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

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

2 5 3 6 2 2 4 2 1 4 0 2 2 7 7 2 0 1 1 1

 

Sample Output

Case #1: 6 4 Case #2: 0 0

 

Author

standy

 

Source

 

#include 
     
      #include 
      
       #include 
       
        #include 
        
         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]
         
          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;}