#include using namespace std; #define _p(...) (void)printf(__VA_ARGS__) #define forr(x,arr) for(auto&& x:arr) #define _overload3(_1,_2,_3,name,...) name #define _rep2(i,n) _rep3(i,0,n) #define _rep3(i,a,b) for(int i=int(a);i=int(a);i--) #define rrep(...) _overload3(__VA_ARGS__,_rrep3,_rrep2,)(__VA_ARGS__) #define all(x) (x).begin(), (x).end() #define bit(n) (1LL<<(n)) #define sz(x) ((int)(x).size()) #define fst first #define snd second using ll=long long;using pii=pair; using vb=vector;using vs=vector; using vi=vector;using vvi=vector;using vvvi=vector; using vl=vector;using vvl=vector;using vvvl=vector; using vd=vector;using vvd=vector;using vvvd=vector; using vpii=vector;using vvpii=vector;using vvvpii=vector; templateT read(){T t;cin>>t;return t;} templateostream&operator<<(ostream&o,const pair&p){o<<'('< 0) return 1; if (ret < 0) return -1; return 0; } inline bool cmp_b(const pii &lhs, decltype(lhs) rhs) { return cmp(lhs.A, lhs.B, rhs.A, rhs.B) <= 0; } // a/(a+b) と足したら1を越えるのが何本あるか // O(log |S|) int bs(ll a, ll b, const vpii &S) { int ok = sz(S); int ng = -1; auto check = [&](int s) -> bool { ll c = S[s].A; ll d = S[s].B; return a*(c+d)+c*(a+b) > (a+b)*(c+d); }; while (abs(ok - ng) > 1) { int mid = (ok + ng) / 2; bool c = check(mid); // cout << "ok: " << ok << ", ng: " << ng << ", "; // cout << "check(" << mid << ") -> " << c << endl; if (c) ok = mid; else ng = mid; } return sz(S) - ok; } void Main() { int n = read(); vpii R[3]; rep(i, n) { int t = read(); int a = read(); int b = read(); int g = __gcd(a, b); a /= g; b /= g; R[t].emplace_back(a, b); } rep(i, 3) sort(all(R[i]), cmp_b); //rep(ii,3) { cout << "R["<= (a+b)*(c+d)) { if (cmp(a, a+b, c, c+d) <= 0) { X.emplace_back(1.0 * a / (a + b)); } else { X.emplace_back(1.0 * c / (c + d)); } { ll e = (a+b)*d + (c+d)*b; ll f = (a+b)*c - (c+d)*b; if (f > 0) { ll g = __gcd(e, f); e /= g; f /= g; if (e <= 4000 && f <= 4000) { CC[e][f]++; //_p("CC[%lld][%lld]++\n", e, f); } } } } } } } // O(N^2 log N^2) TLE? sort(all(X)); //cout << "X:"; rep(ii,sz(X)) cout << ' ' << X[ii]; cout << endl; // 3本 // O(n * log(N^2)) forr(ab, R[2]) { ll a = ab.A; ll b = ab.B; double bab = 1.0 * b / (a + b); // b/(a+b) 以上の X を数える // X は単調増加 // OK な下限を探すパターン int ok = sz(X); int ng = -1; auto check = [&](int x) -> bool { //_p("%d,%d,%lld,%lld\n",X[x].A, X[x].A + X[x].B, b, a + b); return X[x] >= bab; //return cmp(X[x].A, X[x].A + X[x].B, b, a + b) >= 0; }; while (abs(ok - ng) > 1) { int mid = (ok + ng) / 2; bool c = check(mid); // cout << "ok: " << ok << ", ng: " << ng << ", "; // cout << "check(" << mid << ") -> " << c << endl; if (c) ok = mid; else ng = mid; } //cout << "ok: "<