CodeForces 300C Beautiful Numbers

枚举，组合数，逆元。

枚举$a$用了$i$个，那么$b$就用了$n-i$个，这个时候和$sum=a*i+b*(n-i)$，判断$sum$是否满足条件，如果满足，那么答案加上$C(n,i)$。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

int n;
long long a,b;
long long mod=1e9+;
long long ans;
long long f[];

bool check(int x)
{
    long long num=(long long)x*a
                    +(long long)(n-x)*b;
    while(num)
    {
        if(num%!=a&&num%!=b) return ;
        num=num/;
    }
    return ;
}

long long extend_gcd(long long a,long long b,long long &x,long long &y)
{
    if(a==&&b==) return -;
    if(b==){x=;y=;return a;}
    long long d=extend_gcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}

long long mod_reverse(long long a,long long n)
{
    long long x,y;
    long long d=extend_gcd(a,n,x,y);
    if(d==) return (x%n+n)%n;
    else return -;
}

long long C(int a,int b)
{
    return f[a]*mod_reverse(f[b]*f[a-b],mod)%mod;
}

int main()
{
    f[]=;
    for(long long i=;i<=;i++) f[i]=i*f[i-]%mod;
    scanf("%lld%lld%d",&a,&b,&n);
    ans=;
    for(int i=;i<=n;i++)
    {
        if(check(i)) ans=(ans+C(n,i))%mod;
    }
    printf("%lld\n",ans);
    return ;
}