Problem:
Score : 300 points

Problem Statement

We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}.

Let B be a sequence of K \times 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 10^9 + 7.

Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

Constraints

All values in input are integers.

1 \leq N \leq 2000

1 \leq K \leq 10^9

1 \leq A_i \leq 2000

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 ... A_{N - 1}

Output

Print the inversion number of B, modulo 10^9 + 7.

Sample Input 1

2 2
2 1

Sample Output 1

3

In this case, B~=~2,~1,~2,~1. We have:

B_0 > B_1

B_0 > B_3

B_2 > B_3

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 10^9 + 7.

Solution in C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)



#include <utility>
#include <cstdio>
#include <math.h>



using namespace std;

    
int main()
{
  	int Nmax = 2000;
    
    int i,j;
    int len;   // 文字列の長さ
    int flag=0;
   
    int A[Nmax];
    int N;
    int K,k;
    double sho=0;  
    double total=0.0;
    int ans;
    int bunbo = 1000000000 + 7;
    double tmp;
    
   
 
    
    // 数列
    scanf("%d %d", &N,&K);
    for(i=0;i<=N-1;i++){
        scanf("%d ",&A[i]);
    }
    

              
    // １セット内での転倒数
    double count1 = 0;
    for(i=0;i<=N-1;i++){
        for(j=i+1;j<= N-1;j++){
            if(A[i]>A[j]) count1++;
        }
    }
    
    // ２セット間での転倒数
    double count2 = 0;
    for(i=0;i <= N-1;i++){
        for(j=0;j <= N-1;j++){
            if(A[i]>A[j]) count2++;
        }
    }
    

    total = K*count1 + count2*K*(K-1)/2;
//  	printf("%f\n",total);
    tmp =  total/(double)bunbo; 
//    printf("%f\n",tmp);
  　sho = floor(tmp);
//    printf("%f\n",tmp);
   
    
    tmp = total - sho*bunbo;
//  	printf("%f\n",tmp);
    ans = (int)tmp; 
    printf("%d\n",ans );
    
    
    
 /*
    total = count1;
    for(k=2;k<=K;k++){
        total = total + count1 + (k-1)*count2;
    }
    
    // int型より大きいのでオーバーフローしないところまではdouble型で計算
    tmp =  total/(double)bunbo; 
 //   printf("%f\n",tmp);
  　sho = floor(tmp);    // 小数点以下切り捨て
//    printf("%f\n",tmp);
   
    
    tmp = total - sho*bunbo;    // int型に収まってきた
//  printf("%f\n",tmp);
    ans = (int)tmp; 
    printf("%d\n",ans );
*/
   
    
   
  
  return 0;
   
}