Submission #490425

#TimeUsernameProblemLanguageResultExecution timeMemory
490425mathking1021Languages (IOI10_languages)C++17
100 / 100
6729 ms27800 KiB
//#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 <unordered_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; //unordered_map<ll, int> x[57]; set<ll> x[57]; //map<pair<ll, int>, int> xxx; //set<pair<ll, int>> x; //set<pair<ll, int>> xxx; //int y[65]; //int cnt = 0; ll a[57][65555]; 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[i].count(t); // ll t2 = x.count({t, i}); cnt += (t2 > 0);// + xxx.count({t2, i}); cnt += (a[i][E[j]]>0); } 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[k][t] += 1; x[k].insert(t); //ll t2 = E[i] * 65536ULL * 65536ULL + E[i + 1] * 65536ULL + E[i + 2]; //xxx.insert({t2, k}); a[k][E[i]]++; } // cnt++; }
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