bzoj3622 ( 容斥，二项式反演)

https://vjudge.d0j1a1701.cc/problem/黑暗爆炸-3622

f(i)表示钦定i个人没有物品，f(i)= $$\left(\begin{array}{c} n \\ i \end{array}\right) \prod_{j=1}^{m}\left(\begin{array}{c} n-i+a_{j}-1 \\ a_{j} \end{array}\right) 。$$

#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false) ,cin.tie(0), cout.tie(0);
//#pragma GCC optimize(3,"Ofast","inline")
#define ll long long
//#define int long long
const int N = 2e3 + 5;
const int M = 1e3 + 5;
const int INF = 0x3f3f3f3f;
const ll LNF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-13;
const int MATN = 102;
ll F[N];ll inv[N];
ll qmi(ll m, ll k) {
ll res = 1;
while(k) {
if(k & 1) res = res * m % mod;
m = m * m % mod;
k >>= 1;
}
return res;
}
void init(int n){
F[0]=inv[0]= 1;
for(int i=1;i<=n;i++)F[i]=F[i-1]*i%mod;//求阶乘
inv[n]=qmi(F[n],mod - 2);//随便求一下n的逆元
for(int i=n-1;i>=1;i--)inv[i]=inv[i+1]*(i+1)%mod;//由最后一个往前推
}
ll C(ll n, ll m) {
return F[n] * inv[m] % mod * inv[n - m] % mod;
}
ll num[1005];
int main() {
IOS
init(2000);
int n, m; cin >> n >> m;
for ( int i = 1; i <= m; ++ i ) {
cin >> num[i];
}
ll ans = 0; ll t = -1;
for ( int j = 0; j <= n; ++ j ) {
if(j & 1) t = -1; else t = 1;
ll tmp = t * C(n, j) % mod;
for ( int i = 1; i <= m; ++ i ) {
tmp = tmp * C(num[i] + n - j - 1, num[i]) % mod;
}
ans = (ans + tmp) % mod;
}
cout << (ans + mod) % mod << '\n';
return 0;
}


posted @ 2022-10-17 17:40  qingyanng  阅读(41)  评论(0编辑  收藏  举报