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[100005],a[100005];
long long k,total;
void merge_sort(int *a,int x,int y,int *t)
{
    if(y-x>1)
    {
        int m=x+(y-x)/2;
        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=0;i<n;i++)
            {
                cin>>a[i];
            }
            total=0;
            merge_sort(a,0,n,t);
            if(total<=k)
                cout<<'0'<<endl;
            else
            printf("%lld
",total-k);
        }
        return 0;
    }