ICM Technex 2018 and Codeforces Round #463 E. Team Work__codeforces

Given N, K, seek ∑c (n,i) *i^k.

Ideas:

1. Derivative method: (1+x) ^n, continuous derivation. When the game thought of this approach, but no pen in hand without grass manuscript, brain pushing the pain, so ... Gave up.

2.DP:DP[I][J] Represents the number of items in the J box where I put something, and every box has something. The conversion of the equation to: ∑dp[k][i]*2^ (n-i)

Personal Appreciation Idea 2, from: http://blog.csdn.net/qq_36797743/article/details/79332455

AC Code:

#include <bits/stdc++.h>
#define R (N) scanf ("%d", &n)
#define RLL (n) scanf ("%lld", &n)
# Define MEM (A,b) memset (A,b,sizeof (a))
using namespace std;

typedef long long LL;
const LL MOD=1E9+7;
const int n=5001;

ll Dp[n][n];
ll Pow (ll X,ll y) {
	if (y<0) return 0;
	ll Ans=1;
	while (y) {
		if (y&1) ans= (ans*x)%mod;
		x= (x*x)%mod;
		y>>=1;
	}
	return ans;
}
int main () {
//	C (n,i) *i^k => dp[i][j]*2^ (n-j) => dp[i][j]=dp[i-1][j]*j+dp[i-1][j-1]* (n-j+1)
	int n,k;
	R (N); R (k);
	Dp[0][0]=1;
	for (int i=1;i<=k;i++) {for
		(int j=1;j<=min (n,i); j + +) {
			dp[i][j]=dp[i-1][j]*j%mod+dp[i-1][j-1]* (n-j +1)%mod;
			Dp[i][j]%=mod
		}
	}
	ll Ans=0;
	for (int i=1;i<=min (n,k); i++) ans= (Ans+dp[k][i]*pow (2LL, (LL) n-i)%mod)%mod;
	printf ("%i64d\n", ans);
	return 0;
}