Devu and Flowers

Devu and Flowers

题意

即给定$n$和$s$，表示有$n$个箱子要取$s$朵花，每个箱子表示一个种类的花，每个箱子中花的数量分别是

$f_{1} , f_{2} , \dots , f_{n}$，求出不同的选取方案

$1 \leq n \leq 20$

$0 \leq s \leq 10^{14}$

$\left.0 \leq f_{i} \leq 10^{12}\right)$

题解

多种集合的组合数

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N=30,mod=1e9+7;
 5 ll f[N];
 6 ll down;
 7 ll quick_pow(ll a, ll b){
 8     ll res=1;
 9     while(b){
10         if(b&1) res=res*a%mod;
11         a=a*a%mod;
12         b>>=1;
13     }
14     return res;
15 }
16 ll C(ll a,ll b){
17     if(a<b) return 0;
18     ll res=1;
19     for(ll i=a;i>a-b;i--) res=i%mod*res%mod;
20     return res*down%mod;
21 }
22 int main(){
23     ll n, s;
24     cin >> n >> s;
25     for (int i = 0; i < n; i ++ ) cin >> f[i];
26     down=1;
27     for (ll j = 1; j <= n - 1; j ++ ) down = j * down % mod;
28     down = quick_pow(down, mod - 2);
29 
30     int res = 0;
31     for(int i=0;i<1<<n;i++){
32         ll sign=1;
33         ll a=n+s-1;
34         for(int k=0;k<n;k++){
35             if(i>>k&1){
36                 sign*=(-1);
37                 a-=f[k]+1;
38             }
39         }
40         res = (res + C(a, n-1) * sign) % mod;
41     }
42     cout<<(res+mod)%mod<<endl;
43 }