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,?...,?ak}, where a1?<?a2?<?a3?<?...?<?ak, 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).

Sample test(s)

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

思路：跟这题完全一样。点击打开链接  数据比HDU的强。

#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;

struct S{
char op[5];
int x,id;
}e[100005];

bool cmpval(S a,S b)
{
    if(a.x==b.x)  return a.op[0]<b.op[0];

    return a.x<b.x;
}

bool cmpid(S a,S b)
{
    return a.id<b.id;
}

int n,cnt,val[100005],num[400005];
long long sum[400005][5];

void build(int idx,int s,int e)
{
    if(s!=e)
    {
        int mid=(s+e)>>1;

        build(idx<<1,s,mid);
        build(idx<<1|1,mid+1,e);
    }

    num[idx]=0;

    for(int i=0;i<5;i++) sum[idx][i]=0;
}

void update(int idx,int s,int e,int pos,int flag)
{
    num[idx]+=flag;

    if(s==e)
    {
        sum[idx][1]+=val[s]*flag;

        return;
    }

    int mid=(s+e)>>1;

    if(pos<=mid) update(idx<<1,s,mid,pos,flag);
    else update(idx<<1|1,mid+1,e,pos,flag);

    for(int i=0;i<5;i++) sum[idx][i]=sum[idx<<1][i]+sum[idx<<1|1][i-num[idx<<1]%5>=0?i-num[idx<<1]%5:i-num[idx<<1]%5+5];//更新对应区间内下标为模5等于i数的和
}

long long query(int idx,int s,int e,int mod)
{
    return sum[idx<<1][mod]+sum[idx<<1|1][mod-num[idx<<1]%5>=0?mod-num[idx<<1]%5:mod-num[idx<<1]%5+5];
}

int main()
{
    int i;
    long long ans;

    while(~scanf("%d",&n))
    {
        map<int,int>mp;//用于离散化过程中判重，有可能存在先add a,再del a，再 add a的情况

        for(i=0;i<n;i++)
        {
            scanf("%s",e[i].op);

            if(e[i].op[0]=='s') e[i].x=0;
            else scanf("%d",&e[i].x);

            e[i].id=i;
        }

        sort(e,e+n,cmpval);

        cnt=1;

        for(i=0;i<n;i++)//离散化
        {
            if(e[i].op[0]=='a')
            {
                if(!mp[e[i].x]) mp[e[i].x]=cnt++;

                val[mp[e[i].x]]=e[i].x;
                e[i].x=mp[e[i].x];

            }
            else if(e[i].op[0]=='d')
            {
                e[i].x=mp[e[i].x];
            }
        }

        sort(e,e+n,cmpid);

        cnt--;

        if(!cnt)//特判
        {
            for(i=0;i<n;i++)
            {
                if(e[i].op[0]=='s')
                {
                    printf("0\n");
                }
            }

            continue;
        }

        build(1,1,cnt);

        for(i=0;i<n;i++)
        {
            if(e[i].op[0]=='a') update(1,1,cnt,e[i].x,1);
            else if(e[i].op[0]=='d') update(1,1,cnt,e[i].x,-1);
            else printf("%I64d\n",query(1,1,n,3));
        }
    }
}

CF-85D-Sum of Medians(线段树)