HDU 4911 归并排序求逆序对

Problem Description
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

Sample Output
1
2

#include<iostream>#include<algorithm>#include<stdlib.h> using namespace std;int a[100010],tmp[100010];long long ans;void ld(int l,int m,int r){    int i=l;    int j=m+1;    int k=l;    while(i<=m&&j<=r)    {        if(a[i]>a[j])        {            tmp[k++]=a[j++];            ans+=m-i+1;        }        if(a[i]<=a[j])        {            tmp[k++]=a[i++];        }    }    while(i<=m) tmp[k++]=a[i++];    while(j<=r) tmp[k++]=a[j++];    for(int i=l;i<=r;i++)       a[i]=tmp[i]; }void sort2(int l,int r){    if(l<r)    {           int m=(l+r)>>1;        sort2(l,m);        sort2(m+1,r);        ld(l,m,r);    }}int main(){    int n,k;    while(cin>>n>>k)    {        for(int i=1;i<=n;i++)        {            cin>>a[i];        }        ans=0;        sort2(1,n);        if(ans-k<0) cout<<'0'<<endl;        else cout<<ans-k<<endl;         }    system("pause");    return 0;}