Codeforces 300C Beautiful Numbers lucas求组合数逆元

最新推荐文章于 2020-08-07 00:08:24 发布

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本文介绍了一个算法问题，即计算特定条件下由两个指定数字组成的长度为n的优秀数的数量，并给出了解决方案。优秀数需满足其各位数字之和也是由指定的两个数字组成。

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Vitaly is a very weird man. He's got two favorite digits a and b. Vitaly calls a positive integer good, if the decimal representation of this integer only contains digits a and b. Vitaly calls a good number excellent, if the sum of its digits is a good number.

For example, let's say that Vitaly's favourite digits are 1 and 3, then number 12 isn't good and numbers 13 or 311 are. Also, number111 is excellent and number 11 isn't.

Now Vitaly is wondering, how many excellent numbers of length exactly n are there. As this number can be rather large, he asks you to count the remainder after dividing it by 1000000007 (109 + 7).

A number's length is the number of digits in its decimal representation without leading zeroes.

Input

The first line contains three integers: a, b, n (1 ≤ a < b ≤ 9, 1 ≤ n ≤ 106).

Output

Print a single integer — the answer to the problem modulo 1000000007 (109 + 7).

        Sample test(s)
      
          input
        
1 3 3

          output
        
1

          input
        
2 3 10

          output
        
165

给你两个数a，b在1~9之间，让这两个数组成长度为n的数，且各个位上的数之和为sum，要求sum各个位上的数也仅由a和b组成，问一共有多少种方式。

枚举每一位，然后来个排列组合即可。

乘法逆元：(a/b)%mod=a*(b^(mod-2)) mod为素数。

//92 ms	7808 KB
#include<stdio.h>
#define mod 1000000007
#define ll long long
ll f[1000007]={1,1};
ll a,b;
bool check(ll x)
{
    while(x)
    {
        ll z=x%10;
        if(z!=a&&z!=b)return false;
        x/=10;
    }
    return true;
}
ll quick_mod(ll x,ll y)
{
    ll ans=1;
    x%=mod;
    while(y)
    {
        if(y&1)ans=(ans*x)%mod,y--;
        y>>=1;
        x=x*x%mod;
    }
    return ans;
}
int main()
{
    int n;
    for(int i=2;i<=1000007;i++)f[i]=f[i-1]*i%mod;
    while(scanf("%lld%lld%d",&a,&b,&n)!=EOF)
    {
        ll ans=0;
        for(int i=0;i<=n;i++)
        {
            ll sum=i*a+(n-i)*b;
            if(check(sum))
            {
                ll c=f[n-i]*f[i];
                ans=(ans+f[n]*quick_mod(c,mod-2))%mod;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}