Submission #363554

#TimeUsernameProblemLanguageResultExecution timeMemory
363554ACmachinePIN (CEOI10_pin)C++17
100 / 100
402 ms14828 KiB
#include <bits/stdc++.h> using namespace std; #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace __gnu_pbds; template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; template<typename T, typename U> using ordered_map = tree<T, U, less<T>, rb_tree_tag, tree_order_statistics_node_update>; typedef long long ll; typedef long double ld; typedef double db; typedef string str; typedef pair<int,int> pii; typedef pair<ll,ll> pll; typedef vector<int> vi; typedef vector<ll> vll; typedef vector<pii> vpii; typedef vector<pll> vpll; typedef vector<str> vstr; #define FOR(i,j,k,in) for(int i=(j); i < (k);i+=in) #define FORD(i,j,k,in) for(int i=(j); i >=(k);i-=in) #define REP(i,b) FOR(i,0,b,1) #define REPD(i,b) FORD(i,b,0,1) #define pb push_back #define mp make_pair #define ff first #define ss second #define all(x) begin(x), end(x) #define rsz resize #define MANY_TESTS int tcase; cin >> tcase; while(tcase--) const double EPS = 1e-9; const int MOD = 1e9+7; // 998244353; const ll INFF = 1e18; const int INF = 1e9; const ld PI = acos((ld)-1); const vi dy = {1, 0, -1, 0, -1, 1, 1, -1}; const vi dx = {0, 1, 0, -1, -1, 1, -1, 1}; #ifdef DEBUG #define DBG if(1) #else #define DBG if(0) #endif #define dbg(x) cout << "(" << #x << " : " << x << ")" << endl; // ostreams template <class T, class U> ostream& operator<<(ostream& out, const pair<T, U> &par) {out << "[" << par.first << ";" << par.second << "]"; return out;} template <class T> ostream& operator<<(ostream& out, const set<T> &cont) { out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; } template <class T, class U> ostream& operator<<(ostream& out, const map<T, U> &cont) {out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; } template<class T> ostream& operator<<(ostream& out, const vector<T> &v){ out << "["; REP(i, v.size()) out << v[i] << ", "; out << "]"; return out;} // istreams template<class T> istream& operator>>(istream& in, vector<T> &v){ for(auto &x : v) in >> x; return in; } template<class T, class U> istream& operator>>(istream& in, pair<T, U> &p){ in >> p.ff >> p.ss; return in; } //searches template<typename T, typename U> T bsl(T lo, T hi, U f){ hi++; T mid; while(lo < hi){ mid = (lo + hi)/2; f(mid) ? hi = mid : lo = mid+1; } return lo; } template<typename U> double bsld(double lo, double hi, U f, double p = 1e-9){ int r = 3 + (int)log2((hi - lo)/p); double mid; while(r--){ mid = (lo + hi)/2; f(mid) ? hi = mid : lo = mid; } return (lo + hi)/2; } template<typename T, typename U> T bsh(T lo, T hi, U f){ lo--; T mid; while(lo < hi){ mid = (lo + hi + 1)/2; f(mid) ? lo = mid : hi = mid-1; } return lo; } template<typename U> double bshd(double lo, double hi, U f, double p = 1e-9){ int r = 3+(int)log2((hi - lo)/p); double mid; while(r--){ mid = (lo + hi)/2; f(mid) ? lo = mid : hi = mid; } return (lo + hi)/2; } // some more utility functions template<typename T> pair<T, int> get_min(vector<T> &v){ typename vector<T> :: iterator it = min_element(v.begin(), v.end()); return mp(*it, it - v.begin());} template<typename T> pair<T, int> get_max(vector<T> &v){ typename vector<T> :: iterator it = max_element(v.begin(), v.end()); return mp(*it, it - v.begin());} template<typename T> bool ckmin(T& a, const T& b){return b < a ? a = b , true : false;} template<typename T> bool ckmax(T& a, const T& b){return b > a ? a = b, true : false;} int main(){ ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n, d; cin >> n >> d; map<string, int> nm; REP(i, n){ string x; cin >> x; REP(j, (1 << 4)){ if(j == 0) continue; string tm = ""; REP(k, 4){ if(j & (1 << k)) tm += x[k]; else tm += '?'; } nm[tm]++; } } array<ll, 5> pc = {0, 0, 0, 0, 0}; // pair count for(auto x : nm){ int cn = 0; REP(j, 4){ if(x.ff[j] == '?') cn++; } pc[4 - cn] += ((ll)x.ss * (x.ss - 1)) / 2; } pc[3] = pc[3] - 4 * pc[4]; pc[2] = pc[2] - pc[3] * 3 - pc[4] * 6; pc[1] = pc[1] - 2 * pc[2] - 3 * pc[3] - 4 * pc[4]; pc[0] = ((ll)n * (n - 1) / 2) - pc[4] - pc[3] - pc[2] - pc[1]; cout << pc[4 - d] << "\n"; return 0; }
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