codeforces 932e Team Work

最新推荐文章于 2024-03-06 21:32:15 发布

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最新推荐文章于 2024-03-06 21:32:15 发布

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分类专栏： dp 数学

本文链接：https://blog.youkuaiyun.com/C_K_Y_/article/details/79389162

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本文介绍了一种解决特定组合数学问题的方法，通过将问题转换为物品放入箱子的方案数问题，并利用动态规划进行求解。文章给出了详细的算法思路及C++实现代码。

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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
inputCopy
1 1
output
1
inputCopy
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.

题意：给定n,k,求

∑nr=1Crnrk

首先这个题可以转化为现在有x种东西，放到k个箱子中的方案数。

我们定义dp[i][j]表示放到第i个箱子，放了j种东西的方案数，那么dp[i][j]=(dp[i-1][j]*j+dp[i-1][j-1]*(n-j+1))%1000000007;

最后答案就等于

∑

min(n,k)i=1dp[k][i]×2n−i

因为我们已经确定了集合中i个数字，剩下的选与不选都无所谓，就是2^(n-i)了。

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll n,k,dp[2][5005],cur,ans;
ll pow(ll x,ll y)
{
    ll res=1,a=x;
    for(;y;y>>=1)
    {
        if(y&1) res=res*a%1000000007;
        a=a*a%1000000007;
    }
    return res;
}
int main()
{
    cin>>n>>k;
    dp[0][0]=1;
    for(int i=1;i<=k;i++)
    {
        cur=!cur;
        for(int j=1;j<=i;j++)
            dp[cur][j]=(dp[!cur][j]*j+dp[!cur][j-1]*(n-j+1))%1000000007;
        dp[0][0]=0;
    }
    for(int i=1;i<=min(n,k);i++)
        ans=(ans+dp[cur][i]*pow(2ll,n-i)%1000000007)%1000000007;
    printf("%I64d",ans);
}