題面

題解

易得答案為

$$ \sum_{i=1}^m\binom{n-\sum_{j=1}^{i-1}w_j}{\sum_{j=1}^iw_j} $$

擴充套件$\text{Lucas}$即可

程式碼

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x))

#define int long long
inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

inline int fastpow(int x, int y, int Mod)
{
	int ans = 1;
	while(y)
	{
		if(y & 1) ans = 1ll * ans * x % Mod;
		x = 1ll * x * x % Mod, y >>= 1;
	}
	return ans;
}

void exgcd(int a, int b, int &d, int &x, int &y)
{
	!b ? d = a, x = 1, y = 0 : (exgcd(b, a % b, d, y, x), y -= x * (a / b));
}

inline int Inv(int i, int Mod)
{
	if(!i) return 0;
	int x, y, d, a = i, b = Mod;
	exgcd(a, b, d, x, y);
	x = (x % b + b) % b;
	if(!x) x += b;
	return x;
}

int mul(int n, int p, int k)
{
	if(!n) return 1;
	int ans = 1;
	for(int i = 2; i <= k; i++)
		if(i % p) ans = ans * i % k;
	ans = fastpow(ans, n / k, k);
	for(int i = 2; i <= n % k; i++)
		if(i % p) ans = ans * i % k;
	return ans * mul(n / p, p, k) % k;
}

inline int C(int n, int m, int Mod, int p, int k)
{
	if(m > n) return 0;
	int a = mul(n, p, k), b = mul(m, p, k), c = mul(n - m, p, k), _k = 0;
	for(int i = n; i; i /= p) _k += i / p;
	for(int i = m; i; i /= p) _k -= i / p;
	for(int i = n - m; i; i /= p) _k -= i / p;
	int ans = a * Inv(b, k) % k * Inv(c, k) % k * fastpow(p, _k, k) % k;
	return ans * (Mod / k) % Mod * Inv(Mod / k, k) % Mod;
}

int n, m, sum, Mod, ans = 1, a[20];
signed main()
{
#ifndef ONLINE_JUDGE
	file(cpp);
#endif
	Mod = read(), n = read(), m = read();
	for(signed i = 1; i <= m; i++) sum += (a[i] = read());
	if(sum > n) return puts("Impossible") & 0;
	for(signed k = 1; k <= m; k++)
	{
		n -= a[k - 1];
		int now = 0, x = Mod;
		for(int i = 2; i * i <= Mod; i++)
			if(!(x % i))
			{
				int _k = 1;
				while(!(x % i)) _k *= i, x /= i;
				now = (now + C(n, a[k], Mod, i, _k)) % Mod;
			}
		if(x > 1) now = (now + C(n, a[k], Mod, x, x)) % Mod;
		ans = ans * now % Mod;
	}
	printf("%lld\n", ans);
	return 0;
}