

#define SZ 100

using namespace std;

typedef unsigned long long ll;

const int M = 100;
const int N = 2;
const int K = 15;
//map<pair<ll, int>, int> x;
//map<pair<ll, int>, int> xxx;
set<pair<ll, int>> x;
set<pair<ll, int>> xxx;
//int y[65];
//int cnt = 0;

void excerpt(int *E)
{
//    int mx2 = -1, mi2 = 0;
//    for(int i = 0; i < 56; i++) y[i] = 0;
//    for(int i = 0; i < 100; i++)
//    {
//        int mx = -1, mi = 0, mx2 = -1;
//        for(int j = 0; j < 56; j++)
//        {
//            if(x[E[i]][j] > mx) mx = x[E[i]][j], mi = j;
//            else if(x[E[i]][j] > mx2) mx2 = x[E[i]][j];
//        }
//        y[mi] += M + N * sqrt(mx - mx2) + K * sqrt(sqrt(mx - mx2));
//    }
//    mx2 = -1;
//    for(int i = 0; i < 56; i++)
//    {
//        if(y[i] > mx2) mx2 = y[i], mi2 = i;
//    }
    int mx = -1, mi = -1;
    for(int i = 0; i < 56; i++)
    {
        int cnt = 0;
        for(int j = 0; j < 97; j++)
        {
            ll t = E[j] * 65536ULL * 65536ULL * 65536ULL + E[j + 1] * 65536ULL * 65536ULL + E[j + 2] * 65536ULL + E[j + 3];
           // ll t2 = E[j] * 65536ULL * 65536ULL + E[j + 1] * 65536ULL + E[j + 2];
//            ll t2 = x[{t, i}];
            ll t2 = x.count({t, i});
            cnt += (t2 > 0);// + xxx.count({t2, i});
        }
        if(mx < cnt) mx = cnt, mi = i;
    }
    int k = language(mi);
    for(int i = 0; i < 97; i++)
    {
        ll t = E[i] * 65536ULL * 65536ULL * 65536ULL + E[i + 1] * 65536ULL * 65536ULL + E[i + 2] * 65536ULL + E[i + 3];
//        x[{t, k}] += 1;
        x.insert({t, k});
        //ll t2 = E[i] * 65536ULL * 65536ULL + E[i + 1] * 65536ULL + E[i + 2];
        //xxx.insert({t2, k});
    }
//    cnt++;
}

• Subtask 1: 30 / 30
• Subtask 2: 67 / 80