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  • Codeforces 932E Team work 【组合计数+斯特林数】

    Codeforces 932E Team work


    You have a team of N people. For a particular task, you can pick any non-empty subset of people. The cost of having x people for the task is xk.

    Output the sum of costs over all non-empty subsets of people.

    Input

    Only line of input contains two integers N (1 ≤ N ≤ 109) representing total number of people and k (1 ≤ k ≤ 5000).

    Output

    Output the sum of costs for all non empty subsets modulo 109 + 7.

    Examples

    input

    1 1

    output

    1

    input

    3 2

    output

    24

    Note

    In the first example, there is only one non-empty subset {1} with cost 11 = 1.
    In the second example, there are seven non-empty subsets.
    - {1} with cost 12 = 1
    - {2} with cost 12 = 1
    - {1, 2} with cost 22 = 4
    - {3} with cost 12 = 1
    - {1, 3} with cost 22 = 4
    - {2, 3} with cost 22 = 4
    - {1, 2, 3} with cost 32 = 9
    The total cost is 1 + 1 + 4 + 1 + 4 + 4 + 9 = 24.


     

    #include<bits/stdc++.h>
    using namespace std;
    #define N 5010
    #define yyf 1000000007
    #define LL long long
    LL S[N][N],inv[N],C[N],J[N];
    LL n,k;
    LL fast_pow(LL a,LL b){
        LL ans=1;
        while(b){
            if(b&1)ans=ans*a%yyf;
            b>>=1;
            a=a*a%yyf;
        }
        return ans;
    }
    int main(){
        cin>>n>>k;
        inv[0]=inv[1]=1;C[1]=n;J[1]=1;
        for(LL i=2;i<=k;i++)J[i]=J[i-1]*i%yyf;
        for(LL i=2;i<=k;i++)inv[i]=(yyf-yyf/i)*inv[yyf%i]%yyf;
        for(LL i=2;i<=k;i++)C[i]=C[i-1]*inv[i]%yyf*(n-i+1)%yyf;
        S[0][0]=1;
        for(LL i=1;i<=k;i++){
            S[i][0]=0;
            for(LL j=1;j<=i;j++)S[i][j]=(j*S[i-1][j]%yyf+S[i-1][j-1])%yyf;
        }
        LL ans=0;
        for(LL i=1;i<=min(k,n);i++)ans=(ans+S[k][i]*J[i]%yyf*C[i]%yyf*fast_pow(2,n-i)%yyf)%yyf;
        printf("%lld",ans%yyf);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dream-maker-yk/p/9676324.html
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