Problem Description

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!

 

Input

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.

 

Output

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.

 

Sample Input

 

Sample Output

 

Author

standy

 

Source

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

/*

hdu 3473

最开始直接算TL，然后发现由于是计算绝对值，在建树的时候需要记录比你查找的值小的和，

然后分开计算即可

*/

#include <functional>

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <Map>

using namespace std;

typedef long long ll;

typedef long double ld;

using namespace std;

const int maxn = 100010;

int tree[20][maxn];

int sorted[maxn];

int toleft[20][maxn];

ll sum[maxn];

ll lsum[20][maxn];

ll tsum;

void build(int l,int r,int dep)  //模拟快排 并记录左树中比i小的个数

{

    if(l == r)

        return;

    int mid = (l+r)>>1;

    int same = mid-l+1;//可能有很多值与中间那个相等，但不一定被分到左边

    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];

            lsum[dep][i] = lsum[dep][i-1] + tree[dep][i];

            same --;

        }

        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];  //所查找区间放在左树中的个数

     if(cnt >= k)

     {   //得到l左边放到左子树的个数，加上L即是开始位置

         int lpos = L+toleft[dep][l-1]-toleft[dep][L-1];

         int rpos = lpos+cnt-1;

         return query(L,mid,lpos,rpos,dep+1,k);

     }

     else

     {   //R-r可以得出后面空出了多少位置

         int rpos = r+toleft[dep][R]-toleft[dep][r];

         int lpos = rpos-(r-l-cnt);

         tsum += lsum[dep][r] - lsum[dep][l-1];

         return query(mid+1,R,lpos,rpos,dep+1,k-cnt);

     }

 }

 int main()

 {

     int n,m,T;

     int cas = 1;

     scanf("%d",&T);

     while(T--)

     {

         scanf("%d",&n);

         for(int i = 1;i <= n;i++)

         {

             scanf("%d",&sorted[i]);

             sum[i] = sum[i-1]+sorted[i];

             tree[0][i] = sorted[i];

         }

         sort(sorted+1,sorted+n+1);

         build(1,n,0);

         scanf("%d",&m);

         int l,r;

         printf("Case #%d:\n",cas++);

         while(m--)

         {

             scanf("%d%d",&l,&r);

             l++;r++;tsum = 0;

             ll k =(r-l)/2+1;

             ll ans = 0;

             ll aver = query(1,n,l,r,0,k);

             ans = sum[r]-sum[l-1]-aver*(r-l+1-k)-tsum-aver;

             ans += (k-1)*aver-tsum;

             printf("%I64d\n",ans);

         }

         puts("");

     }

     return 0;

 }