Problem:
Balls and Boxes 9

Balls

Boxes

Any way

At most one ball

At least one ball

Distinguishable

Distinguishable

1

2

3

Indistinguishable

Distinguishable

4

5

6

Distinguishable

Indistinguishable

7

8

9

Indistinguishable

Indistinguishable

10

11

12

Problem

You have $n$ balls and $k$ boxes. You want to put these balls into the boxes.

Find the number of ways to put the balls under the following conditions:

Each ball is distinguished from the other.

Each box is not distinguished from the other.

Each ball can go into only one box and no one remains outside of the boxes.

Each box must contain at least one ball.

Note that you must print this count modulo $10^9+7$.

Input

$n$ $k$

The first line will contain two integers $n$ and $k$.

Output

Print the number of ways modulo $10^9+7$ in a line.

Constraints

$1 \le n \le 1000$

$1 \le k \le 1000$

Sample Input 1

4 3

Sample Output 1

6

Sample Input 2

10 5

Sample Output 2

42525

Sample Input 3

100 30

Sample Output 3

203169470

Solution in JAVA (judge verdict: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.util.*;
 
public class Main implements Runnable {
	
    public static void main(String[] args) {
    	new Thread(null, new Main(), "", 16 * 1024 * 1024).start();
    }
    
    public void run() {
        FastScanner sc = new FastScanner();
        
        int n = sc.nextInt();
        int k = sc.nextInt();
        
        Combinatrics c = new Combinatrics(k);

    	System.out.println(c.stirling(n,k));

    }
}



class Combinatrics {
	
	static long mod = 1000000007; //逆元を取るので素数前提
	static long[] fact; //階乗のテーブル
	static long[] inv;	//逆元のテーブル
	static long[] finv; //階乗の逆元のテーブル
	
	public Combinatrics(int n){
		int N = Math.max(2,n+1);
		fact = new long[N];
		inv = new long[N];
		finv = new long[N];
	    fact[0] = fact[1] = 1;
	    finv[0] = finv[1] = 1;
	    inv[1] = 1;
	    for(int i=2;i<n+1; i++){
	        fact[i] = fact[i-1] * i % mod;
	        inv[i] = mod - inv[(int)mod%i] * (mod / i) % mod;
	        finv[i] = finv[i-1] * inv[i] % mod;
	    }
	}
	
	//逆元
	int inv(int n){
		return (int)inv[n];
	}
	
	//n!
	int factorial(int n){
		return (int)fact[n];
	}
	
	//n!の逆元
	int factInv(int n){
		return (int)finv[n];
	}
	
	//nCr（n個からr個選ぶ）
    int combination(int n, int r) {
        if(n<r|| r<0){
        	return 0;
        }
        return (int)(fact[n]*finv[r]%mod*finv[n-r]%mod);
    }

    //nPr (=n*(n-1)*...*(n-r+1))（n個からr個選んで並べる）
    int permutation(int n, int r) {
        if(n<r || r<0){
        	return 0;
        }
        return (int)(fact[n]*finv[n-r]%mod);
    }

    //nHr (=n+r-1Cr)（n個から重複を許してk個とる）
    //（テーブルがn+r-1まで必要な事に注意！）
    int homogeneous(int n, int r) {
        if(n==0 && r==0){
        	return 1;
        }
        return combination(n+r-1,r);
    }
    
    int surjection(int n, int k){
    	long ans = 0;
    	int sign = 1; //符号
    	if(k%2 == 0){
    		sign = -1;
    	}
    	for(int i=1;i<=k;i++){
    		if(sign==1){
    			ans += (long)combination(k,i)*iterationPow(i,n);
    			ans %= mod;
    		}
    		else{
    			ans -= (long)combination(k,i)*iterationPow(i,n);
    			ans %= mod;
    		}
    		sign *= -1;
    	}
    	ans = mod(ans);
    	return (int)ans;
    }
	//x^nをmodのもとで繰り返し2乗法で求める。
	static long iterationPow(long x, long n){
	    long sum = 1;
	    while(n>0){
	        if((n&1)== 1){  //ビットが立っていたら
	            sum *= x;        //x^2^k1 * x^2^k2 * x^2^k3 * ・・・
	            sum %= mod;
	        }
	        x *= x;              //((x^2)^2)^2・・・
	        x %= mod;
	        n >>= 1;
	    }
	    return sum;
	}
	//負の数に対応したmod
	static long mod(long a) {
		return (a % mod + mod) % mod;
	}
	
	//スターリング数（テーブルはkまで必要）
	//n個の区別可能なボールをちょうどk個の区別不可能な箱に入れる組合せの数
	int stirling(int n, int k){
		//全射の個数をk!で割ったものに等しい
		return (int) ((long)surjection(n,k)*(long)factInv(k) % mod);
	}

}

class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;
	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}
	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}
	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}
	public boolean hasNext() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
		return hasNextByte();
	}
	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
			throw new NumberFormatException();
		return (int) nl;
	}
	public int[] nextIntArray(int n){
		int[] a = new int[n];
		for(int i=0;i<n;i++){
			a[i] = nextInt();
		}
		return a;
	}
	public Integer[] nextIntegerArray(int n){
		Integer[] a = new Integer[n];
		for(int i=0;i<n;i++){
			a[i] = nextInt();
		}
		return a;
	}
	public double nextDouble() {
		return Double.parseDouble(next());
	}
}