HDU 4288 Coder

線段樹。數據不算多（1e5），但是有點大（1e9），就要離散化一下，又是動態插入刪除，所以必須先離線才能離散化。換句話說就是先把所有可能出現的數字都存起來，然後再把它們映射到0到num，就叫做離線+離散化。

利用sum數組記錄當前節點有多少子節點，ans分別記錄當前節點的5種餘數的和。每次詢問時輸出ans[2][1]，就是根節點餘2（第三個）的數的和，會超過int。然後pushup操作中，每個ans都是左右兩部分的和，注意右半部分，需要判斷一下。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<queue>
#include<cmath>
#include<map>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("\n")
#define PI(n) printf("%d\n", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s\n", s)
#define PSS(s) printf("%s ", s)
#define PC(n) printf("Case %d: ", n)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 9901;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int N = 111111;
const int M = 66;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

struct OP
{
    int op, v;
}p[N];
int X[N], sum[N << 2], num, k;
LL ans[5][N << 2];

void pushup(int rt)
{
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    REP(i, 5)ans[i][rt] = ans[i][rt << 1] + ans[(i - sum[rt << 1] % 5 + 5) % 5][rt << 1 | 1];
}

void update(int p, int v, int l, int r, int rt)
{
    if(l == r)
    {
        ans[0][rt] = X[p] * v;
        sum[rt] += 2 * v - 1;
        return;
    }
    int m = (l + r) >> 1;
    if(p <= m)update(p, v, lson);
    else update(p, v, rson);
    pushup(rt);
}

int main()
{
    //freopen("input.txt", "r", stdin);

    int n, m;
    char s[5];
    while(RI(n) != EOF)
    {
        num = 0;
        CLR(sum, 0);
        CLR(ans, 0);
        REP(i, n)
        {
            RS(s);
            if(s[0] == 's')p[i].op = 2;
            else
            {
                RI(p[i].v);
                if(s[0] == 'd')p[i].op = 0;
                else p[i].op = 1;
                X[num++] = p[i].v;
            }
        }
        sort(X, X + num);
        num = unique(X, X + num) - X;
        REP(i, n)
        {
            if(p[i].op == 2)printf("%I64d\n", ans[2][1]);
            else
            {
                m = lower_bound(X, X + num, p[i].v) - X;
                update(m, p[i].op, 1, num, 1);
            }
        }
    }
    return 0;
}