#pragma GCC optimize("O3") #include "bits/stdc++.h" using namespace std; #define rep(i,a,b) for(int i=(a); i<(b); ++i) #define all(x) x.begin(),x.end() #define sz(x) int(x.size()) typedef long long ll; typedef unsigned long long ull; typedef vector<int> vi; typedef vector<vi> vvi; // maybe make B one smaller for large table const int B = 19; map<ll,ll> mp[B]; vector<pair<ll,ll>> prec[B]; typedef array<ll,10> A; map<ll,ll> odp[B]; vector<pair<ll,ll>> preco[B]; vector<ll> coords; const int N = 36101; ll F(ll n){ if (n<10) return n; ll a = 1; while(a and n){ a *= n%10; n/=10; } return F(a); } const int PB = 1e6; int table[PB]; inline auto digit(auto n){ return n - (n / 10) * 10; } inline int f(ll n) { while (n >= 10) { if (n<PB) return table[n]; ll a = 1; while (n and a) { ll tmp = n / 10; a *= n - tmp * 10; n = tmp; } n = a; } return n; } /* 1 1000000000000000000 */ const int NZ = 8968; size_t nonzero[NZ]; A precFT[B][N]; A precOFT[B]; int id(ll n){ return lower_bound(all(coords),n)-begin(coords); } A calc(ll n){ ++n; A ans; rep(i,0,10) ans[i] = 0; for (ll blok = 1, k=0;; blok*=10, k++){ ll big = n/blok; ll mxd = big%10; big/=10; bool domul = big>0; ll alwaysmul = 1; while(big and alwaysmul){ alwaysmul *= big%10; big/=10; } rep(d,0,mxd){ if (d!=0 or domul) { int to; if ((alwaysmul*d)%10 == 0) to = 0; else to = id(alwaysmul*d); rep(i,0,10){ ans[i] += precFT[k][to][i]; } } else { rep(i,0,10) ans[i] += precOFT[k][i]; } } if (n/10 < blok) break; } return ans; } void solve(){ ll n; cin >> n; auto ans = calc(n); rep(i,0,9) cout << ans[i] << ' '; cout << ans[9]; cout << '\n'; } int main(){ cin.tie(NULL),cin.sync_with_stdio(false); // auto t0 = chrono::high_resolution_clock::now(); rep(i,0,PB) table[i] = F(i); // auto t1 = chrono::high_resolution_clock::now(); int t; cin >> t; // precompute target nums mp[0][1] = 1; rep(k,0,B-1){ for (auto [v,c] : mp[k]){ rep(d,0,10){ mp[k+1][v*d] += c; } } } // rep(i,0,B) cout << i << ' ' << sz(mp[i]) << '\n'; rep(i,0,B) prec[i] = vector<pair<ll,ll>>(all(mp[i])); // precompute other nums rep(i,1,10) odp[1][i] = 1; rep(i,1,B-1){ // nonzero in front for(auto [k,v] : odp[i]){ rep(d,0,10){ odp[i+1][k*d] += v; } } } // take pref sums rep(i,0,B-1){ for(auto [k,v] : odp[i]){ odp[i+1][k] += v; } } // rep(i,0,B) cout << i << ' ' << sz(odp[i]) << '\n'; rep(i,0,B) preco[i] = vector<pair<ll,ll>>(all(odp[i])); rep(i,0,B){ for(auto [k,v] : preco[i]) coords.push_back(k); } rep(i,0,B) for(auto [k,v] : prec[i]) coords.push_back(k); sort(all(coords)); coords.erase(unique(all(coords)),end(coords)); // left to optimize: // nonzero k, // d zero // store sum of nonzero k/v // only loop over nonzero d vector<size_t> zero; int cc = 0; rep(i,0,N){ if (digit(coords[i])==0) zero.push_back(i); else nonzero[cc++] = i; } // auto t2 = chrono::high_resolution_clock::now(); rep(i,0,B){ ll nulto = 0; ll sumnonzerokv = 0; for(auto [k,v] : prec[i]){ if (digit(k) == 0){ nulto += v; continue; } ll lim = 1000000000000000002ll/k; sumnonzerokv += v; for(auto d : nonzero){ if (coords[d]>lim) break; ll prod = k * coords[d]; int to = digit(prod) ? f(prod) : 0; precFT[i][d][to] += v; } } rep(d,0,N) precFT[i][d][0] += nulto; for(auto d : zero) precFT[i][d][0] += sumnonzerokv; for(auto [k,v] : preco[i]) { precOFT[i][f(k)] += v; } } // auto t3 = chrono::high_resolution_clock::now(); while(t--) solve(); // auto t4 = chrono::high_resolution_clock::now(); // cerr << chrono::duration_cast<chrono::milliseconds>(t1-t0).count() << '\n'; // cerr << chrono::duration_cast<chrono::milliseconds>(t2-t1).count() << '\n'; // cerr << chrono::duration_cast<chrono::milliseconds>(t3-t2).count() << '\n'; // cerr << chrono::duration_cast<chrono::milliseconds>(t4-t3).count() << '\n'; }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 | #pragma GCC optimize("O3") #include "bits/stdc++.h" using namespace std; #define rep(i,a,b) for(int i=(a); i<(b); ++i) #define all(x) x.begin(),x.end() #define sz(x) int(x.size()) typedef long long ll; typedef unsigned long long ull; typedef vector<int> vi; typedef vector<vi> vvi; // maybe make B one smaller for large table const int B = 19; map<ll,ll> mp[B]; vector<pair<ll,ll>> prec[B]; typedef array<ll,10> A; map<ll,ll> odp[B]; vector<pair<ll,ll>> preco[B]; vector<ll> coords; const int N = 36101; ll F(ll n){ if (n<10) return n; ll a = 1; while(a and n){ a *= n%10; n/=10; } return F(a); } const int PB = 1e6; int table[PB]; inline auto digit(auto n){ return n - (n / 10) * 10; } inline int f(ll n) { while (n >= 10) { if (n<PB) return table[n]; ll a = 1; while (n and a) { ll tmp = n / 10; a *= n - tmp * 10; n = tmp; } n = a; } return n; } /* 1 1000000000000000000 */ const int NZ = 8968; size_t nonzero[NZ]; A precFT[B][N]; A precOFT[B]; int id(ll n){ return lower_bound(all(coords),n)-begin(coords); } A calc(ll n){ ++n; A ans; rep(i,0,10) ans[i] = 0; for (ll blok = 1, k=0;; blok*=10, k++){ ll big = n/blok; ll mxd = big%10; big/=10; bool domul = big>0; ll alwaysmul = 1; while(big and alwaysmul){ alwaysmul *= big%10; big/=10; } rep(d,0,mxd){ if (d!=0 or domul) { int to; if ((alwaysmul*d)%10 == 0) to = 0; else to = id(alwaysmul*d); rep(i,0,10){ ans[i] += precFT[k][to][i]; } } else { rep(i,0,10) ans[i] += precOFT[k][i]; } } if (n/10 < blok) break; } return ans; } void solve(){ ll n; cin >> n; auto ans = calc(n); rep(i,0,9) cout << ans[i] << ' '; cout << ans[9]; cout << '\n'; } int main(){ cin.tie(NULL),cin.sync_with_stdio(false); // auto t0 = chrono::high_resolution_clock::now(); rep(i,0,PB) table[i] = F(i); // auto t1 = chrono::high_resolution_clock::now(); int t; cin >> t; // precompute target nums mp[0][1] = 1; rep(k,0,B-1){ for (auto [v,c] : mp[k]){ rep(d,0,10){ mp[k+1][v*d] += c; } } } // rep(i,0,B) cout << i << ' ' << sz(mp[i]) << '\n'; rep(i,0,B) prec[i] = vector<pair<ll,ll>>(all(mp[i])); // precompute other nums rep(i,1,10) odp[1][i] = 1; rep(i,1,B-1){ // nonzero in front for(auto [k,v] : odp[i]){ rep(d,0,10){ odp[i+1][k*d] += v; } } } // take pref sums rep(i,0,B-1){ for(auto [k,v] : odp[i]){ odp[i+1][k] += v; } } // rep(i,0,B) cout << i << ' ' << sz(odp[i]) << '\n'; rep(i,0,B) preco[i] = vector<pair<ll,ll>>(all(odp[i])); rep(i,0,B){ for(auto [k,v] : preco[i]) coords.push_back(k); } rep(i,0,B) for(auto [k,v] : prec[i]) coords.push_back(k); sort(all(coords)); coords.erase(unique(all(coords)),end(coords)); // left to optimize: // nonzero k, // d zero // store sum of nonzero k/v // only loop over nonzero d vector<size_t> zero; int cc = 0; rep(i,0,N){ if (digit(coords[i])==0) zero.push_back(i); else nonzero[cc++] = i; } // auto t2 = chrono::high_resolution_clock::now(); rep(i,0,B){ ll nulto = 0; ll sumnonzerokv = 0; for(auto [k,v] : prec[i]){ if (digit(k) == 0){ nulto += v; continue; } ll lim = 1000000000000000002ll/k; sumnonzerokv += v; for(auto d : nonzero){ if (coords[d]>lim) break; ll prod = k * coords[d]; int to = digit(prod) ? f(prod) : 0; precFT[i][d][to] += v; } } rep(d,0,N) precFT[i][d][0] += nulto; for(auto d : zero) precFT[i][d][0] += sumnonzerokv; for(auto [k,v] : preco[i]) { precOFT[i][f(k)] += v; } } // auto t3 = chrono::high_resolution_clock::now(); while(t--) solve(); // auto t4 = chrono::high_resolution_clock::now(); // cerr << chrono::duration_cast<chrono::milliseconds>(t1-t0).count() << '\n'; // cerr << chrono::duration_cast<chrono::milliseconds>(t2-t1).count() << '\n'; // cerr << chrono::duration_cast<chrono::milliseconds>(t3-t2).count() << '\n'; // cerr << chrono::duration_cast<chrono::milliseconds>(t4-t3).count() << '\n'; } |