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, ... , ak} satisfying a1 < a2 < a3 < ... < ak

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);
}