Codeforces 85D Sum of Medians[线段树]

题意：给了一个set，有n个操作，有三种操作

1. add  x 把 x 放入set中；

2. del  x 把 x 从set中删去；

3. sum 求set中，第n大的数，n%5==3，的总和。

分析：因为x有[0,1e9]，所以肯定要离散化处理，刚开始我的思路是线段树存当前区间的个数，然后每个sum去二分[1,m]有余5是3的个数的值，发现还是会T。

正解是：每个线段树节点维护5个值，即当前区间中%5==0,1,2,3,4的总和，这样的话，左右儿子合并的时候，左儿子直接加到父节点上，右儿子根据左儿子中的个数加到父节点上，这样的话，我们还需要一个siz数组记录下当前节点中有几个数，具体的话是

siz[id] = siz[id * 2] + siz[id * 2 + 1];
for (int i = 0; i < 5; ++i)Tree[id][i] = 0;
for (int i = 0; i < 5; ++i)
{
Tree[id][i] += Tree[id * 2][i];
Tree[id][(i + siz[id * 2]) % 5] += Tree[id * 2 + 1][i];
}

这样的话，应对每个sum查询，答案就是Tree[1][3]；

以下是代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cstring>
#include<string>
#include<vector>
#include<unordered_set>
#include<unordered_map>
#include<cmath>
using namespace std;

#define ll long long
#define lson l,mid,id<<1
#define rson mid+1,r,id<<1|1

typedef pair<int, int>pii;
typedef pair<ll, ll>pll;

const int MAXN = 100005;
const int MAXM = 1000005;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const double FINF = 1e30;

struct lx
{
string s;
ll num;
}que[MAXN];
vector<ll>v;
int getid(ll x) { return lower_bound(v.begin(), v.end(), x) - v.begin() + 1; }
ll Tree[MAXN * 4][5];
int siz[MAXN * 4];
void update(int l, int r, int id, int x, int val)
{
if (l == r && x == l)
{
for (int i = 0; i < 5; ++i)Tree[id][i] = 0;
Tree[id][1] = v[x - 1] * val;
siz[id] = val;
return;
}
int mid = (l + r) / 2;
if (x <= mid)update(lson, x, val);
else update(rson, x, val
4000
);
siz[id] = siz[id * 2] + siz[id * 2 + 1]; for (int i = 0; i < 5; ++i)Tree[id][i] = 0; for (int i = 0; i < 5; ++i) { Tree[id][i] += Tree[id * 2][i]; Tree[id][(i + siz[id * 2]) % 5] += Tree[id * 2 + 1][i]; }
}
int main()
{
int n;
scanf("%d", &n);
memset(Tree, 0, sizeof(Tree));
for (int i = 0; i < n; ++i)
{
cin >> que[i].s;
if (que[i].s == "sum")que[i].num = -1;
else cin >> que[i].num;
if (que[i].s == "add")v.push_back(que[i].num);
}
sort(v.begin(), v.end());
int len = v.size();
for (int i = 0; i < n; ++i)
{
if (que[i].s == "add")update(1, len, 1, getid(que[i].num), 1);
else if (que[i].s == "del")update(1, len, 1, getid(que[i].num), 0);
else
{
printf("%I64d\n", Tree[1][3]);
}
}
}