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\n",total-k);
}
return 0;
}