A - Inversion 归并排序求逆序数

A - Inversion

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

Submit  Status  Practice  HDU 4911

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≤ii>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

#include 
#define N  150000

int a[N],tmp[N];
long long ans;//

void Merge(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;  //这里加上相应的数
        }
        else
        {
            tmp[k++]=a[i++];
        }
    }
    while(j<=r) tmp[k++]=a[j++];
    while(i<=m) tmp[k++]=a[i++];
    for(int i=l;i<=r;i++)
        a[i]=tmp[i];
}

void Merge_sort(int l,int r)
{
    if(l>1;
        Merge_sort(l,m);
        Merge_sort(m+1,r);
        Merge(l,m,r);
    }
}

int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        for(int i=0;i