You are given N positive integers, denoted as x0, x1 ... xN-1. Then give you some intervals [l, r]. For each interval, you need to find a number x to make as small as possible!

The first line is an integer T (T <= 10), indicating the number of test cases. For each test case, an integer N (1 <= N <= 100,000) comes first. Then comes N positive integers x (1 <= x <= 1,000, 000,000) in the next line. Finally, comes an integer Q (1 <= Q <= 100,000), indicting there are Q queries. Each query consists of two integers l, r (0 <= l <= r < N), meaning the interval you should deal with.

For the k-th test case, first output “Case #k:” in a separate line. Then output Q lines, each line is the minimum value of  . Output a blank line after every test case.

2

5
3 6 2 2 4
2
1 4
0 2

2
7 7
2
0 1
1 1

Case #1:
6
4

Case #2:
0
0

standy

2010 ACM-ICPC Multi-University Training Contest（4）——Host by UESTC

解题思路：显然x是中位数，用划分树找出来，用lsum[20][i]维护每一层从1到i划分到左子树的数的和,查询区间的时候找到区间内划分到左子树的数的个数lnum和他们的和suml，最后公式是mid*lnum-suml+mid*rnum-sumr。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
const int Maxn=100010;
int tree[20][Maxn];
LL lsum[20][Maxn],suml;
int sorted[Maxn];
int toleft[20][Maxn];
LL sum[Maxn];
int lnum;
void build(int l,int r,int dep)
{
	if(l==r)
	    return ;
	int mid=(l+r)>>1;
	int same=mid-l+1;
	for(int i=l;i<=r;i++)
	{
		if(tree[dep][i]<sorted[mid])
			same--;
	}
	int lpos=l;
	int rpos=mid+1;
	for(int i=l;i<=r;i++)
	{
		if(tree[dep][i]<sorted[mid])
		{
			tree[dep+1][lpos++]=tree[dep][i];
			lsum[dep][i]=lsum[dep][i-1]+tree[dep][i];
		}
		else if(tree[dep][i]==sorted[mid]&&same>0)
		{
			tree[dep+1][lpos++]=tree[dep][i],same--;
			lsum[dep][i]=lsum[dep][i-1]+tree[dep][i];
		}
		else
		{
			tree[dep+1][rpos++]=tree[dep][i];
			lsum[dep][i]=lsum[dep][i-1];
		}
		toleft[dep][i]=toleft[dep][l-1]+lpos-l;
	}
	build(l,mid,dep+1);
	build(mid+1,r,dep+1);
}
int query(int L,int R,int l,int r,int dep,int k)
{
	if(l==r)
		return tree[dep][l];
	int mid=(L+R)>>1;
	int cnt=toleft[dep][r]-toleft[dep][l-1];
	if(cnt>=k)
	{
		int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
		int newr=newl+cnt-1;
		return query(L,mid,newl,newr,dep+1,k);
	}
	else
	{
		lnum+=cnt;
		suml+=lsum[dep][r]-lsum[dep][l-1];
		int newr=r+toleft[dep][R]-toleft[dep][r];
		int newl=newr-(r-l-cnt);
		return query(mid+1,R,newl,newr,dep+1,k-cnt);
	}
}
int main()
{
	int t,n,m,q,a,b,ncase=1;
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		memset(tree,0,sizeof(tree));
		sum[0]=0;
		for(int i=1;i<=n;i++)
			scanf("%d",&tree[0][i]),sorted[i]=tree[0][i],sum[i]=sum[i-1]+sorted[i];
		sort(sorted+1,sorted+1+n);
		build(1,n,0);
		scanf("%d",&m);
		printf("Case #%d:\n",ncase++);
		while(m--)
		{
			scanf("%d%d",&a,&b);
			a++,b++;
			int k=(b-a)/2+1;
			lnum=suml=0;
			//suml是所有中位数左边的数的和
			int tmp=query(1,n,a,b,0,k);
			int rnum=b-a+1-lnum;
			LL sumr=sum[b]-sum[a-1]-suml;//sumr是所有中位数开始的右边的数的和(包括中位数)
			LL ans=sumr-tmp*rnum+tmp*lnum-suml;
			printf("%I64d\n",ans);
		}
		printf("\n");
	}
	return 0;
}