// Author: Bartek Knapik #include <cstdio> const int MAX_MEM = 10000000; long long ans[10]; long long mem[MAX_MEM]; int t; long long n; long long mno(long long n) { if (0 <= n && n <= 9) return n; if (n < MAX_MEM && mem[n] != -1LL) return mem[n]; int n_ = (int) n; long long p = 1LL; while (n) { p *= (n % 10LL); n /= 10LL; } mem[n_] = mno(p); return mem[n_]; } void solve() { scanf("%lld", &n); for (int i = 0; i < 10; ++i) ans[i] = 0LL; for (long long i = 1; i <= n; ++i) ans[mno(i)]++; for (int i = 0; i < 10; ++i) printf("%lld%c", ans[i], i == 9 ? '\n' : ' '); } int main() { scanf("%d", &t); for (int i = 0; i < MAX_MEM; ++i) mem[i] = -1LL; while (t--) solve(); return 0; }
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 | // Author: Bartek Knapik #include <cstdio> const int MAX_MEM = 10000000; long long ans[10]; long long mem[MAX_MEM]; int t; long long n; long long mno(long long n) { if (0 <= n && n <= 9) return n; if (n < MAX_MEM && mem[n] != -1LL) return mem[n]; int n_ = (int) n; long long p = 1LL; while (n) { p *= (n % 10LL); n /= 10LL; } mem[n_] = mno(p); return mem[n_]; } void solve() { scanf("%lld", &n); for (int i = 0; i < 10; ++i) ans[i] = 0LL; for (long long i = 1; i <= n; ++i) ans[mno(i)]++; for (int i = 0; i < 10; ++i) printf("%lld%c", ans[i], i == 9 ? '\n' : ' '); } int main() { scanf("%d", &t); for (int i = 0; i < MAX_MEM; ++i) mem[i] = -1LL; while (t--) solve(); return 0; } |