好久不见数论的题，碰到组合数+取模 马上想到卢卡斯定理 （其实组合数的递推式也可以做

Ac代码

#include <iostream>
#include<algorithm>

using namespace std;
typedef long long ll;
ll f[101];
ll mod = 1000000007;

void init() {
    f[0] = 1;
    for (ll i = 1; i <= 100; i++)
        f[i] = (i * f[i - 1]) % mod;
}

long long pow1(long long n, long long m) {
    long long ans = 1;
    while (m > 0) {
        if (m & 1)ans = (ans * n) % mod;
        m = m >> 1;
        n = (n * n) % mod;
    }
    return ans;
}

long long lucas(ll n, ll m) {
    ll ans = 1;
    while (n && m) {
        ll x, y;
        x = n % mod;
        y = m % mod;
        if (x < y)
            return 0;
        // printf("%lld %lld %lld\n",f[x],f[y],f[x-y]);
        ans = ans * f[x] * pow1(f[y] * f[x - y] % mod, mod - 2) % mod;
        n /= mod;
        m /= mod;
    }
    return ans % mod;
}

ll quick_multi(ll a, ll b) {
    ll ans = 0;
    while (b) {
        if (b & 1)
            ans = (ans + a)%mod;
        a = (a + a)%mod;
        b >>=1;
    }
    return ans;
}

int main() {

    ll K, X, A, Y, B;
    init();
    while (cin >> K) {
        cin >> A >> X >> B >> Y;
        ll ans = 0;
        for (int i = 0; i <= X; i++)
            for (int j = 0; j <= Y; j++) {
                if ((i * A + j * B) == K) {
                    ans += quick_multi(lucas(X, i) % mod , lucas(Y, j) % mod) % mod;
                }
            }
        cout << ans%mod << endl;
    }
    return 0;
}