B. Emordnilap

原题链接
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). There are n!=n⋅(n−1)⋅(n−2)⋅…⋅1 different permutations of length n.

Given a permutation p of n numbers, we create an array a consisting of 2n numbers, which is equal to p concatenated with its reverse. We then define the beauty of p as the number of inversions【颠倒】 in a.

The number of inversions in the array a is the number of pairs of indices i, j such that iaj.

For example, for permutation p=[1,2], a would be [1,2,2,1]. The inversions in a are (2,4) and (3,4) (assuming 1-based indexing). Hence, the beauty of p is 2.

Your task is to find the sum of beauties of all n! permutations of size n. Print the remainder we get when dividing this value by 1000000007 (109+7).

Input
Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤105). The description of the test cases follows.

Each test case has only one line — the integer n (1≤n≤105).

It is guaranteed that the sum of n over all test cases does not exceed 105.

Output
For each test case, print one integer — the sum of beauties of all permutations of size n modulo 1000000007 (109+7).

Example
input
3
1
2
100
output
0
4
389456655
Note
For the first test case of the example, p=[1] is the only permutation. a=[1,1] has 0 inversions.

For the second test case of the example, the permutations are [1,2] and [2,1]. Their respective a arrays are [1,2,2,1] and [2,1,1,2], both of which have 2 inversions.

题意

• 给出一个n，求长度为n的排列经过反转拼接后得到的数组中颠倒的数目，求出n!种排列的情况得到的总和取模1000000007的值

思路

1. 无论是哪一种排列情况颠倒数目都是相同的(\(n*(n-1)\))
   因此无论怎么排列，每两个不同的数\(i，j\)总会是\(2\)的颠倒贡献值
   \(2*\frac{n*(n-1)}{2} = n*(n-1)\)

• 为什么？
  我们可以只看排列中的两个元素，假设\(i<j\)，则反转得到的新下标\(i_{reverse} > j_{reverse}\)
  即\(i<j < j_{reverse} < i_{reverse}\)
  ①如果\(p_i < p_j\)，得到贡献为2
  ②如果\(p_i > p_j\)，得到贡献为2
  因此无论怎么排列，每两个不同的数\(i，j\)总会是\(2\)的颠倒贡献值

2. 总共有n!中排列情况
3. 原题答案为\(n!*n*(n+1) mod 1000000007\)

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;

#define X first
#define Y second

typedef long long LL;
const char nl = '\n';
const int N = 1e6+10;
LL n;
//int a[N];
int cnt[N];
int p = 1e9+7;
void solve(){
	cin >> n;
	LL res = n*(n-1)%p;
	if(n == 1)cout << 0 << nl;
	else{
		for(int i = 1; i <= n; i ++){
			res *= i;
			res %= p;
		}

		//res *= res;
		res %= p;
		cout << res << nl;
	}
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);

	int T;
	cin >> T;
	while(T --){
		solve();
	}
	//solve();
}

陌生单词

1. indices【下标】