题意：

三种操作：

1. add x – add the element x to the set;

2. del x – remove the element x from the set;

3. sum – find the digest sum of the set.

The digest sum should be understood by：sum(ai) where i % 5 ==3 

where the set S is written as {a₁, a₂, ... , a_k} satisfying a₁ < a₂ < a₃ < ... < a_k 

 

数据规格：

N ( 1 <= N <= 10⁵ )

1 <= x <= 10⁹.

 

思路：

BC做过一题和这个一模一样的，，，哎，，，这题早就做过了，可是BC却没有做出来，，，

这个还要离线处理，因为x很大

 

代码：

int const maxn=1e5+5;struct node{    int cnt;    ll sum[5];}tree[maxn<<2];int n,tot;int q[maxn],a[maxn];char ope[maxn][15];void build(int l,int r,int rt){    tree[rt].cnt=0;    memset(tree[rt].sum,0,sizeof(tree[rt].sum));    if(l==r) return;    int m=(l+r)>>1;    build(lson);    build(rson);}void pushUp(int rt){    rep(i,0,4)        tree[rt].sum[i]=tree[rt<<1].sum[i]+tree[rt<<1|1].sum[((5-tree[rt<<1].cnt%5)%5+i)%5];}void update(int k,int pos,int num,int l,int r,int rt){    tree[rt].cnt+=k;    if(l==r){        tree[rt].sum[0]+=(k*num);        return;    }    int m=(l+r)>>1;    if(pos<=m)        update(k,pos,num,lson);    else        update(k,pos,num,rson);    pushUp(rt);}int main(){    //freopen("test.in","r", stdin);    while(scanf("%d",&n)!=EOF){        tot=0;        rep(i,1,n){            scanf("%s",ope[i]);            if(ope[i][0]!='s'){                scanf("%d",&q[i]);                a[tot++]=q[i];            }        }        sort(a,a+tot);        tot=unique(a,a+tot)-a;        //rep(i,0,tot-1) cout<
     
      <<" "; cout<