hdu 4288 Coder（单点操作，查询）

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<<a[i]<<" "; cout<<endl;
        if(tot==0)
            memset(tree[1].sum,0,sizeof(tree[1].sum));
        else
            build(1,tot,1);
        rep(i,1,n){
            int pos=lower_bound(a,a+tot,q[i])-a+1;
            if(ope[i][0]=='a'){
                update(1,pos,q[i],1,tot,1);
                continue;
            }
            if(ope[i][0]=='d'){
                update(-1,pos,q[i],1,tot,1);
                continue;
            }
            printf("%I64d\n",tree[1].sum[2]);
        }
    }
    //fclose(stdin);
}