题目大意
链接：CF932E
求(sumlimits_{i=0}^ninom nii^k)。
题目分析
[egin{split}
sum_{i=0}^ninom nii^k&=sum_{i=0}^ninom nisum_{j=0}^kinom ijegin{Bmatrix}k\jend{Bmatrix}j!\
&=sum_{j=0}^kegin{Bmatrix}k\jend{Bmatrix}j!sum_{i=0}^ninom niinom ij\
end{split}
]
根据常识，有
[inom niinom ij=inom njinom {n-j}{n-i}
]
可得
[egin{split}
sum_{j=0}^kegin{Bmatrix}k\jend{Bmatrix}j!sum_{i=0}^ninom niinom ij&=sum_{j=0}^kegin{Bmatrix}k\jend{Bmatrix}j!sum_{i=0}^ninom njinom {n-j}{n-i}\
&=sum_{j=0}^kegin{Bmatrix}k\jend{Bmatrix}j!inom njsum_{i=0}^ninom {n-j}{n-i}\
&=sum_{j=0}^kegin{Bmatrix}k\jend{Bmatrix}j!inom njsum_{i=0}^{n-j}inom {n-j}i\
&=sum_{j=0}^kegin{Bmatrix}k\jend{Bmatrix}j!inom njcdot 2^{n-j}
end{split}
]
代码实现
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#define MAXN 0x7fffffff
typedef long long LL;
const int N=5005,mod=1e9+7;
using namespace std;
inline int Getint(){register int x=0,f=1;register char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}return x*f;}
int s[N][N];
int ksm(int x,int k){
int ret=1;
while(k){
if(k&1)ret=(LL)ret*x%mod;
x=(LL)x*x%mod,k>>=1;
}
return ret;
}
int fac[N],inv[N];
int C(int n,int m){
if(n<m)return 0;
return (LL)fac[m]*inv[m]%mod;
}
int main(){
int n=Getint(),K=Getint();
s[0][0]=1;
for(int i=1;i<N;i++)
for(int j=1;j<N;j++)
s[i][j]=((LL)s[i-1][j]*j+s[i-1][j-1])%mod;
fac[0]=1;for(int i=1;i<=K;i++)fac[i]=(LL)fac[i-1]*(n-i+1)%mod;
inv[0]=1;for(int i=1;i<=K;i++)inv[i]=(LL)inv[i-1]*ksm(i,mod-2)%mod;
int ans=0;
for(int i=0,t=1,lim=min(n,K);i<=lim;i++,t=(LL)t*i%mod){
ans=(ans+(LL)s[K][i]*t%mod*C(n,i)%mod*ksm(2,n-i)%mod)%mod;
}
cout<<(ans+mod)%mod;
return 0;
}
P.S
升级版：【BZOJ5093】图的价值