Atcoder 800 Painting Machine Math (Counting)

Question meaning: n white balls. n-1 machines, the i-th machine will turn the (i, i+1)-th ball into black.
An operation sequence is [1..n-1] permutation, its value Defined as the time when n balls turn black for the first time.
2<=n<=1e6. Given n, ask all permutations of operations, the sum of their values ​​modulo 1e9+7 ?

After reading the solution.. first film under rng_58.

How many permutations are there with the calculated value <= k? Then the value of k is d[k]-d[k-1]
First, the two machines 1, n-1 must be selected, otherwise 1, n will not become black .Assume
1=a[1]<a[2]<.....a[k] =n-1 Then the interval of a[i]-a[i-1] is 1 or 2. Let these two The number is a, b.
You can get a+b=k-1 , a+2b=n-2 to solve b=nk-1 Then choose b from k-1 intervals as interval 2, and you can determine a plan .The number of methods is C(k-1,nk-1)

Then the total number of permutations of value k is: C(k-1,nk-1) * k! * (n-1-k) !.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+5,mod=1e9+7;
ll n,f[N];
ll powmod(ll x,ll n)
{
	ll s=1;
	while(n)
	{
		if(n&1)
			s=(s*x)%mod;
		n>>=1;
		x=(x*x)%mod;
	}
	return s;
}
ll C(ll n,ll m)
{
	if(n<m)
		return 0;
	ll a=f[n],b=(f[n-m]*f[m])%mod;
	return (a*powmod(b,mod-2))%mod;
}
intmain()
{
	cin>>n;
	f[0]=1;
	for(int i=1;i<=n;i++)
		f[i]=(f[i-1]*i)%mod;
	ll res=0,pre=0;
	for(int k=1;k<n;k++)
	{
		ll tk=(f[k]*f[n-1-k])%mod;
		tk = (tk * C (k-1, nk-1))% mod;
		ll num = (tk-pre + mod)% mod;
		res = (res + (num * k)% mod)% mod;
		pre = tk;
	}
	cout<<res<<'\n';
	return 0;
}

When counting, you can first assume order and calculate its plan. Then arrange