說好的省選題怎麼就輕易地變成CF難度了呢
答案顯然是C(n,m)*D(n-m),D(x)爲x個元素的錯排
由容斥原理可得D(n)=n!*sigma(k=0->n)(-1)^k/k!
線性預處理一下就可以O(1)回答詢問了
(竟然沒卡帶log的逆元預處理)
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
#include<map>
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define per(i,r,l) for(int i=r;i>=l;i--)
#define mmt(a,v) memset(a,v,sizeof(a))
#define tra(i,u) for(int i=head[u];i;i=e[i].next)
using namespace std;
typedef long long ll;
const int p=(1e9)+7;
const int N=(1e6)+5;
ll qmul(ll a,ll b){
ll ans=1;
for(;b;b>>=1,a=a*a%p)if(b&1)ans=ans*a%p;
return ans;
}
ll inv(ll x){
return qmul(x,p-2);
}
ll f[N],g[N],s[N];
int main(){
//freopen("a.in","r",stdin);
f[0]=1;
rep(i,1,1000000)f[i]=f[i-1]*i%p;
g[1000000]=qmul(f[1000000],p-2);
per(i,999999,1)g[i]=g[i+1]*(i+1)%p;g[0]=g[1];
s[0]=1;
rep(i,1,1000000)s[i]=((s[i-1]+(i&1?-1:1)*g[i])%p+p)%p;
int T;scanf("%d",&T);
while(T--){
int n,m;scanf("%d%d",&n,&m);
printf("%lld\n",f[n]*g[m]%p*s[n-m]%p);
}
return 0;
}