HDU 4288线段树

//2012成都网赛a题

思路： 关键是想到直接 将所有出现过的数字作为叶子节点建一棵空的树（就是树中节点暂时不保存任何值）。然后add就添加，del就删，直接模板水之~~~

这道题stl的vector也可以水过 ，真没天理呀。弱菜表示不懂肿么算stl的复杂度了~~~

#include<stdio.h>
#include<string.h>
#include <string>
#include <cmath>
#include <iostream>
#include <map>
#include<vector>
#include<queue>
#include<algorithm>
#define fr(i,s,n) for(int i=s;i<n;i++)
#define pf printf
#define sf scanf
#define sfv1(a) scanf("%d",&a)
#define sfv2(n,m) scanf("%d%d",&n,&m)
#define sfv3(u,v,w) scanf("%d%d%d",&u,&v,&w)
#define sfstr(c) scanf("%s",c)
#define pfv1(a) printf("%d\n",a)
#define fi freopen("in.txt","r",stdin)
#define fo freopen("out.txt","w",stdout)
#define cl(a) memset(a,0,sizeof(a))
#define me(a,x) memset(a,x,sizeof(a))

#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define Mid (L+R)>>1
using namespace std;
typedef __int64 ll;

const int N=200010;

struct Stree{
int sum;
ll ans[5];
}stree[N*4];

struct Op{
char c[10];
int x;
}op
;

int num
;
void up(int pos){
stree[pos].sum=stree[LL(pos)].sum+stree[RR(pos)].sum;
fr(i,0,5){
stree[pos].ans[i]=stree[LL(pos)].ans[i]+stree[RR(pos)].ans[(i-stree[LL(pos)].sum%5+5)%5];
}
}

int m;
void build(int pos,int L,int R){
stree[pos].sum=0;
fr(i,0,5){
stree[pos].ans[i]=0;
}
if (L==R){
return;
}
m=Mid;
build(LL(pos),L,m);
build(RR(pos),m+1,R);
}

int flag;
void update(int pos,int L,int R,int k){
if (L==R&&L==k){
stree[pos].sum+=flag;
stree[pos].ans[0]+=(ll(flag*num[k]));
return ;
}
m=Mid;
if (m<k) update(RR(pos),m+1,R,k);
else update(LL(pos),L,m,k);
up(pos);
}
int n;
int main(){
//fi;
char c[5];
while(sfv1(n)!=EOF){
int tot=0,cnt;
fr(i,0,n){
sfstr(op[i].c);
if (op[i].c[0]!='s'){
sfv1(op[i].x);
num[tot++]=op[i].x;
}
}
if (tot>0){
sort(num,num+tot);
cnt=unique(num,num+tot)-num;
build(1,0,cnt-1);
}
int pos;
fr(i,0,n){
if (op[i].c[0]=='s'){
pf("%I64d\n",stree[1].ans[2]);
}else{
pos=lower_bound(num,num+cnt,op[i].x)-num;
if (op[i].c[0]=='a') flag=1;
else flag=-1;
update(1,0,cnt-1,pos);
}
}
}
return 0;
}