fzu 2282（错位排列+逆元+快速幂函数）

 Problem 2282 Wand

Accept: 30    Submit: 94

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

2
1 1
3 1

 Sample Output

1
4
题意：给你一个n代表编号为1-n的有序的n个人，一个k，重新排序之后，询问至少有k人在自己原来位置的方案数。
思路：首先n个人排序，每个人都不在自己原来位置的可能有F(n)=(i-1)*(F(n-1)+F(n-2)) （n>=2） 特殊的，F(0)=1，F(1)=0;
然后考虑每次错排的组合数：C（n，i）= i！/（i！*（n-i）！），对于（A/B）mod C，找到B的逆元b=B^-1，求出（A*b）%C即可。
由费马小定理：B
关于 P 的逆元为  B^(p-2);

费马小定理(Fermat Theory)是数论中的一个重要定理，其内容为：
假如p是质数，且gcd(a,p)=1，那么 a(p-1)≡1（mod p）。即：假如a是整数，p是质数，且a,p互质(即两者只有一个公约数1)，那么a的(p-1)次方除以p的余数恒等于1。所以，a^-1*a=1=a^(p-1),所以：a^-1=a^(p-2);

代码：

#include<cstdio>
#include<vector>
#include<iostream>
const int mod = 1e9 + 7;
const int maxn = 10005;
typedef long long LL;
using namespace std;
LL dp[maxn];
LL inv[maxn];
int n,k;
LL pri[maxn];
LL ni[maxn];
LL pow(LL a,int b)
{
LL ans=1,base=a;
while (b>0)
{
if (b%2==1)
ans=(base*ans)%mod;
base=(base*base)%mod;
b/=2;
}
return ans;
}
void s()   //打表
{
pri[0]=1;
ni[0]=1;
for (int i=1;i<=maxn ;i++)
{
pri[i]=pri[i-1]*i%mod;  //N！
ni[i]=pow(pri[i],mod-2);
}
}
int main()
{
s();
int T;
scanf("%d", &T);
dp[0]=1,dp[1] = 0;
for(int i = 2; i <= maxn; i++)
{
dp[i] = ((i - 1) * (dp[i - 1] + dp[i - 2])) % mod;
}
while(T--)
{
LL ans=1;
scanf("%d%d", &n, &k);
int cnt = n - k;
if(cnt==0)
{
puts("1");
continue;
}
if(cnt==1)
{
puts("0");
continue;
}
for(int i=2;i<=cnt;i++)
{
LL sum=((pri
*ni[i]%mod)*ni[n-i])%mod;
//printf("%lld\n",sum);
ans+=(sum*dp[i])%mod;
}
printf("%lld\n",ans%mod);
}
return 0;
}