CF451E Devu and Flowers
題意
Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All flowers in a single box are of the same color (hence they are indistinguishable). Also, no two boxes have flowers of the same color.
Now Devu wants to select exactly s flowers from the boxes to decorate his garden. Devu would like to know, in how many different ways can he select the flowers from each box? Since this number may be very large, he asks you to find the number modulo (109?+?7).
Devu considers two ways different if there is at least one box from which different number of flowers are selected in these two ways.
The first line of input contains two space-separated integers n and s (1?≤?n?≤?20, 0?≤?s?≤?1014).
The second line contains n space-separated integers f1,?f2,?... fn (0?≤?fi?≤?1012).
Output a single integer — the number of ways in which Devu can select the flowers modulo (109?+?7).
2 3Output
1 3
2Input
2 4Output
2 2
1Input
3 5Output
1 3 2
3
Sample 1. There are two ways of selecting 3 flowers: {1,?2} and {0,?3}.
Sample 2. There is only one way of selecting 4 flowers: {2,?2}.
Sample 3. There are three ways of selecting 5 flowers: {1,?2,?2}, {0,?3,?2}, and {1,?3,?1}.
分析
多重集組合問題。根據結論，答案為
\[
\binom{N+M-1}{N-1}-\sum_{i=1}^N \binom{N+M-A_i-2}{N-1} + \sum_{1 \le i < J \le N} \binom{N+M-A_i-A_j-3}{N-1}-\dots+(-1)^N \binom{N+M-\sum_{i=1}^N A_i -(N+1)}{N-1}
\]
由於N很小，所以直接枚舉子集即可。算組合數的時候用for循環算。時間復雜度\(O(2^N N)\)
代碼
#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
rg T data=0,w=1;
rg char ch=getchar();
while(!isdigit(ch)){
if(ch=='-') w=-1;
ch=getchar();
}
while(isdigit(ch))
data=data*10+ch-'0',ch=getchar();
return data*w;
}
template<class T>il T read(rg T&x){
return x=read<T>();
}
typedef long long ll;
co int mod=1e9+7;
ll a[22],m,ans=0;
int inv[22],n;
int power(int a,int b){
int c=1;
for(;b;b>>=1){
if(b&1) c=(ll)c*a%mod;
a=(ll)a*a%mod;
}
return c;
}
int C(ll y,int x){
if(y<0||x<0||y<x) return 0;
y%=mod;
if(y==0||x==0) return 1;
int ans=1;
for(int i=0;i<x;++i)
ans=(ll)ans*(y-i)%mod;
for(int i=1;i<=x;++i)
ans=(ll)ans*inv[i]%mod;
return ans;
}
int main(){
// freopen(".in","r",stdin),freopen(".out","w",stdout);
for(int i=1;i<=20;++i) inv[i]=power(i,mod-2);
read(n),read(m);
for(int i=0;i<n;++i) read(a[i]);
for(int x=0;x<1<<n;++x){
ll t=n+m;
int p=0;
for(int i=0;i<n;++i)
if(x>>i&1) ++p,t-=a[i];
t-=p+1;
if(p&1) ans=(ans-C(t,n-1))%mod;
else ans=(ans+C(t,n-1))%mod;
}
printf("%lld\n",(ans+mod)%mod);
return 0;
}CF451E Devu and Flowers