数据结构（线段树）：CodeForces 85D Sum of Medians

D. Sum of Medians

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In one well-known algorithm of finding the k-th order statistics we should divide all elements into groups of five consecutive elements and find the median of each five. A median is called the middle element of a sorted array (it's the third largest element for a group of five). To increase the algorithm's performance speed on a modern video card, you should be able to find a sum of medians in each five of the array.

A sum of medians of a sorted k-element set S = {a₁, a₂, ..., a_k}, where a₁ < a₂ < a₃ < ... < a_k, will be understood by as

The operator stands for taking the remainder, that is stands for the remainder of dividing x by y.

To organize exercise testing quickly calculating the sum of medians for a changing set was needed.

Input

The first line contains number n (1 ≤ n ≤ 10⁵), the number of operations performed.

Then each of n lines contains the description of one of the three operations:

• add x — add the element x to the set;
• del x — delete the element x from the set;
• sum — find the sum of medians of the set.

For any add x operation it is true that the element x is not included in the set directly before the operation.

For any del x operation it is true that the element x is included in the set directly before the operation.

All the numbers in the input are positive integers, not exceeding 10⁹.

Output

For each operation sum print on the single line the sum of medians of the current set. If the set is empty, print 0.

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams (also you may use the %I64d specificator).

Examples

Input

6
add 4
add 5
add 1
add 2
add 3
sum

Output

3

Input

14
add 1
add 7
add 2
add 5
sum
add 6
add 8
add 9
add 3
add 4
add 10
sum
del 1
sum

Output

5
11
13

　　这道题目不难，注意去重，还要防止爆int。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=100010;
int hsh[maxn],tot,Q;
long long ans[maxn<<2][5];
int sum[maxn<<2],tp[maxn],num[maxn];

void Push_up(int x){
    int l=x<<1,r=x<<1|1;
    sum[x]=sum[l]+sum[r];
    for(int i=0;i<=4;i++)
        ans[x][i]=ans[l][i]+ans[r][((i-sum[l])%5+5)%5];
}

void Insert(int x,int l,int r,int g,int d){
    if(l==r){
        ans[x][0]+=hsh[l]*d;
        sum[x]+=d;
        return;
    }
    int mid=(l+r)>>1;
    if(mid>=g)Insert(x<<1,l,mid,g,d);
    else Insert(x<<1|1,mid+1,r,g,d);
    Push_up(x);
}

char op[10];
int main(){
    scanf("%d",&Q);
    for(int q=1;q<=Q;q++){
        scanf("%s",op);
        if(op[0]=='a')tp[q]=1;
        else if(op[0]=='d')tp[q]=-1;
        else continue;
        scanf("%d",&num[q]);
        if(tp[q]==1){++tot;hsh[tot]=num[q];}
    }
    
    sort(hsh+1,hsh+tot+1);
    tot=unique(hsh+1,hsh+tot+1)-hsh-1;

    for(int q=1;q<=Q;q++){
        if(tp[q]==1){
            int p=lower_bound(hsh+1,hsh+tot+1,num[q])-hsh;
            Insert(1,1,tot,p,1);
        }
        else if(tp[q]==-1){
            int p=lower_bound(hsh+1,hsh+tot+1,num[q])-hsh;
            Insert(1,1,tot,p,-1);
        }
        else
            printf("%I64d
",ans[1][2]);
    }
    return 0;
}