cf932E(第二类斯特林数)

题目链接

https://codeforces.com/contest/932/problem/E

题意

题解

发现 $n$ 和 $k$ 的差距很大,所以要尽可能去掉 $n$ ,利用斯特林数的性质,得

$k$ 那么小直接 $k^2$ 暴力预处理就够了




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<complex>
#include<cstdlib>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define mid (x+y>>1)
#define sqr(x) ((x)*(x))
#define NM 140005
#define nm 400005
const double pi=acos(-1);
const ll inf=1e9+7;
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}


int n,m;
ll d[NM],ans;

ll qpow(ll x,ll t){
ll s=1;
for(;t;t>>=1,x=sqr(x)%inf)if(t&1)s=s*x%inf;
return s;
}

int main(){
n=read();m=read();
d[1]=1;
inc(i,2,m)dec(j,i,1)d[j]=(j*d[j]+d[j-1])%inf;
ll t=1;
m=min(n,m);
inc(i,1,m){
t=t*(n-i+1)%inf;
ans+=d[i]*qpow(2,n-i)%inf*t%inf;
ans%=inf;
}
return 0*printf("%lld\n",ans);
}