Codeforces 932E - Team Work

 

Problem Link:

 

http://codeforces.com/problemset/problem/932/E

 

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Problem Statement:

 

E. Team Work

 

time limit per test: 2 seconds

memory limit per test: 256 megabytes

input: standard input

output: standard output

 

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 x^k.

 

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.

 

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Analysis:

 

At the first sight, the cost of having x people for the task seems very intriguing: x^k. To help make it more understandable, let us give the cost a combinatorial meaning: the ways of choosing k elements (repetitions are allowed) from a set of size x. In order to avoid duplicate counting, when we say we choose k elements from a set of size x, we must use exactly every element of the set at least once.

 

How do we calculate the number of ways of choosing i from j? We can either choose an element that has already appeared in choosing i - 1 from j, in which case we have j options, or choose an element that has not appeared in choosing i - 1 from j - 1, in which case we have n - j + 1 options.

 

Then, to calculate the sum of costs for all non empty subsets, we need to count the number of subsets in which a specific combination appears. Assume we choose k people with x distinct people used, then there are n - x people unused. These people can either belong to the set or not, so the total number of subsets from which we can obtain the current combination, i.e. a specific cost, equals the product of the number of ways of choosing k from x (as stated above) and 2^{n - x}.

 

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The DP Equations:

 

dp[i][j] = dp[i - 1][j] * j + dp[i - 1][j - 1] * (n - j + 1)

 

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Time Complexity:

 

O(k²)

 

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AC Code:

 

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <memory.h>
#include <functional>
#include <string.h>
#include <iomanip>
#include <unordered_set>
#include <unordered_map>
using namespace std;

typedef long long ll;
typedef pair<int,int> pi; 
typedef double db;

const int mod=1e9+7;
ll qpow(ll x,ll k){return k==0?1:1ll*qpow(1ll*x*x%mod,k>>1)*(k&1?x:1)%mod;}
#define MP make_pair
#define FF first
#define SS second
#define LB lower_bound
#define UB upper_bound
#define PB push_back
#define lc ((p<<1)+1)
#define rc ((p<<1)+2)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rrep(i,b,a) for(int i=b;i>=a;i--)
#define all(v) (v).begin(),(v).end()
#define clean(v,a) memset(v,a,sizeof(v))
#define get(a) scanf("%d",&a)
#define get2(a,b) scanf("%d%d",&a,&b)
#define get3(a,b,c) scanf("%d%d%d",&a,&b,&c)

/*
 * high precision: printf("%.12lf",(db)ans);
 *
 * clear output buffer: fflush(stdout);
 *
 * sync off: ios::sync_with_stdio(false);
 *
 */

const int md=1e9+7;
const int Maxn=5005;

int n,k;
ll dp[Maxn][Maxn];//dp[i][j]: the ways of choosing a combination of size i with repetition among j distinct people (must choose every people at least once)

int main()
{
    scanf("%d%d",&n,&k);
    dp[0][0]=1;
    for(int i=1;i<=k;i++)
    {
        for(int j=1;j<=i;j++)
        {
            dp[i][j]=(dp[i-1][j]*j+dp[i-1][j-1]*(n-j+1))%md;
        }
    }
    ll ans=0;
    for(int i=1;i<=min(n,k);i++)
    {
        ans=(ans+dp[k][i]*qpow(2,n-i))%md;
    }
    printf("%lld\n",ans);
    return 0;
}