Counting such number whose digit product multiple to itself smaller than given number

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The problem

Lets $$$f(x) = $$$ product of digits of $$$x$$$. Example: $$$f(123) = 1 * 2 * 3 = 6$$$, $$$f(990) = 9 * 9 * 0 = 0$$$, $$$f(1) = 1$$$

The statement is, given such limitation $$$N$$$, count the number of positive $$$x$$$ that $$$1 \leq x * f(x) \leq N$$$

Example: For $$$N = 20$$$, there are 5 valid numbers $$$1, 2, 3, 4, 11$$$

The limitation

Subtask 1: $$$N \leq 10^6$$$
Subtask 2: $$$N \leq 10^{10}$$$
Subtask 3: $$$N \leq 10^{14}$$$
Subtask 4: $$$N \leq 10^{18}$$$

My approach for subtask 1

If $$$(x > N)$$$ or $$$(f(x) > N)$$$ then $$$(x * f(x) > N)$$$. So we will only care about $$$x \leq N$$$ that $$$x * f(x) \leq N$$$

Calculating x * f(x) - O(log10(x))

ll cal(ll x, ll lim)
{
    ll res = x;
    do {
        int t = x % 10;
        if (res > lim / t) return -1; /// (x * f(x) > N)
        res *= t;
    } while (x /= 10);
    return res;
}

Counting - O(n log10(n))

ll brute(ll n)
{
    ll res = 0;
    for (ll x = 1; x <= n; ++x)
        res += cal(x, n) > 0; /// 1 <= x * f(x) <= N

    return res;
}

My approach for subtask 2

If $$$x$$$ contains $$$0$$$ then $$$f(x) = 0 \Rightarrow x \times f(x) < 1$$$. We only care about such $$$x$$$ without $$$0$$$ digit

Building function - O(result + log10(n))

/// Run 2e13 under 1 secs in (-O2) flag
/// X: current X
/// Y: f(X)
/// T: X * f(X)

ll N;
int res = 0;
vector<int> d;
void build(ll X = 0, ll Y = 1, ll T = 0)
{
    if (T >= 1) ++res; /// 1 <= x * f(x) <= N
    for (int v = 1; v <= 9; ++v) /// if (v = 0) then f(x) = 0
    {
        ll NX = X * 10 + v; /// Insert rightmost digits
        ll NY = Y * v;      /// Calculate digits production
        ll NT = NX * NY;
        if (NT > N) break;  /// x * f(x) > N
        build(NX, NY, NT);
    }
}

Here is the solution:

Let takes some $$$x$$$ satisfy $$$1 \leq x * f(x) \leq N$$$

We can easily prove that $$$f(x) \leq x$$$, and because $$$x * f(x) \leq N$$$, we have $$$f(x) \leq \sqrt{N}$$$ (notice that $$$x$$$ might bigger than $$$\sqrt{N}$$$)

Since $$$f(x)$$$ is product of digits of $$$x$$$, which can be obtain by such digits {$$$1, 2, \dots, 9$$$}. So $$$f(x) = 2^a \times 3^b \times 5^c \times 7^d$$$

So we can bruteforces all possible tuple of such $$$(a, b, c, d)$$$ satisfy ($$$P = 2^a \times 3^b \times 5^c \times 7^d \leq \sqrt{N}$$$). There are small amount of such tuples (493 tuples for $$$N = 10^9$$$ and 5914 tuples for $$$N = 10^{18}$$$)

Find all possible tuples - O(quartic_root(N))

/// [L, R] = [1, N]
/// lim = sqrt(N)
ll solve(int p2 = 0, int p3 = 0, int p5 = 0, int p7 = 0, ll P = 1)
{
    if (P > lim) return 0; /// Dont care such tuples whose P > sqrt(N)

    VL = (L + val - 1) / val; ///  ceil(L / P)
    VR = R / val;             /// floor(R / P)
    ll res = magic(0, 18, p2, p3, p5, p7); /// Calculating subproblem

    /// By doing these if-condition, it is guarantee that all tuples generated are all unique
    if (!p3 && !p5 && !p7) res += solve(p2 + 1, p3, p5, p7, P * 2); /// Continue increasing a
    if (       !p5 && !p7) res += solve(p2, p3 + 1, p5, p7, P * 3); /// Continue increasing b
    if (              !p7) res += solve(p2, p3, p5 + 1, p7, P * 5); /// Continue increasing c
                           res += solve(p2, p3, p5, p7 + 1, P * 7); /// Continue increasing d

    return res;
}

For each tuples, we need to counting the numbers of such $$$x$$$ that $$$1 \leq x \times f(x) \leq N$$$ and $$$f(x) = P$$$.

We have the value $$$P$$$, so $$$\lceil \frac{1}{P} \rceil \leq x \leq \lfloor \frac{N}{P} \rfloor$$$.
We have the value $$$f(x) = P$$$, so $$$x$$$ can be made by digits having the product exactly $$$P$$$, so we can do some DP-digit

So now we have to solve this DP-digit problem: Calculate numbers of such $$$x$$$ ($$$L \leq x \leq R$$$) whose $$$f(x) = P$$$

Solving Subproblem

We try to build each digits by digits for $$$X$$$. Because $$$X \leq N$$$, so we have to build about $$$18$$$ digits.

Lets make a recursive function $$$magic(X, N, p2, p3, p5, p7)$$$

Lets make some definition

Notice that

Ugly precalculation code - O(1)

const int l2 = 60, l3 = 37, l5 = 25, l7 = 21, l10 = 19;
ll        pw2[l2], pw3[l3], pw5[l5], pw7[l7], pw10[l10];
int       cost[10][10];
void precal()
{
    pw2[0] = pw3[0] = pw5[0] = pw7[0] = pw10[0] = 1; 
    for (int i2  = 1; i2  < l2 ; ++i2 ) pw2 [i2]  =  2 * pw2 [i2  - 1];
    for (int i3  = 1; i3  < l3 ; ++i3 ) pw3 [i3]  =  3 * pw3 [i3  - 1];
    for (int i5  = 1; i5  < l5 ; ++i5 ) pw5 [i5]  =  5 * pw5 [i5  - 1];
    for (int i7  = 1; i7  < l7 ; ++i7 ) pw7 [i7]  =  7 * pw7 [i7  - 1];
    for (int i10 = 1; i10 < l10; ++i10) pw10[i10] = 10 * pw10[i10 - 1];
                                                                                /// 2^a * 3^b * 5^c * 7^d = val
    cost[1][2] = 0;    cost[1][3] = 0;    cost[1][5] = 0;    cost[1][7] = 0;    ///  0     0     0     0  =  1
    cost[2][2] = 1;    cost[2][3] = 0;    cost[2][5] = 0;    cost[2][7] = 0;    ///  1     0     0     0  =  2
    cost[3][2] = 0;    cost[3][3] = 1;    cost[3][5] = 0;    cost[3][7] = 0;    ///  0     1     0     0  =  3
    cost[4][2] = 2;    cost[4][3] = 0;    cost[4][5] = 0;    cost[4][7] = 0;    ///  2     0     0     0  =  4
    cost[5][2] = 0;    cost[5][3] = 0;    cost[5][5] = 1;    cost[5][7] = 0;    ///  0     0     1     0  =  5
    cost[6][2] = 1;    cost[6][3] = 1;    cost[6][5] = 0;    cost[6][7] = 0;    ///  1     1     0     0  =  6
    cost[7][2] = 0;    cost[7][3] = 0;    cost[7][5] = 0;    cost[7][7] = 1;    ///  0     0     0     1  =  7
    cost[8][2] = 3;    cost[8][3] = 0;    cost[8][5] = 0;    cost[8][7] = 0;    ///  3     0     0     0  =  8
    cost[9][2] = 0;    cost[9][3] = 2;    cost[9][5] = 0;    cost[9][7] = 0;    ///  0     2     0     0  =  9
}

magic function - O(18*p2*p3*p5*p7)

ll VL, VR;
ll f[l10][l2][l3][l5][l7];
ll magic(ll X, int N, int p2, int p3, int p5, int p7)
{
    if (p2 < 0 || p3 < 0 || p5 < 0 || p7 < 0) return 0;
    if (N == 0) return (p2 + p3 + p5 + p7 == 0) && (VL <= X && X <= VR);

    ll mn = X * pw10[N];
    ll mx = mn + pw10[N] - 1;
    if (mx < VL || mn > VR) return 0;

    ll &save = f[N][p2][p3][p5][p7];
    bool memo = (VL <= mn) && (mx <= VR);
    if (memo) if (save != -1) return save;
    
    ll res = 0;
    if (X == 0) res = magic(0, N - 1, p2, p3, p5, p7);
    for (int v = 1; v <= 9; ++v)
    {
        int c2 = cost[v][2];
        int c3 = cost[v][3];
        int c5 = cost[v][5];
        int c7 = cost[v][7];
        res += magic(X * 10 + v, N - 1, p2 - c2, p3 - c3, p5 - c5, p7 - c7);
    }
    if (memo) save = res;
    
    return res;
}

About the proving stuff

1) $$$\forall x \in \mathbb{N}, f(x) \leq x$$$

Proof: $$$x = \overline{\dots dcba} = \dots + d \times 10^3 + c \times 10^2 + b \times 10^1 + a \times 10^0 \geq \dots \times d \times c \times b \times a = f(x)$$$

2) If $$$x$$$ satisfy then $$$f(x) \leq \sqrt{N}$$$ must be satisfied

Proof: $$$x \times f(x) \leq N \Rightarrow f(x) \times f(x) \leq N \Rightarrow f(x) \leq \sqrt{N}$$$

3) $$$\exists\ a, b, c, d \in \mathbb{N} \rightarrow f(x) = 2^a \times 3^b \times 5^c \times 7^d$$$

Since $$$x = \overline{\dots dcba} \Rightarrow (0 \leq \dots, d, c, b, a \leq 9)$$$ and $$$f(x) = \dots \times d \times c \times b \times a$$$
And we also have $$$\forall$$$ digit $$$v$$$ ($$$v \in \mathbb{N}, 0 \leq v \leq 9$$$) $$$\rightarrow \exists\ a, b, c, d \in \mathbb{N} \rightarrow v = 2^a \times 3^b \times 5^c \times 7^d$$$
And because $$$f(x)$$$ is the product of digits of $$$x$$$, hence the statement is correct

4) If we know $$$f(x)$$$ we can find such $$$x$$$ satisfy $$$x \in [L, R]$$$

Proof: Since $$$f(x)$$$ is created from factors of digits of $$$x$$$, so $$$x$$$ can also be generated using the factors

5) Number of tuples $$$(a, b, c, d)$$$ satisfy $$$P = 2^a \times 3^b \times 5^c \times 7^d \leq \sqrt{N}$$$ is very small

Lets $$$O(k(x)) = O(log_2(x) \times log_3(x) \times log_5(x) \times log_7(x))$$$
Since each value $$$x$$$ have the upper bound of $$$log_x(\sqrt{N})$$$. So the complexity is about $$$O(log_2(\sqrt{N}) \times log_3(\sqrt{N}) \times log_5(\sqrt{N}) \times log_7(\sqrt{N})) = O(k(\sqrt{N})) \leq O(log_2(\sqrt{N})^4)$$$
But actually for $$$R \leq 10^{18}$$$, the complexity is just about $$$O(k(\sqrt[4]{N}))$$$

Weak proof - N = 10^k

With (N = 1) -> Found 1 valid tuples || upper_bound 1
With (N = 10) -> Found 3 valid tuples || upper_bound 3
With (N = 100) -> Found 10 valid tuples || upper_bound 10
With (N = 1000) -> Found 22 valid tuples || upper_bound 24
With (N = 10000) -> Found 46 valid tuples || upper_bound 50
With (N = 100000) -> Found 83 valid tuples || upper_bound 95
With (N = 1000000) -> Found 141 valid tuples || upper_bound 166
With (N = 10000000) -> Found 225 valid tuples || upper_bound 269
With (N = 100000000) -> Found 338 valid tuples || upper_bound 414
With (N = 1000000000) -> Found 493 valid tuples || upper_bound 612
With (N = 10000000000) -> Found 694 valid tuples || upper_bound 874
With (N = 100000000000) -> Found 951 valid tuples || upper_bound 1212
With (N = 1000000000000) -> Found 1273 valid tuples || upper_bound 1640
With (N = 10000000000000) -> Found 1670 valid tuples || upper_bound 2172
With (N = 100000000000000) -> Found 2155 valid tuples || upper_bound 2825
With (N = 1000000000000000) -> Found 2736 valid tuples || upper_bound 3614
With (N = 10000000000000000) -> Found 3427 valid tuples || upper_bound 4558
With (N = 100000000000000000) -> Found 4246 valid tuples || upper_bound 5676
Error: 5701

Weak proof - N = 2^k

With (N = 1) -> Found 1 valid tuples || upper_bound 1
With (N = 2) -> Found 1 valid tuples || upper_bound 1
With (N = 4) -> Found 2 valid tuples || upper_bound 2
With (N = 8) -> Found 2 valid tuples || upper_bound 3
With (N = 16) -> Found 4 valid tuples || upper_bound 4
With (N = 32) -> Found 5 valid tuples || upper_bound 6
With (N = 64) -> Found 8 valid tuples || upper_bound 8
With (N = 128) -> Found 10 valid tuples || upper_bound 11
With (N = 256) -> Found 14 valid tuples || upper_bound 14
With (N = 512) -> Found 17 valid tuples || upper_bound 19
With (N = 1024) -> Found 23 valid tuples || upper_bound 24
With (N = 2048) -> Found 28 valid tuples || upper_bound 30
With (N = 4096) -> Found 36 valid tuples || upper_bound 38
With (N = 8192) -> Found 43 valid tuples || upper_bound 47
With (N = 16384) -> Found 53 valid tuples || upper_bound 58
With (N = 32768) -> Found 63 valid tuples || upper_bound 71
With (N = 65536) -> Found 77 valid tuples || upper_bound 85
With (N = 131072) -> Found 89 valid tuples || upper_bound 102
With (N = 262144) -> Found 106 valid tuples || upper_bound 121
With (N = 524288) -> Found 122 valid tuples || upper_bound 143
With (N = 1048576) -> Found 143 valid tuples || upper_bound 167
With (N = 2097152) -> Found 164 valid tuples || upper_bound 195
With (N = 4194304) -> Found 190 valid tuples || upper_bound 225
With (N = 8388608) -> Found 216 valid tuples || upper_bound 260
With (N = 16777216) -> Found 248 valid tuples || upper_bound 298
With (N = 33554432) -> Found 279 valid tuples || upper_bound 339
With (N = 67108864) -> Found 317 valid tuples || upper_bound 386
With (N = 134217728) -> Found 355 valid tuples || upper_bound 437
With (N = 268435456) -> Found 400 valid tuples || upper_bound 492
With (N = 536870912) -> Found 446 valid tuples || upper_bound 553
With (N = 1073741824) -> Found 498 valid tuples || upper_bound 620
With (N = 2147483648) -> Found 553 valid tuples || upper_bound 692
With (N = 4294967296) -> Found 614 valid tuples || upper_bound 770
With (N = 8589934592) -> Found 679 valid tuples || upper_bound 855
With (N = 17179869184) -> Found 749 valid tuples || upper_bound 946
With (N = 34359738368) -> Found 825 valid tuples || upper_bound 1045
With (N = 68719476736) -> Found 905 valid tuples || upper_bound 1152
With (N = 137438953472) -> Found 991 valid tuples || upper_bound 1266
With (N = 274877906944) -> Found 1083 valid tuples || upper_bound 1388
With (N = 549755813888) -> Found 1181 valid tuples || upper_bound 1520
With (N = 1099511627776) -> Found 1286 valid tuples || upper_bound 1660
With (N = 2199023255552) -> Found 1398 valid tuples || upper_bound 1810
With (N = 4398046511104) -> Found 1517 valid tuples || upper_bound 1970
With (N = 8796093022208) -> Found 1646 valid tuples || upper_bound 2140
With (N = 17592186044416) -> Found 1780 valid tuples || upper_bound 2321
With (N = 35184372088832) -> Found 1924 valid tuples || upper_bound 2513
With (N = 70368744177664) -> Found 2074 valid tuples || upper_bound 2717
With (N = 140737488355328) -> Found 2235 valid tuples || upper_bound 2933
With (N = 281474976710656) -> Found 2402 valid tuples || upper_bound 3161
With (N = 562949953421312) -> Found 2581 valid tuples || upper_bound 3403
With (N = 1125899906842624) -> Found 2767 valid tuples || upper_bound 3658
With (N = 2251799813685248) -> Found 2966 valid tuples || upper_bound 3928
With (N = 4503599627370496) -> Found 3174 valid tuples || upper_bound 4212
With (N = 9007199254740992) -> Found 3393 valid tuples || upper_bound 4511
With (N = 18014398509481984) -> Found 3625 valid tuples || upper_bound 4826
With (N = 36028797018963968) -> Found 3866 valid tuples || upper_bound 5157
With (N = 72057594037927936) -> Found 4121 valid tuples || upper_bound 5505
With (N = 144115188075855872) -> Found 4386 valid tuples || upper_bound 5870
With (N = 288230376151711744) -> Found 4665 valid tuples || upper_bound 6253
With (N = 576460752303423488) -> Found 4955 valid tuples || upper_bound 6655
With (N = 1152921504606846976) -> Found 5260 valid tuples || upper_bound 7076
Error: 23112

Code

#include <iostream>
#include <cmath>
#include <cstdio>

using namespace std;

typedef long long ll;

ll L;
ll R;
int lim;
ll solve(int p2 = 0, int p3 = 0, int p5 = 0, int p7 = 0, ll val = 1)
{
    if (val > lim) return 0;

    ll res = 1;
    if (!p3 && !p5 && !p7) res += solve(p2 + 1, p3, p5, p7, val * 2);
    if (       !p5 && !p7) res += solve(p2, p3 + 1, p5, p7, val * 3);
    if (              !p7) res += solve(p2, p3, p5 + 1, p7, val * 5);
                           res += solve(p2, p3, p5, p7 + 1, val * 7);

    return res;
}

const double EPS = 0.0188704;
int main()
{
    freopen("Test.OUT", "w", stdout);

    double error = 0;
    ll mul = 2;
    ll over = (1LL << 62) / mul;
    for (ll t = 1; t < over; t *= mul)
    {
        L = 1;
        R = t;
        lim = sqrt(R);
        int res = solve();
        double base = log(sqrt(sqrt(R))) + 1;
        int complexity = ceil(EPS + (base / log(2)) * (base / log(3)) * (base / log(5)) * (base / log(7)));

        cout << "With (N = " << R << ") -> Found " << res << " valid tuples || upper_bound " << complexity << endl;
        int delta = complexity - res;
        if (delta < 0)
        {
            cout << "Program failed at (t = " << t << ") -> ";
            cout << "Returning " << res << " compared to " << complexity << endl;
            exit(0);
        }
        else error += delta;
    }
    cout << "Error: " << error;

    return 0;
}

About the complexity:

$$$O(h(x)) = O(log_{10}(N))$$$ is number of digits we have to build
$$$O(k(x)) = O(log_2(N) \times log_3(N) \times log_5(N) \times log_7(N)) = O(log(N)^4)$$$ is product of all prime digits $$$p$$$ with its maximum power $$$k$$$ satisfy $$$p^k \leq N$$$
$$$O(g(x)) = O(k(\sqrt{N}))$$$ is number of such valid tuples, but for $$$1 \leq N \leq 10^{18}$$$ it is about $$$\approx O(k(\sqrt[4]{N})) \leq O(log_2^4{\sqrt[4]{N}})$$$
The space complexity is $$$O(SPACE) = O(h(x) \times k(x)) = O(log_2(N) \times log_3(N) \times log_5(N) \times log_7(N) \times log_{10}(N)) = O(log(N)^5)$$$
The time complexity is $$$O(TIME) = O(SPACE) + O(g(x) \times k(x)) \approx O(log(N)^4 \times log_2^4{\sqrt[4]{N}})$$$

Other

Here is the Vietnamese version of the problem where we need to count such valid numbers in range $$$[A..B]$$$ where $$$1 \leq A \leq B \leq 10^{18}$$$. (Vietnamese Editorial)

About the real complexity for $$$O(k(x))$$$, do you find a better/closer upper bound of counting such tuples of $$$(a, b, c, d \in \mathbb{N})$$$ that $$$P = 2^a \times 3^b \times 5^c \times 7^d \leq N$$$ ? Since my $$$O(log_2(N))$$$ complexity seem very far than the real counting and $$$O(log_2(\sqrt{N}))$$$ is closer but just correct for $$$N \leq 10^{18}$$$
Thanks for reading and sorry for updating this blog publicly so many times (most is for the proving path and correcting the complexity)

Comments (7)

Write comment?

Nezzar

4 years ago, # |

Not really sure, but isnt that f(x)<x, therefore we only need to consider f(x)<=sqrt(10^9)?

→ Reply

SPyofgame

4 years ago, # ^ |

← Rev. 2 →

Yes your way is correct ^^, thank sir. I have just got accepted from the problem and I am writting the solution

I have just done the problem and wrote the solution, thanks for reading ^^

If you have a better explaination/implementation or better complexity, please comment in this post too ^^.

← Rev. 3 →

By the way, if you guys are interesting, here is the Vietnamese version of the problem where we need to count such valid numbers in range $$$[A..B]$$$ where $$$1 \leq A \leq B \leq 10^{18}$$$. And here is the Editorial in Vietnamese

MonsieurV

Wow, that's an amazing problem

_FireGhost_

That's your problem...

No, that's just the same name

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