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;

}