划分树,统计每层移到左边的数的和.

Minimum Sum

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2959    Accepted Submission(s): 684

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!

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

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn=100010;

typedef long long int LL;

int tree[18][maxn];

LL sumL[18][maxn];

int sorted[maxn];

int toleft[18][maxn];

void build(int l,int r,int dep)

{

    if(l==r) return ;

    int mid=(l+r)/2;

    int same=mid-l+1;

    for(int i=l;i<=r;i++)

        if(tree[dep][i]<sorted[mid]) same--;

    int lpos=l,rpos=mid+1;

    for(int i=l;i<=r;i++)

    {

        if(tree[dep][i]<sorted[mid])

        {

            tree[dep+1][lpos++]=tree[dep][i];

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

        }

        else if(tree[dep][i]==sorted[mid]&&same>0)

        {

            tree[dep+1][lpos++]=tree[dep][i];

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

            same--;

        }

        else

        {

            tree[dep+1][rpos++]=tree[dep][i];

            sumL[dep][i]=sumL[dep][i-1];

        }

        toleft[dep][i]=toleft[dep][l-1]+lpos-l;

    }

    build(l,mid,dep+1); build(mid+1,r,dep+1);

}

LL SUMOFLEFT,NUMOFLEFT;

LL query(int L,int R,int l,int r,int dep,int k)

{

    if(l==r) return tree[dep][l];

    int mid=(L+R)/2;

    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

    {

        SUMOFLEFT+=sumL[dep][r]-sumL[dep][l-1];

        NUMOFLEFT+=cnt;

        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 n,m;

LL sum[maxn];

int main()

{

    int T_T,cas=1;

    scanf("%d",&T_T);

    while(T_T--)

    {

        scanf("%d",&n);

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

        {

            scanf("%d",sorted+i);

            tree[0][i]=sorted[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",cas++);

        while(m--)

        {

            int l,r,k;

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

            l++; r++;

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

            SUMOFLEFT=0;NUMOFLEFT=0;

            LL ave=query(1,n,l,r,0,k);

            printf("%I64d\n",(sum[r]-sum[l-1]-SUMOFLEFT)-SUMOFLEFT+(NUMOFLEFT-(r-l+1-NUMOFLEFT))*ave);

        }

        putchar(10);

    }

    return 0;

}