Beautiful Numbers

https://codeforces.com/problemset/problem/300/C

就是数位上的是好数，然后数位求和后的数，数位上也得是好数

满足关系

考虑n<=1e6,枚举x,找到符合条件的sum

得到x个a,(n-x)个b可以组合成极好数，组合方法数有

总组合方法数

// Problem: C. Beautiful Numbers
// Contest: Codeforces - Codeforces Round 181 (Div. 2)
// URL: https://codeforces.com/problemset/problem/300/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int N=9e6+9;
ll f[N],invf[N];
int a,b,n;
ll qmi(ll a,ll b){
	ll res=1;
	while(b){
		if(b&1){
			res=res*a%mod;
		}
		a=a*a%mod;
		b>>=1;
	}
	return res;
}
ll inv(ll x){return qmi(x,mod-2)%mod;}
ll C(ll n,ll m){return (((f[n]*invf[m])%mod)*invf[n-m]%mod)%mod;}
bool check(ll x){
	while(x){
		ll t=x%10;
		if(t!=a && t!=b){
			return false;
		}
		x/=10;
	}
	return true;
}
void init(){
	f[0]=1;
	for(int i=1;i<=1000002;i++){
		f[i]=f[i-1]*i%mod;
	}
	invf[1000001]=inv(f[1000001])%mod;
	for(int i=1000000;i>=0;i--){
		invf[i]=invf[i+1]*(i+1)%mod;//
	}
}
int main(){
	init();
	cin>>a>>b>>n;
	ll ans=0;	
	for(int i=0;i<=n;i++){
		if(check(a*i+b*(n-i))){
			ans+=(C(n,i));
			ans%=mod;
		}
	}
	cout<<ans%mod<<'\n';
	return 0;
}