F - Beautiful Numbers CodeForces - 300C（组合数）

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, number 111 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).

Examples
Input
1 3 3
Output
1
Input
2 3 10
Output
165

题意：
给出a,b。数位只包含a,b的是好数。数位和为好数的好数是好好数。求多少个长度为n的好好数。

思路：
直接枚举有多少个a即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <stack>
#include <vector>
#include <set>

using namespace std;

typedef long long ll;
const int maxn = 1e6 + 7;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;

ll fac[maxn],inv[maxn];

ll qpow(ll x,ll n) {
    ll res = 1;
    while(n) {
        if(n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

void init() {
    fac[0] = inv[0] = 1;
    for(int i = 1;i < maxn;i++) {
        fac[i] = fac[i - 1] * i % mod;
        inv[i] = qpow(fac[i],mod - 2);
    }
}

ll C(ll n,ll m) {
    if(m > n || m < 0) {
        return 0;
    }
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
}

bool check(int x,int a,int b) {
    while(x) {
        int num = x % 10;
        if(num != a && num != b) return false;
        x /= 10;
    }
    return true;
}

int main() {
    init();
    int a,b,n;scanf("%d%d%d",&a,&b,&n);
    ll ans = 0;
    for(int i = 0;i <= n;i++) { //a的数量
        ll num = a * i + b * (n - i);
        if(check(num,a,b)) {
            ans = (ans + C(n,i)) % mod;
        }
    }
    printf("%lld\n",ans);
    return 0;
}