【LG2183】[国家集训队]礼物
【LG2183】[国家集训队]礼物
题面
题解
插曲:不知道为什么,一看到这个题目,我就想到了这个人。。。
如果不是有\(exLucas\),这题就是\(sb\)题。。。
首先,若\(\sum_{i=1}^mw_i>n\)就直接\(Impossible\)了
然后我们考虑怎么求方案,其实很简单啊。。。
就是
\[ans=\prod_{i=1}^m(n-\sum_{j=1}^{i-1}w_j)
\]
因为模数小,要用\(exLucas\)
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) return x = 1, y = 0, a;
ll res = exgcd(b, a % b, x, y), t;
t = x, x = y, y = t - a / b * y;
return res;
}
ll fpow(ll x, ll y, ll Mod) {
ll res = 1;
while (y) {
if (y & 1ll) res = res * x % Mod;
x = x * x % Mod;
y >>= 1ll;
}
return res;
}
ll fac(ll n, ll pi, ll pk) {
if (!n) return 1;
ll res = 1;
for (ll i = 2; i <= pk; i++)
if (i % pi) res = res * i % pk;
res = fpow(res, n / pk, pk);
for (ll i = 2; i <= n % pk; i++)
if (i % pi) res = res * i % pk;
return res * fac(n / pi, pi, pk) % pk;
}
ll inv(ll n, ll Mod) {
ll x, y;
exgcd(n, Mod, x, y);
return (x + Mod) % Mod;
}
ll CRT(ll b, ll p, ll Mod) { return b * inv(p / Mod, Mod) % p * (p / Mod) % p; }
ll C(ll n, ll m, ll pi, ll pk) {
ll fz = fac(n, pi, pk), fm1 = fac(m, pi, pk), fm2 = fac(n - m, pi, pk);
ll k = 0;
for (ll i = n; i; i /= pi) k += i / pi;
for (ll i = m; i; i /= pi) k -= i / pi;
for (ll i = n - m; i; i /= pi) k -= i / pi;
return fz * inv(fm1, pk) % pk * inv(fm2, pk) % pk * fpow(pi, k, pk) % pk;
}
ll exlucas(ll n, ll m, ll Mod) {
ll res = 0, tmp = Mod;
for (int i = 2; 1ll * i * i <= Mod; i++)
if (tmp % i == 0) {
ll pk = 1; while (tmp % i == 0) pk *= i, tmp /= i;
res = (res + CRT(C(n, m, i, pk), Mod, pk)) % Mod;
}
if (tmp > 1) res = (res + CRT(C(n, m, tmp, tmp), Mod, tmp)) % Mod;
return res;
}
ll N, M, Mod;
ll sum, w[10];
int main () {
cin >> Mod >> N >> M;
for (int i = 1; i <= M; i++) cin >> w[i], sum += w[i];
if (N < sum) return puts("Impossible") & 0;
ll ans = 1;
for (int i = 1; i <= M; i++) {
ans = ans * exlucas(N, w[i], Mod) % Mod;
N -= w[i];
}
printf("%lld\n", ans);
return 0;
}