CF932 E. Team Work 结题报告

CF932 E. Team Work

题意

求

\[\sum_{i=0}^n\binom{n}{i}i^k \]

其中\(n\le 10^9,k\le 5000\)，对\(mod=998244353\)取模

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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

\[x^n=\sum_{i=1}^x\binom{x}{i}{n \brace i}i! \]

然后带进去推一波式子就完事了

\[\begin{aligned} \sum_{i=0}^n\binom{n}{i}i^k&=\sum_{i=0}^n\binom{n}{i}\sum_{j=1}^k\binom{i}{j}{k\brace j}j!\\ &=\sum_{j=1}^k{k \brace j}\sum_{i=0}^n\binom{n}{i}\binom{i}{j}j!\\ &=\sum_{j=1}^k{k \brace j}\sum_{i=0}^n\frac{n!}{(n-i)!(i-j)!}\\ &=\sum_{j=1}^k{k \brace j}\sum_{i=0}^n\frac{n!(n-j)!}{(n-i)!(i-j)!(n-j)!}\\ &=\sum_{j=1}^k{k \brace j}n^{\underline j}\sum_{i=0}^n\binom{n-j}{n-i}\\ &=\sum_{j=1}^k{k \brace j}n^{\underline j}2^{n-j} \end{aligned} \]

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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\n",ans);
	return 0;
}

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2019.3.27

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