这题不就是个组合数加上错排公式吗?
数论稍稍学过一点的人都会啊.
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<string>
#include<iomanip>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
#define FILE "dealing"
#define up(i,j,n) for(int i=j;i<=n;++i)
#define db double
#define eps 1e-10
#define pii pair<int,int>
int read(){
int x=0,f=1,ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return f*x;
}
const int maxn=1000200,mod=(int)(1e9+7+0.1),limit=(int)(1e6+1);
LL fac[maxn],inv[maxn],f[maxn];
LL C(int n,int m){
return fac[n]*inv[m]%mod*inv[n-m]%mod;
}
int main(){
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
int T=read();
fac[0]=1;
up(i,1,limit)fac[i]=fac[i-1]*i%mod;
inv[0]=inv[1]=1;
up(i,2,limit)inv[i]=(-mod/i)*inv[mod%i]%mod;
up(i,1,limit)inv[i]=inv[i]*inv[i-1]%mod;
f[0]=1,f[1]=0,f[2]=1;
up(i,3,limit)f[i]=((LL)i-1)*((f[i-1]+f[i-2])%mod)%mod;
while(T--){
int n=read(),m=read();
printf("%lld
",(C(n,m)*f[n-m]%mod+mod)%mod);
}
return 0;
}