题目链接：Codeforces 85D - Sum of Medians
题目大意：N个操作，add x：向集合中添加x；del x：删除集合中的x；sum：将集合排序后，将集合中所有下标i % 5 = 3的元素累加求和。
解题思路：线段树单点更新，每个点维护5个值，分别表示从该段区间中i % 5 = t的和。然后两端区间合并时只需要根据左孩子中元素的个数合并。所以有一个c表示区间上元素的个数。因为有相同的数，所以要离线操做，将所有的数映射成位置，但是对于del则不需要映射，因为集合中肯定有才能减掉。那么add和sum操作都是可以搞定了，只剩下del操作，对于del x，x肯定在集合中出现过，所以每次删除第一个x即可，如果快速查找，要借助map和一个辅助数组，因为删除一个后要重新映射，所以借助辅助数组。
#include <cstdio>#include <cstring>#include <map>#include <vector>#include <algorithm>using namespace std;typedef long long ll;const int mod = 5;const int maxn = 1e5+5;int N, M, pos[maxn], v[maxn];map<ll, int> G;#define lson(x) ((x)<<1)#define rson(x) (((x)<<1)|1)int lc[maxn << 2], rc[maxn << 2], c[maxn << 2];ll s[maxn << 2][6];inline void maintain (int u, int d) {c[u] += d;memset(s[u], 0, sizeof(s[u]));s[u][0] = (c[u] ? pos[lc[u]] : 0);}inline void pushup(int u) {int t = c[lson(u)] % mod;c[u] = c[lson(u)] + c[rson(u)];for (int i = 0; i < mod; i++)s[u][i] = s[lson(u)][i] + s[rson(u)][(i + mod - t) % mod];}void build (int u, int l, int r) {lc[u] = l;rc[u] = r;c[u] = 0;memset(s[u], 0, sizeof(s[u]));if (l == r)return;int mid = (l + r) / 2;build(lson(u), l, mid);build(rson(u), mid + 1, r);pushup(u);}void modify (int u, int x, int d) {if (lc[u] == x && rc[u] == x) {maintain(u, d);return;}int mid = (lc[u] + rc[u]) / 2;if (x <= mid)modify(lson(u), x, d);elsemodify(rson(u), x, d);pushup(u);}struct OP {int p, k, id;OP (int k = 0, int p = 0, int id = 0) {this->k = k;this->p = p;this->id = id;}friend bool operator < (const OP& a, const OP& b) {if (a.k == 0)return false;if (b.k == 0)return true;if (a.p != b.p)return a.p < b.p;return a.id < b.id;}};inline bool cmp (const OP& a, const OP& b) {return a.id < b.id;}vector<OP> vec;void init () {scanf("%d", &N);char op[5];int x;for (int i = 1; i <= N; i++) {scanf("%s", op);if (op[0] == 's')vec.push_back(OP(0, 0, i));else {scanf("%d", &x);vec.push_back(OP(op[0] == 'a' ? 1 : -1, x, i));}}M = 1;sort(vec.begin(), vec.end());for (int i = 0; i < N; i++) {if (vec[i].k < 0)continue;if (vec[i].k == 0)break;pos[M] = vec[i].p;vec[i].p = M++;}build(1, 1, M);}void solve () {sort(vec.begin(), vec.end(), cmp);for (int i = 0; i < N; i++) {//printf("%d %d!\n", vec[i].k, pos[vec[i].p]);if (vec[i].k == 0)printf("%lld\n", s[1][2]);else if (vec[i].k == -1) {int tmp = vec[i].p;v[G[tmp]] = 0;modify(1, G[tmp], -1);if (G[tmp] <= N && v[G[tmp]+1] && pos[G[tmp]] == pos[G[tmp]+1])G[tmp]++;elseG[tmp] = 0;} else {int tmp = pos[vec[i].p];v[vec[i].p] = 1;modify(1, vec[i].p, 1);if (G[tmp] == 0)G[tmp] = vec[i].p;}}}int main () {init();solve();return 0;}
