Problem Description

bobo has a sequence a₁,a₂,…,a_n. He is allowed to swap two adjacent numbers for no more than k times.



Find the minimum number of inversions after his swaps.



Note: The number of inversions is the number of pair (i,j) where 1≤i<j≤n and a_i>a_j.

 

Input

The input consists of several tests. For each tests:



The first line contains 2 integers n,k (1≤n≤10⁵,0≤k≤10⁹). The second line contains n integers a₁,a₂,…,a_n (0≤a_i≤10⁹).

 

Output

For each tests:



A single integer denotes the minimum number of inversions.

 

Sample Input

3 1

2 2 1

3 0

2 2 1

 

Sample Output

1

2

 

题意：给出n个数，每次能够交换相邻的两个数。最多交换k次，求交换后最小的逆序数是多少。

分析：假设逆序数大于0。则存在1 ≤ i < n，使得交换ai和ai+1后逆序数减1。所以最后的答案就是max((inversion-k), 0)。

利用归并排序求出原序列的逆序对数就能够解决这个问题了。

#include<stdio.h>

#include<string.h>

#define N 100005

__int64 cnt, k;

int a[N],c[N];

//归并排序的合并操作

void merge(int a[], int first, int mid, int last, int c[])

{

    int i = first, j = mid + 1;

    int m = mid, n = last;

    int k = 0;

    while(i <= m || j <= n)

    {

        if(j > n || (i <= m && a[i] <= a[j]))

            c[k++] = a[i++];

        else

        {

            c[k++] = a[j++];

            cnt += (m - i + 1);

        }

    }

    for(i = 0; i < k; i++)

        a[first + i] = c[i];

}

//归并排序的递归分解和合并

void merge_sort(int a[], int first, int last, int c[])

{

    if(first < last)

    {

        int mid = (first + last) / 2;

        merge_sort(a, first, mid, c);

        merge_sort(a, mid+1, last, c);

        merge(a, first, mid, last, c);

    }

}

int main()

{

    int n;

    while(~scanf("%d%I64d",&n,&k))

    {

        memset(c, 0, sizeof(c));

        cnt = 0;

        for(int i = 0; i < n; i++)

            scanf("%d", &a[i]);

        merge_sort(a, 0, n-1, c);

        if(k >= cnt) cnt = 0;

        else cnt -= k;

        printf("%I64d\n",cnt);

    }

}