# Sum of Medians

## 题目描述

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 $S={a_1, a_2, \ldots, a_k}$, where $a_1 \lt a_2 \lt a_3 \lt \cdots \lt a_k$, will be understood by as $\sum_{i \bmod 5=3} a_i$.
The $\bmod$ operator stands for taking the remainder, that is $x \bmod y$ 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.

## 代码1

#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;
long long n, x;
string o;
struct treap {
int tot, root;
struct node_type {
int child[2], rank;
long long val, size, mod[5];
} node[100001];
void update(int t) {
node[t].size = node[node[t].child[0]].size + node[node[t].child[1]].size + 1;
for (int i = 0; i < 5; ++i)
node[t].mod[i] = node[node[t].child[0]].mod[i] + node[node[t].child[1]].mod[(i + 99999 - node[node[t].child[0]].size) % 5];
node[t].mod[(node[node[t].child[0]].size + 1) % 5] += node[t].val;
}
void rotate(int &t, int dire) {
int p = node[t].child[!dire];
node[t].child[!dire] = node[p].child[dire], node[p].child[dire] = t;
update(t), update(p), t = p;
}
void insert(int &t, long long val) {
if (!t) {
t = ++tot, node[t].rank = ((rand() & ((1 << 16) - 1)) << 10) ^ rand(), node[t].val = node[t].mod[1] = val, node[t].size = 1;
return;
}
++node[t].size;
if (val < node[t].val) {
insert(node[t].child[0], val);
if (node[node[t].child[0]].rank > node[t].rank) rotate(t, 1);
} else {
insert(node[t].child[1], val);
if (node[node[t].child[1]].rank > node[t].rank) rotate(t, 0);
}
update(t);
}
void remove(int &t, long long val) {
if (!node[t].child[0] && !node[t].child[1]) { t = 0; return; }
if (val == node[t].val) {
if (node[node[t].child[0]].rank > node[node[t].child[1]].rank) rotate(t, 1);
else rotate(t, 0);
}
if (val < node[t].val) remove(node[t].child[0], val);
else remove(node[t].child[1], val);
update(t);
}
long long query() { return node[root].mod[3]; }
} tree;
int main()
{
srand(time(NULL));
cin >> n;
while (n--) {
cin >> o;
if (o[0] == 'a') { cin >> x; tree.insert(tree.root, x); }
else if (o[0] == 'd') { cin >> x; tree.remove(tree.root, x); }
else cout << tree.query() << endl;
}
return 0;
}


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