CF932 E. Team Work 结题报告

CF932 E. Team Work

题意

求

[sum_{i=0}^ninom{n}{i}i^k ]

其中(nle 10^9,kle 5000)，对(mod=998244353)取模

---

事实证明我斯特林数学到狗身上去了...

关于斯特林数的一个常用公式是

[x^n=sum_{i=1}^xinom{x}{i}{n race i}i! ]

然后带进去推一波式子就完事了

[egin{aligned} sum_{i=0}^ninom{n}{i}i&=sum_{i=0}^ninom{n}{i}sum_{j=1}^kinom{i}{j}{krace j}j!\ &=sum_{j=1}^k{k race j}sum_{i=0}^ninom{n}{i}inom{i}{j}j!\ &=sum_{j=1}^k{k race j}sum_{i=0}^nfrac{n!}{(n-i)!(i-j)!}\ &=sum_{j=1}^k{k race j}sum_{i=0}^nfrac{n!(n-j)!}{(n-i)!(i-j)!(n-j)!}\ &=sum_{j=1}^k{k race j}n^{underline j}sum_{i=0}^ninom{n-j}{n-i}\ &=sum_{j=1}^k{k race j}n^{underline j}2^{n-j} end{aligned} ]

---

Code:

#include <cstdio>
const int mod=1e9+7;
inline int add(int x,int y){return x+y>=mod?x+y-mod:x+y;}
#define mul(x,y) (1ll*(x)*(y)%mod)
inline int qp(int d,int k){int f=1;while(k){if(k&1)f=mul(f,d);d=mul(d,d),k>>=1;}return f;}
int str[5010][5010],n,k,ans;
int main()
{
	scanf("%d%d",&n,&k);
	str[0][0]=1;
	for(int i=1;i<=k;i++)
		for(int j=1;j<=i;j++)
			str[i][j]=add(str[i-1][j-1],mul(str[i-1][j],j));
	for(int i=1,f=1;i<=n&&i<=k;i++)
	{
		f=mul(f,n-i+1);
		ans=add(ans,mul(f,mul(str[k][i],qp(2,n-i))));
	}
	printf("%d
",ans);
	return 0;
}

---

2019.3.27

---