Wand FZU - 2282 (组合数 + 错排公式)

N wizards are attending a meeting. Everyone has his own magic wand. N magic wands was put in a line, numbered from 1 to n(Wand_i owned by wizard_i). After the meeting, n wizards will take a wand one by one in the order of 1 to n. A boring wizard decided to reorder the wands. He is wondering how many ways to reorder the wands so that at least k wizards can get his own wand.

For example, n=3. Initially, the wands are w1 w2 w3. After reordering, the wands become w2 w1 w3. So, wizard 1 will take w2, wizard 2 will take w1, wizard 3 will take w3, only wizard 3 get his own wand.

Input

First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.

For each test case: Two number n and k.

1<=n <=10000.1<=k<=100. k<=n.

Output

For each test case, output the answer mod 1000000007(10^9 + 7).

Sample Input

2
1 1
3 1

Sample Output

1
4

题意：t 组数据，输入一个 n，k， n个人分配东西，问至少k个人分配到属于自己的编号的方案数。 

思路：组合数 + 错排，中间要用到求逆元；n个数字里要求至少k个数字位置不变，其余进行错排的方案数，容易想到不变的数字从k枚举到n，每次取i(k <= i < n)个出来，对剩下的n-i个进行错排；

注意输入 n k，为 long long 时，一定要用 Visual C++

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define mod 1000000007
#define ll long long
#define Max 10010

ll a[Max+10],fic[Max+10];

void fff()
{
	//int i;
	a[0] = 1;
	a[1] = 0;
	a[2] = 1;
	fic[1] = 1;
	fic[2] = 2;
	for(ll i = 3;i<Max;i++)
	{
		a[i] = ((i-1)*a[i-1]%mod + (i-1)*a[i-2]%mod)%mod;
		fic[i] = (fic[i-1] * i)%mod; 
	}
		
}

ll p_pow(ll a,ll b)
{
	ll res = 1;
	while(b)
	{
		if(b&1) res = res*a%mod;
		b>>=1;
		a = a*a%mod;
	}
	return res;
}

ll C(ll n,ll m)
{
	if(m>n) return 0;
	if(m==0) return 1;
	if(m>n-m)
		m = n-m;
	ll sum = 1;
	ll sum1 = 1;
	ll k = n;
	for(ll i = 1;i<=m;i++)
	{
		sum = sum*k%mod;
		k--;
		sum1 = sum1 * i %mod;     // 因为 mod 是个素数 所以分母可以先取余，再求逆元； 
	}	
	return sum*p_pow(sum1,mod-2)%mod;	
}
int main()
{
	ll n,m;
	fff();
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld",&n,&m);
		ll res = 0;
		for(ll i = 0;i<m;i++)
		{
			res = (res + C(n,i)*a[n-i]%mod) %mod;
		}
		printf("%lld\n",(fic[n] - res + mod)%mod );
	}
	return 0;
}