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≤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 12 2 13 02 2 1

Sample Output

12

#include <stdio.h>#define N  150000int 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<r)    {        int m=(l+r)>>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<n;i++)            scanf("%d",&a[i]);        ans=0;        Merge_sort(0,n-1);        ans-=k;        if(ans<0)ans=0;        printf("%I64d\n",ans);    }    return 0;}