Combinatorial mathematics codefoeces300c Beautiful Numbers

... At first did not see test instructions, the original premise of the excellent is good ...

We think that the number of the n-bit length for each of the a,a is called B.

First of all, we need to discuss it in two different situations.

In the first case, when a equals B, it means that the n digits in a are all a.

That is, the value of B equals a*n, and we just need to verify that the number of each digit in B is equal to a.

If all equals, then the answer is 1, otherwise the answer is 0.

In the second case, when a is not equal to B

Set a appears x times in A, and b appears in a For Y times.

Because the length of a is n, there must be x+y = n

B is the sum of each digit of a, so B can only be 9 * N, that is, every bit of n digits is 9

However, B can only have A and B two numbers, stating that the number of B is not too much, up to thousands of, so it is possible to direct violence to cite b

Also know that a in a in the appearance of x times, b in a in the occurrence of Y. B is the sum of each number in a, and a is only A and b ...

Speaking of which, it's obvious, b=a*x+b*y.

So, X+y = N,b=a*x+b*y, you can solve the x and y directly

If you find that x and y are not integers or have a less than 0, then it means that this type of B is not sufficient to ignore the situation.

Well, now we know that when B is a case, the number of x and Y, how to ask for the number of species?

In fact, in the n position to choose X position to place the number A, the other position must also only put the number B, so is C (n,x)

It's a big circle, and the final idea is clear.

1. Special break

2.DFS Enumeration Number B

3. Solution equation, get x and Y

4. C (n,x) for all conditions of satisfaction, and add modulus is the answer

So how to ask C,, this can Baidu a bit inverse

Here I directly affixed to the template, and later encountered the number of combinations, even if the provisional understanding of the principle, you can directly hit the template

const int MX = 1000000 + 5;//n size const int mod = 1e9 + 7;//If the modulo is not prime, this template is not available!! LL F[MX], invf[mx]; ll Power (ll A, ll b) {    LL ret = 1;    while (b) {        if (b & 1) ret = (ret * a)% mod;        A = (A * a)% mod;        b >>= 1;    }    return ret;} void init () {//This is used for initialization, remember    f[0] = 1;    for (int i = 1; i < MX; i++) {        F[i] = (f[i-1] * i)% mod;    }    INVF[MX-1] = Power (f[mx-1], mod-2);    for (int i = MX-2; I >= 0; i--) {        Invf[i] = invf[i + 1] * (i + 1)% mod;    }} LL C (int n, int m) {    if (N < 0 | | m < 0 | | m > N) return 0;    if (m = = 0 | | m = = N)    return 1;    return F[n] * invf[n-m]% mod * invf[m]% MoD;} LL A (int n, int m) {    if (N < 0 | | m < 0 | | m > N) return 0;    return F[n] * invf[n-m]% MoD;}

Here is the code to apply the template AC this topic

#include <map> #include <set> #include <cmath> #include <stack> #include <queue> #include <cstdio> #include <string> #include <vector> #include <cstring> #include <iostream># Include<algorithm> #include <functional>using namespace std;typedef long long ll;typedef pair<int, int > pii;const int MX = 1000000 + 5;const int mod = 1e9 + 7; LL F[MX], invf[mx];    ll Power (ll A, ll b) {ll ret = 1;        while (b) {if (b & 1) ret = (ret * a)% mod;        A = (A * a)% mod;    b >>= 1; } return ret;}    void init () {//This is used for initialization, remember f[0] = 1;    for (int i = 1; i < MX; i++) {F[i] = (f[i-1] * i)% mod;    } Invf[mx-1] = Power (f[mx-1], mod-2);    for (int i = MX-2; I >= 0; i--) {Invf[i] = invf[i + 1] * (i + 1)% MoD;    }}ll C (int n, int m) {if (N < 0 | | m < 0 | | m > N) return 0;    if (m = = 0 | | m = = n) return 1; return F[n] * invf[n-m]% mod * invf[m]% MoD;} LL A (int n,int m) {if (N < 0 | | m < 0 | | m > N) return 0; return F[n] * invf[n-m]% MoD;}        BOOL Check (int n, int x) {while (n) {if (n%! = x) return false;    n/= 10; } return true; int A, b, N;    LL solve (int z) {if ((Z-A * n)% (b-a)) return 0;    int y = (Z-A * N)/(B-A); Return C (n, y);}    LL DFS (int x) {if (x > 9 * N) return 0;    LL ans = solve (x);    Ans + = DFS (x + a);    Ans + = DFS (x + b);    Ans%= MoD; return ans;}    int main () {//freopen ("Input.txt", "R", stdin);    Init ();    scanf ("%d%d%d", &a, &b, &n);        if (a = = b) {if (check (A * n, a)) printf ("1\n");        else printf ("0\n");    return 0;    } printf ("%i64d\n", DFS (0)); return 0;}

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Combinatorial mathematics codefoeces300c Beautiful Numbers