A题

A - A

Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

 

Description

bobo has a sequence a ₁,a ₂,…,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≤10 ⁵,0≤k≤10 ⁹). The second line contains n integers a ₁,a ₂,…,a _n (0≤a _i≤10 ⁹).

 

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

 

题解：同归并排序，只需要逆序数减去交换次数。但是要注意交换次数的大小是否大于原本逆序数。

 

#include<iostream>
#include<cstdio>
using namespace std;
int n,t[],a[];
long long k,total;
void merge_sort(int *a,int x,int y,int *t)
{
    if(y-x>)
    {
        int m=x+(y-x)/;
        int p=x,q=m,i=x;
        merge_sort(a,x,m,t);
        merge_sort(a,m,y,t);
        while(p<m||q<y)
        {
            if(q>=y||(p<m&&a[p]<=a[q]))
                t[i++]=a[p++];
            else
                {
                    t[i++]=a[q++];
                    total+=m-p;
                }
        }
        for(i=x;i<y;i++)
            a[i]=t[i];
    }
}
    int main()
    {
        while(cin>>n>>k)
        {
            for(int i=;i<n;i++)
            {
                cin>>a[i];
            }
            total=;
            merge_sort(a,,n,t);
            if(total<=k)
                cout<<''<<endl;
            else
            printf("%lld\n",total-k);
        }
        return ;
    }