#include "bits/stdc++.h"
using namespace std;
#define all(x) x.begin(),x.end()
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << p.first << " " << p.second; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = " "; return os; }
#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#define ASSERT(...) 42
#endif
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pi;
const int oo = 1e9;
unordered_map<ll,array<ll,10>> dp[19] = {};
unordered_map<ll,short> dig;
short getdig(ll v) {
if(v<10) return v;
if(dig.count(v)) return dig[v];
ll cur = 1;
for(auto c : to_string(v)) {
cur*=c-'0';
}
return dig[v] = getdig(cur);
}
auto operator+(array<ll,10> a, const array<ll,10>& b) {
for(int j=0;j<10;++j) a[j]+=b[j];
return a;
}
array<ll,10> solve(ll mult, int left) {
if(left==0) {
array<ll,10> res = {};
res[getdig(mult)]++;
return res;
}
auto it = dp[left].find(mult);
if(it!=dp[left].end()) return it->second;
auto& res = dp[left][mult];
for(int d=0;d<10;++d) {
res = res+ solve(mult*d,left-1);
}
return res;
}
unordered_map<int,array<ll,10>> precomp;
auto solvenonzero(int x) {
if(x==0) return array<ll,10>{};
auto it = precomp.find(x);
if(it!=precomp.end()) return it->second;
auto& res = precomp[x];
res=solvenonzero(x-1);
for(int j=1;j<10;++j) res = res +solve(j,x-1);
return res;
}
void solve() {
ll x; cin >> x;
x++; // strictly smaller
string n = to_string(x);
for(auto& i : n) i-='0';
// take product of those, then bruteforce
// next digit, then have arbitrary digits...
ll mult=1;
array<ll,10> ans = {};
for(int i=0;i<n.size();++i) {
for(int j=0;j<n[i];++j) {
if(i==0 and j==0) continue; // no leading zeros.
ans = ans+solve(mult*j,n.size()-1-i);
}
mult*=n[i];
}
ans = ans+solvenonzero(n.size()-1);
cout << ans << '\n';
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t; cin >> t;
while(t--) solve();
}
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 | #include "bits/stdc++.h" using namespace std; #define all(x) x.begin(),x.end() template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << p.first << " " << p.second; } template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = " "; return os; } #ifdef LOCAL #include "debug.h" #else #define debug(...) 42 #define ASSERT(...) 42 #endif typedef long long ll; typedef vector<int> vi; typedef vector<vi> vvi; typedef pair<int,int> pi; const int oo = 1e9; unordered_map<ll,array<ll,10>> dp[19] = {}; unordered_map<ll,short> dig; short getdig(ll v) { if(v<10) return v; if(dig.count(v)) return dig[v]; ll cur = 1; for(auto c : to_string(v)) { cur*=c-'0'; } return dig[v] = getdig(cur); } auto operator+(array<ll,10> a, const array<ll,10>& b) { for(int j=0;j<10;++j) a[j]+=b[j]; return a; } array<ll,10> solve(ll mult, int left) { if(left==0) { array<ll,10> res = {}; res[getdig(mult)]++; return res; } auto it = dp[left].find(mult); if(it!=dp[left].end()) return it->second; auto& res = dp[left][mult]; for(int d=0;d<10;++d) { res = res+ solve(mult*d,left-1); } return res; } unordered_map<int,array<ll,10>> precomp; auto solvenonzero(int x) { if(x==0) return array<ll,10>{}; auto it = precomp.find(x); if(it!=precomp.end()) return it->second; auto& res = precomp[x]; res=solvenonzero(x-1); for(int j=1;j<10;++j) res = res +solve(j,x-1); return res; } void solve() { ll x; cin >> x; x++; // strictly smaller string n = to_string(x); for(auto& i : n) i-='0'; // take product of those, then bruteforce // next digit, then have arbitrary digits... ll mult=1; array<ll,10> ans = {}; for(int i=0;i<n.size();++i) { for(int j=0;j<n[i];++j) { if(i==0 and j==0) continue; // no leading zeros. ans = ans+solve(mult*j,n.size()-1-i); } mult*=n[i]; } ans = ans+solvenonzero(n.size()-1); cout << ans << '\n'; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int t; cin >> t; while(t--) solve(); } |
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