Submission #773795

#	Time	Username	Problem	Language	Result	Execution time	Memory
773795		lollipop	Languages (IOI10_languages)	C++17	0 / 100	54 ms	39932 KiB

lang

#include "grader.h"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
//#define int long long
#define pb push_back
#define s second
#define f first
#define pf push_front
#define inf 100000000000000000
#define bitebi __builtin_popcountll
#define FOR( i , n ) for( int i = 0 ; i < n ; i ++ )
#define YES cout <<"YES\n"
#define NO cout << "NO\n"
#define debug cout << "Here Fine" << endl ;
#define pr pair < int , int >
#define fbo find_by_order // returns iterator
#define ook order_of_key // returns strictly less numbers than key
using namespace std ;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
const double Pi=acos(-1.0);
const double EPS=1E-8;
const int mod =  1000000007 ;
const int mod1 = 998244353 ;
const int N = 2e5 + 10 ;
mt19937 R(time(0));
map < int , int > ma , ma1 ;
int sm[ 56 ][ 123990 ] ;
int PP = 123989 ;
int hsh[ 66000 ] ; 
void excerpt( int E[ 100 ] ){
    pair < int , int > p = { 0 , 0 } ;        
    int pip = 55 , pp = 1 ;  
    if( hsh[ 0 ] == 0 ){
        hsh[ 0 ] = 1 ;
        for( int j = 1 ; j <= 65535 ; j ++ ){
            hsh[ j ] = j * pp % PP ;
            pp = pp * pip % PP ; 
        }
    }
    FOR( i, 56 ){
        int cur = 0 ;
        FOR( j , 97 ) {
            int cur1 = hsh[ E[ j ] ] + pip * hsh[ E[ j + 1 ] ] % PP ;
            cur1 = cur1 + hsh[ E[ j + 2 ] ] * pip * pip ;
            cur1 %= PP ; 
            cur1 = cur1 + (hsh[ E[ j + 3 ] ] * pip * pip % PP )* pip ;
            cur1 %= PP ; 
            if( sm[ i ][ cur1 ] == 1 ) cur ++ ; 
        } 
        if( p.f < cur ) p = { cur , i } ;
    }
    int x = language( p.s ) ;
    FOR( j , 97 ){
        int cur = hsh[ E[ j ] ] + pip * hsh[ E[ j + 1 ] ] % PP ;
        cur = cur + hsh[ E[ j + 2 ] ] * pip * pip % PP ;
        cur %= PP ; 
        cur = cur + ( hsh[ E[ j + 3 ] ] * pip * pip % PP )* pip ;
        cur %= PP ; 
        sm[ x ][ cur ] = 1 ; 
    }
}

• Subtask 1: 0 / 30
• Subtask 2: 0 / 80