D. Sum of Medians

Time Limit: 1 Sec Memory Limit: 256 MB

题目连接

http://codeforces.com/problemset/problem/85/D

Description

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

codeforces 85D D. Sum of Medians Vector的妙用

The codeforces 85D D. Sum of Medians Vector的妙用 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 Input

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

Sample Output

HINT

题意

给你一堆数，然后排序，然后让你输出下标mod5=3的和

题解：

用vector来做，虽然感觉有很多数据结构都可以把这道题秒了

代码：

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 200001

#define mod 10007

#define eps 1e-9

int Num;

char CH[];

//const int inf=0x7fffffff;   //нчоч╢С

const int inf=0x3f3f3f3f;

/*

inline void P(int x)

{

    Num=0;if(!x){putchar('0');puts("");return;}

    while(x>0)CH[++Num]=x%10,x/=10;

    while(Num)putchar(CH[Num--]+48);

    puts("");

}

*/

//**************************************************************************************

inline ll read()

{

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

inline void P(int x)

{

    Num=;if(!x){putchar('');puts("");return;}

    while(x>)CH[++Num]=x%,x/=;

    while(Num)putchar(CH[Num--]+);

    puts("");

}

vector<int> s;

int main()

{

    char a[];

    int b;

    int n=read();

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

    {

        scanf("%s",a);

        if(a[]=='a')

        {

            scanf("%d",&b);

            s.insert(lower_bound(s.begin(),s.end(),b),b);

        }

        if(a[]=='d')

        {

            scanf("%d",&b);

            s.erase(lower_bound(s.begin(),s.end(),b));

        }

        if(a[]=='s')

        {

            ll sum=;

            for(int j=;j<s.size();j+=)

                sum+=s[j];

            printf("%lld\n",sum);

        }

    }

}

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