学步园

题意略。

离线处理，离散化。然后就是简单的线段树了。需要根据mod 5的值来维护。具体看代码了。

/*
    线段树+离散化+离线处理
*/

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
#define N 100010
ll sum[N<<2][5];
int a[N], n, m, cnt[N<<2];

struct node {
    char c;
    int x;
} q[N];
void Up(int rt) {
    cnt[rt] = cnt[rt<<1] + cnt[rt<<1|1];
    for (int i=0; i<5; i++)
        sum[rt][i] = sum[rt<<1][i] + sum[rt<<1|1][(i-cnt[rt<<1]%5+5)%5];
}
void add(int idx, int val, int l, int r, int rt) {
    if (l == r) {
        cnt[rt] = 1;
        sum[rt][1] = val;
        return ;
    }
    int mid = (l + r) >> 1;
    if (idx <= mid) add(idx, val, l, mid, rt<<1);
    else add(idx, val, mid+1, r, rt<<1|1);
    Up(rt);
}
void del(int idx, int l, int r, int rt) {
    if (l == r) {
        sum[rt][1] = cnt[rt] = 0;
        return ;
    }
    int mid = (l + r) >> 1;
    if (idx <= mid) del(idx, l, mid, rt<<1);
    else del(idx, mid+1, r, rt<<1|1);
    Up(rt);
}
int main() {

    char s[10];
    while (scanf("%d", &n) == 1) {
        m = 0;
        for (int i=0; i<n; i++) {
            scanf(" %s", s);
            q[i].c = s[0];
            if (s[0] != 's') {
                scanf("%d", &q[i].x);
                a[m++] = q[i].x;
            }
        }
        sort(a, a+m);
        m = unique(a, a+m) - a;
        memset(cnt, 0, sizeof(cnt));
        memset(sum, 0, sizeof(sum));
        int pos;
        for (int i=0; i<n; i++) {
            if (q[i].c == 's') printf("%I64d\n", sum[1][3]);
            else {
                pos = lower_bound(a, a+m, q[i].x) - a + 1;
                if (q[i].c == 'a') add(pos, q[i].x, 1, m, 1);
                else del(pos, 1, m, 1);
            }
        }
    }


    return 0;
}