题目链接：
Sum of Medians
Time Limit:3000MSMemory Limit:262144KB
#### 问题描述
> 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 = {a1, a2, ..., ak}, where a1
>
> 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.
#### 输入
> 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.
#### 输出
> 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).
#### 样例
> **sample 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
>
> **sample output**
> 5
> 11
> 13
题意
求当前有序集合中所有下标%5==3的数字的和。
题解
对于一个区间，我们可以维护相对的%5=x的位置，比如对于区间[a,a]，那它%5为1的数为val[a]，其他的都为0，这样我们在合并的时候左边的%5==x的位置是不会变的，右边的只要看左边有多少个数就知道要用x’（偏移）取和x合并了。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define lson (o<<1)
#define rson ((o<<1)|1)
#define M l+(r-l)/2
using namespace std;
const int maxn = 1e5 + 10;
typedef __int64 LL;
LL sumv[maxn << 2][5];
int cntv[maxn << 2];
int n;
char cmd[maxn][11];
int val[maxn];
vector<int> ha;
void maintain(int o) {
cntv[o] = cntv[lson] + cntv[rson];
for (int i = 0; i < 5; i++) {
sumv[o][i] = sumv[lson][i] + sumv[rson][((i - cntv[lson]) % 5 + 5) % 5];
}
}
int _p,_type;
void update(int o, int l, int r) {
if (l==r) {
if (_type =='d') {
sumv[o][1] = 0;
cntv[o] = 0;
}
else {
sumv[o][1] = ha[l - 1];
//printf("(%d,%d):%d", l);
cntv[o] = 1;
}
}
else {
if (_p <= M) update(lson, l, M);
else update(rson, M + 1, r);
maintain(o);
}
}
int main() {
scanf("%d", &n);
memset(sumv, 0, sizeof(sumv));
memset(cntv, 0, sizeof(cntv));
for (int i = 0; i < n; i++) {
scanf("%s", cmd[i]);
if (cmd[i][0] != 's') {
scanf("%d", &val[i]);
ha.push_back(val[i]);
}
}
sort(ha.begin(), ha.end());
ha.erase(unique(ha.begin(), ha.end()),ha.end());
for (int i = 0; i < n; i++) {
if (cmd[i][0] !='s') {
_p = lower_bound(ha.begin(), ha.end(), val[i]) - ha.begin() + 1;
_type = cmd[i][0];
update(1, 1, n);
}
else {
printf("%I64d\n", sumv[1][3]);
}
}
return 0;
}
乱起八糟
线段树是递归的思想，所以它区间保存的数据不能是绝对的！只能是相对的！是只对当前这个区间定义的！而不是对整个区间定义的！所以，如果你想用子节点来表示在全局中%5是什么情况，可能就会比较麻烦了吧。