Discription
You have a team of N people. For a particular task, you can pick any non-empty subset of people. The cost of having x people for the task is xk.
Output the sum of costs over all non-empty subsets of people.
Input
Only line of input contains two integers N (1?≤?N?≤?109) representing total number of people and k (1?≤?k?≤?5000).
Output
Output the sum of costs for all non empty subsets modulo 109?+?7.
Example
1 1
1
3 2
24
Note
In the first example, there is only one non-empty subset {1} with cost 11?=?1.
In the second example, there are seven non-empty subsets.
- {1} with cost 12?=?1
- {2} with cost 12?=?1
- {1,?2} with cost 22?=?4
- {3} with cost 12?=?1
- {1,?3} with cost 22?=?4
- {2,?3} with cost 22?=?4
- {1,?2,?3} with cost 32?=?9
The total cost is 1?+?1?+?4?+?1?+?4?+?4?+?9?=?24.
题目大意就是要你求一下ΣC(n,i)*i^k。
然后直接上我推的式子了(就是用第二类斯特林数代换一下)
(怎么这个图这么大。。。。不管了)
然后就开开心心A了。
不过好像k再大一点也可以做，，，就是要用FFT 在N log N的时间求出某一行的斯特林数了(反正我也不会hhhh)，不能再N^2递推斯特林数了。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#define ll long long
#define maxn 5005
using namespace std;
const int ha=1000000007;
const int inv=ha/2+1;
int S[maxn],n,m;
int ans=0,tmp,ci;
inline void init(){
S[1]=1;
for(int i=2;i<=m;i++)
for(int j=i;j;j--){
S[j]=(S[j]*(ll)j+(ll)S[j-1])%ha;
}
}
inline int ksm(int x,int y){
int an=1;
for(;y;y>>=1,x=x*(ll)x%ha) if(y&1) an=an*(ll)x%ha;
return an;
}
inline void solve(){
tmp=1,ci=ksm(2,n);
for(int i=1;i<=m;i++){
tmp=tmp*(ll)(n-i+1)%ha;
ci=ci*(ll)inv%ha;
ans=((ll)ans+S[i]*(ll)tmp%ha*(ll)ci)%ha;
}
}
int main(){
scanf("%d%d",&n,&m);
init();
solve();
printf("%d\n",ans);
return 0;
}
