显然每次交换都能降低1
所以求出逆序数对数，然后-=k就好了。。
。
_(:зゝ∠)_
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<set>
#include<map>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 100005
#define ll long long
ll c[N+100000], maxn;
inline ll Lowbit(ll x){return x&(-x);}
void change(ll i, ll x)//i点增量为x
{
while(i <= maxn)
{
c[i] += x;
i += Lowbit(i);
}
}
ll sum(ll x){//区间求和 [1,x]
ll ans = 0;
for(ll i = x; i >= 1; i -= Lowbit(i))
ans += c[i];
return ans;
}
ll a[N], n, k;
set<ll>s;
set<ll>::iterator p;
map<ll,ll>mp;
int main(){
ll i;
while(cin>>n>>k){
s.clear(); mp.clear();
for(i = 1; i <= n; i++)scanf("%I64d",&a[i]), s.insert(a[i]);
maxn = n+100;
for(p = s.begin(), i = 2; p!=s.end(); p++, i++)
{
mp[*p] = i;
}
for(i = 1; i <= n; i++)a[i] = mp[a[i]];
memset(c, 0, sizeof c);
ll ans = 0;
for(i = n; i >= 1; i--)
{
ans += sum(a[i]-1);
change(a[i], 1);
}
ans -= k;
cout<< max(0ll, ans) <<endl;
}
return 0;
}