Kleene Inversion AtCoder - 5165

Problem Statement
We have a sequence of N integers A₌A0,_A1,…,~AN−1.

Let B be a sequence of K×N integers obtained by concatenating K copies of A. For example, if A₌1,_3,2 and K₌2, B₌1,_3,2,_1,3,~2.

Find the inversion number of B, modulo 109+7.

Here the inversion number of B is defined as the number of ordered pairs of integers (i,_j)(0≤i<j≤K×N−1) such that Bi>Bj.

Constraints
All values in input are integers.
1≤N≤2000
1≤K≤109
1≤Ai≤2000
Input
Input is given from Standard Input in the following format:

N K
A0 A1 … AN−1
Output
Print the inversion number of B, modulo 109+7.

Sample Input 1
2 2
2 1
Sample Output 1
3
In this case, B₌2,_1,2,~1. We have:

B0>B1
B0>B3
B2>B3
Thus, the inversion number of B is 3.

Sample Input 2
3 5
1 1 1
Sample Output 2
0
A may contain multiple occurrences of the same number.

Sample Input 3
10 998244353
10 9 8 7 5 6 3 4 2 1
Sample Output 3
185297239
Be sure to print the output modulo 109+7.

题意:
给你一串数字，一个k；让你把这一串数字复制k次，求出来每个数的右边有多少个比他小的数，并且输出总和；
思路；

遍历最初的数组（），找出来每个数右边有几个比他小的数，求出总数cot；然后再找出来每个数左边比它小的数有几个，求出总数cnt；根据数学推导：复制k次后，cot会计算k*（1+k）/2次；cnt会计算（1-k）*k/2次；

#include <bits/stdc++.h>

using namespace std;
typedef unsigned long long ll;
const int mod=1e9+7;
int a[20010];
int main()
{
	ll n,k;
	scanf("%lld %lld",&n,&k);
	ll cot=0; 
	for(int i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
	}
	for(int i=0;i<n;i++)
	{
		for(int j=i+1;j<n;j++)
		{
			if(a[i]>a[j]) cot++;
		}
	}
	ll cnt=0;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<i;j++)
		{
			if(a[j]<a[i]) cnt++;
		}
	} 
	ll sum=0;
	ll kk=k*(k+1)/2;
	if(kk==1) cout <<cot;
	else
	{
		ll kkk=(k-1)*(k)/2;
		sum=cot%mod*(kk%mod)%mod+cnt%mod*(kkk%mod)%mod;
		cout <<sum%mod<<endl;
	}
	return 0;
}