Minimum Sum

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

Total Submission(s): 2235    Accepted Submission(s): 512

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

 

 

Recommend

zhengfeng

 

 

 

 

 

 

其实就是找到中间那个数。

 

然后需要记录和，右边的减掉左边的

 

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;const int MAXN = 100010;int tree[20][MAXN];int sorted[MAXN];int toleft[20][MAXN];long long sum[20][MAXN];void build(int l,int r,int dep){    if(l == r)    {        sum[dep][l] = sum[dep][l-1]+tree[dep][l];        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--;        sum[dep][i] += sum[dep][i-1]+tree[dep][i];    }    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];        else if(tree[dep][i] == sorted[mid] && same > 0)        {            tree[dep+1][lpos++] = tree[dep][i];            same--;        }        else            tree[dep+1][rpos++] = tree[dep][i];        toleft[dep][i] = toleft[dep][l-1] + lpos - l;    }    build(l,mid,dep+1);    build(mid+1,r,dep+1);}long long ans;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 ee = r-L+1-(toleft[dep][r]-toleft[dep][L-1])+mid;        int ss = l-L-(toleft[dep][l-1]-toleft[dep][L-1])+mid;        ans += sum[dep+1][ee]-sum[dep+1][ss];        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    {        int s = L + toleft[dep][l-1] - toleft[dep][L-1];        int e = s + cnt - 1;        ans -= sum[dep+1][e] - sum[dep+1][s-1];        int newr = r + toleft[dep][R] - toleft[dep][r];        int newl = newr - (r-l+1-cnt) + 1;        return query(mid+1,R,newl,newr,dep+1,k-cnt);    }}int main(){    int T;    int n;    scanf("%d",&T);    int iCase = 0;    while(T--)    {        iCase++;        scanf("%d",&n);        memset(tree,0,sizeof(tree));        memset(sum,0,sizeof(sum));        for(int i = 1;i <= n;i++)        {            scanf("%d",&tree[0][i]);            sorted[i] = tree[0][i];        }        sort(sorted+1,sorted+n+1);        build(1,n,0);        printf("Case #%d:\n",iCase);        int m,l,r;        scanf("%d",&m);        while(m--)        {            scanf("%d%d",&l,&r);            l++;r++;            ans = 0;            int tmp = query(1,n,l,r,0,(l+r)/2-l+1);            if((l+r)%2)ans-=tmp;            printf("%I64d\n",ans);        }        printf("\n");    }    return 0;}