C++11(clang++ 3.9) 解法, 执行用时: 367ms, 内存消耗: 36480K, 提交时间: 2020-10-09 21:44:37

#include<iostream>
#include<cstdio>
using namespace std;
const int M=1000000;
const int MOD=1e9+7;
long long t,n,m,d[M+1],jc[M+1],inv[M+1],invjc[M+1];
void pre(){
	d[0]=invjc[0]=invjc[1]=inv[1]=d[2]=jc[0]=jc[1]=1;
	for(register int i=2;i<=M;i++){
		inv[i]=((MOD-MOD/i)*inv[MOD%i])%MOD;
		jc[i]=(i*jc[i-1])%MOD;
		invjc[i]=(invjc[i-1]*inv[i])%MOD;
	}
	for(register int i=3;i<=M;i++){
		d[i]=(i-1)*(d[i-1]+d[i-2])%MOD;
	}
}
int main()
{
	pre();
	scanf("%lld",&t);
	while(t--){
		scanf("%lld%lld",&n,&m);
		printf("%lld\n",(((jc[n]*invjc[m]%MOD)*invjc[n-m]%MOD)*d[n-m]%MOD));
	}
	return 0;
}