#663 (Div. 2)C. Cyclic Permutations（数学）

题目描述

A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
Consider a permutation p of length n, we build a graph of size n using it as follows:
For every 1≤i≤n, find the largest j such that 1≤j< i and pj >pi, and add an undirected edge between node i and node j
For every 1≤i≤n, find the smallest j such that ipi, and add an undirected edge between node i and node j
In cases where no such j exists, we make no edges. Also, note that we make edges between the corresponding indices, not the values at those indices.
For clarity, consider as an example n=4, and p=[3,1,4,2]; here, the edges of the graph are (1,3),(2,1),(2,3),(4,3).
A permutation p is cyclic if the graph built using p has at least one simple cycle.
Given n, find the number of cyclic permutations of length n. Since the number may be very large, output it modulo 109+7.
Please refer to the Notes section for the formal definition of a simple cycle

Input

The first and only line contains a single integer n (3≤n≤106).

Output

Output a single integer 0≤x<109+7, the number of cyclic permutations of length n modulo 109+7.

Examples

input
4
output
16
input
583291
output
135712853

Note

There are 16 cyclic permutations for n=4. [4,2,1,3] is one such permutation, having a cycle of length four: 4→3→2→1→4.
Nodes v1, v2, …, vk form a simple cycle if the following conditions hold:
k≥3.
vi≠vj for any pair of indices i and j. (1≤i vi and vi+1 share an edge for all i (1≤i

题目大意

给你一个数n，n的全排列中每一个序列根据如下规则建图：
1<=i<=n，找到最大的j使得1<=j<=i，p[j]>p[i]，在i和j之间连一条无向边。
1<=i<=n，找到最小的j使得i<=j<=n,p[j]>p[i]，在i和j之间连一条无向边。
问：n的全排列中有多少按照上面的规则建图，图中是有环的。

题目分析

首先我们可以确定：n的全排列一共有n!种情况，也就是说总共有n!种情况。
直接算有多少种情况是有环的不好算，我们可以先算出有多少种情况是没有环的，再用n!减去就行了。
什么样的序列是没有环的呢？根据题意：当一个序列满足先增后减、单调递增或者单调递减时，这个序列构成的图不会含有环。
然后以n（最大的数）为分界线，剩下的n-1个数每个都有两种选择：放在n前面和放在n后面。这样一共就有2^n-1种情况，答案即为n!-2^n-1。

补：对于这n-1个数，我们只能选择它们是放在n前还是n后，因为n前面的数必须满足单调递增，而n后面的数必须满足单调递减。

代码如下

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define PII pair
using namespace std;
const int N=1e5+5,mod=1e9+7;
int qmi(int a,int k,int p)		//快速幂模板
{
    int res=1;
    while(k)
    {
        if(k&1) res=(LL)res*a%p;
        k>>=1;
        a=(LL)a*a%p;
    }
    return res;
}
int f(int x)			//求x的阶乘
{
	LL res=1;
	for(int i=2;i<=x;i++)
		res=res*i%mod;
	return res;
}
int main()
{
	int n;
	cin>>n;
	cout<<(f(n)-qmi(2,n-1,mod)+mod)%mod<<endl;
    return 0;
}