015 HUAS Summer Contest#3~A

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

解题思路：题目的意思是输入一组数，交换k次后求出它的逆序数来。但是需要注意的是题目中并没有说直到输入0为止，所以不需要&&n，在每一次重新输入数据之后要记得将求得的和清零。利用归并思想是很好的方法。

程序代码：

#include<cstdio>
 #include<iostream>
 using namespace std;
long long k;
const int maxn=100005;
long long a[maxn],v[maxn];
int mer_sort(long long x,long long y)
{
	if(x<y)
	{
		long long m=(y+x)/2;
		mer_sort(x,m);
		mer_sort(m+1,y);
		int p=0,q=m+1,i=x;

		while(i<=m&&q<=y)
		{
			if(a[i]>a[q])
                {v[p++]=a[q++];k+=m+1-i;}
			else   v[p++]=a[i++];
		}
		while(i<=m)
             v[p++]=a[i++];
         while(q<=y)
             v[p++]=a[q++];
        for(i=0;i<p;++i)
             a[x+i]=v[i];
	}
	return 0;
}
int main()
{
	long long n,t;
	while(scanf("%lld%lld",&n,&t)!=EOF)
	{k=0;
		for(int i=0;i<n;i++)
			scanf("%lld",&a[i]);


		mer_sort(0,n-1);
		if(k<=t)printf("0
");
		else
		printf("%lld
",k-t);
	}
	return 0;
}