bobo has a sequence a1,a2,…,an. 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 ai>aj.

 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,…,an (0≤ai≤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

Output

 1
2

原题链接：传送门     

题意：求逆序对数

求逆序对数，这题有两种很好的做法一个是树状数组另一个就是归并排序。

最朴素的做法是O(n^2)的纯暴力，这里推荐归并排序的求法时间复杂度为O(nlogn)

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 9;
int n, m, ans, q[N], t[N];

void csort(int l, int r) {
	if (l == r) return;
	int mid = l + r >> 1;
	csort(l, mid), csort(mid + 1, r);
	int k = 0, i = l, j = mid + 1;
	while (i <= mid && j <= r) {
		while (q[i] <= q[j] && i <= mid) t[k++] = q[i++];
		while (q[i] > q[j] && j <= r) t[k++] = q[j++], ans += mid - i + 1;
	}
	while (i <= mid) t[k++] = q[i++];
	while (j <= r) t[k++] = q[j++];
	for (int i = l, j = 0; j < k; i++, j++) q[i] = t[j];
}

signed main() {
	while (~scanf("%lld%lld", &n, &m)) {
		ans = 0; 
		for (int i = 1; i <= n; i++) scanf("%lld", &q[i]);
		csort(1, n);
		if (ans <= m) printf("0\n");
		else printf("%lld\n", ans - m);
	}
	return 0;
}

可以看到时间复杂度还是很可观的。