【FZU

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

公式很明显，就是这个错排不好弄。
错排是有公式的 cp[i]=(i-1)*(cp[i-1]+cp[i-2]) ;
因为组合中 有除法，所以要求逆元，这里我们可以用费马小定理 求逆元。
代码

#include<cstring>#include<cstdio>#include<algorithm>using namespace std;  typedef long long LL ;const int MAXN = 1e4+10;const LL  mod  = 1e9+7;LL cp[MAXN]={0,0,1},k[MAXN]={1,1};void init(){    for(LL i=3;i<MAXN;i++)  //  错排         cp[i]=(i-1)*(cp[i-1]+cp[i-2])%mod;     for(LL i=2;i<MAXN;i++)         k[i]=k[i-1]*i%mod;}LL qp(LL a,LL b,LL c){    LL base=a%c,s=1;    while(b){        if(b&1)  s=s*base%c;        base=base*base%c;        b>>=1;    }    return s%c;}LL getC(LL n,LL m){    return (( k[n]*qp(k[m]*k[n-m],mod-2,mod))%mod);}int main(){    init();    int t;scanf("%d",&t);    while(t--){        int n,K;scanf("%d%d",&n,&K);        LL ans=0;        for(LL i=0;i<K;i++)            ans=(ans%mod+(getC((LL)n,i)*cp[n-i])%mod)%mod;         printf("%lld\n",(mod+k[n]%mod-(ans%mod))%mod);// 注意这里要 加上一个mod ，因为可能 是负的      }     return 0;}