FZU 2282 Wand（错排+费马小定理）

 Problem 2282 Wand

Accept: 68    Submit: 250

Time Limit: 1000 mSec    Memory Limit : 262144 KB

 Problem Description

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

21 13 1

 Sample Output

14

 Source

第八届福建省大学生程序设计竞赛-重现赛（感谢承办方厦门理工学院）

题意：
给出n个人，每个人有自己的膜杖，求出只要有k个人拿到自己的膜杖的数量数。

POINT：
错排+组合数，组合数没有很大，只要用费马小定理就行了。

如果当时看过错排就可以AC了呀！！哎！

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 10000+2;
#define  LL long long
const LL p = 1e9+7;
LL cp
;
LL k
;
void init()
{
cp[1]=0;
cp[2]=1;
for(int i=3;i<=10000;i++)
{
(cp[i]=(i-1)*(cp[i-1]+cp[i-2]))%=p;
}
k[0]=1;
for(int i=1;i<=10000;i++)
{
(k[i]=k[i-1]*i)%=p;
}
}
LL qkm(LL base,LL mi)
{
LL ans=1;
while(mi)
{
if(mi&1)
(ans*=base)%=p;
(base*=base)%=p;
mi>>=1;
}
return ans;
}
LL C(int i,int n)
{
return ((k
*qkm((k[n-i]*k[i])%p,p-2))%p);//取模，两次！很关键k[n-i]*k[i]也会爆LL
}
int main()
{
int n,k;
int T;
init();
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&n,&k);
LL ans=1;
for(int i=2;i<=n-k;i++)
{
(ans+=cp[i]*C(i,n))%=p;
}
printf("%lld\n",ans);
}
}