CF932E

在此题中。

#include <algorithm>
#include <bitset>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <cmath>
#include <stack>
#include <set>
#include <queue>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define sz(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

const int mod=1e9+7;

ll powmod(ll a,ll b){
    ll res=1;
    for(;b>0;b>>=1,a=a*a%mod)if(b&1)res=res*a%mod;
    return res;
}

ll n,k,s[(int)5e3+9][(int)5e3+9];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>k;
    s[0][0]=1;
    for(int i=1;i<=k;++i){
        for(int j=1;j<=k;++j){
            s[i][j]=(s[i-1][j-1]+s[i-1][j]*j)%mod;
        }
    }
    ll xjm=1,ans=0;
    for(int i=0;i<=k;++i){
        ans=(ans+s[k][i]*xjm%mod*powmod(2ll,n-i))%mod,xjm=xjm*(n-i)%mod;
    }
    
    cout<<ans<<"\n";
    return 0;
}