再来单点更新我就呵呵了

http://codeforces.com/problemset/problem/85/D

题意：有三种类型的操作，1."add x"表示往集合里添加数x。2.“del x”表示将集合中数x删除。3.“sum”求出从小到大排列的集合中下标模5为3的数的和。集合中的数都是唯一的。

线段树。在线段树中维护当前这个集合中数的个数cnt，和所有的数模5为0……4内的数的和设为mod[0...4]。在进行区间合并的时候，父区间里的mod[0...4]首先等于左子区间里的mod[0...4]，设要加入的右子区间的数为mod[i]，则它应该加到父区间的mod[左子区间的cnt+i]。

#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <algorithm>#include <vector>using namespace std;typedef long long LL;#define LL(x) (x<<1)#define RR(x) (x<<1|1)const int N=100005;char str[5];vector<int> y;map<int,int> imap;struct OP{int type,valu;void input(){scanf("%s",str);if(strcmp(str,"add")==0) type=0;else if(strcmp(str,"del")==0) type=1;else type=2;if(type==2) return;scanf("%d",&valu);y.push_back(valu);}}op[N];struct node{int lft,rht,cnt;LL mod[5];int mid(){return lft+(rht-lft)/2;}void clear(){memset(mod,0,sizeof(mod));}void change(){if(cnt==0) mod[1]=0;else mod[1]=y[lft];}};struct Segtree{node tree[N*4];void relax(int ind){int c=tree[LL(ind)].cnt;for(int i=0;i<5;i++) tree[ind].mod[i]=tree[LL(ind)].mod[i];for(int i=0;i<5;i++) tree[ind].mod[(i+c)%5]+=tree[RR(ind)].mod[i];}void build(int lft,int rht,int ind){tree[ind].lft=lft;tree[ind].rht=rht;tree[ind].cnt=0;tree[ind].clear();if(lft!=rht){int mid=tree[ind].mid();build(lft,mid,LL(ind));build(mid+1,rht,RR(ind));}}void updata(int pos,int ind,int valu){tree[ind].cnt+=valu;if(tree[ind].lft==tree[ind].rht) tree[ind].change();else {int mid=tree[ind].mid();if(pos<=mid) updata(pos,LL(ind),valu);else updata(pos,RR(ind),valu);relax(ind);}}}seg;int main(){int n;while(scanf("%d",&n)!=EOF){int sc=0;y.clear(); imap.clear();for(int i=0;i<n;i++) op[i].input();sort(y.begin(),y.end());y.erase(unique(y.begin(),y.end()),y.end());for(int i=0;i<(int)y.size();i++) imap.insert(make_pair(y[i],sc++));seg.build(0,sc,1);for(int i=0;i<n;i++){if(op[i].type==0) seg.updata(imap[op[i].valu],1,1);else if(op[i].type==1) seg.updata(imap[op[i].valu],1,-1);else printf("%I64d\n",seg.tree[1].mod[3]);}}return 0;}

我忽略了从小到大排列的错误代码

#include <iostream>#include <cstdio>#include <string>#include <cstring>using namespace std;#define lz 2*u,l,mid#define rz 2*u+1,mid+1,rint n, l, r,t;const int maxn=100007;int sum[4*maxn], a[maxn];int cnt=0;void build(int u, int l, int r){    //cout<<"haha"<<endl;    sum[u]=0;    if(l==r)    {        sum[u]=a[l];        return ;    }    int mid=(l+r)>>1;    build(lz);    build(rz);    sum[u]=sum[2*u]+sum[2*u+1];}void Update(int u, int l, int r, int pos, int c){    //cout<<"haha"<<endl;    if(l==r) //(l%5==3)    {        //cout<<"haha"<<endl;        if(l%5==3)           sum[u]+=c;        return ;    }    int mid=(l+r)>>1;    if(pos<=mid) Update(lz,pos,c);    else Update(rz,pos,c);    sum[u]=sum[2*u]+sum[2*u+1];}int Query(int u, int l, int r, int tl, int tr){    if(tl<=l&&r<=tr) return sum[u];    int mid=(l+r)>>1;    if(tr<=mid) return Query(lz,tl,tr);    else if(tl>mid) return Query(rz,tl,tr);    else    {        int t1=Query(lz,tl,mid);        int t2=Query(rz,mid+1,tr);        return t1+t2;    }}int main(){    freopen("in.txt","r",stdin);    memset(a,0,sizeof(a));    cin>>t;    n=100007;    build(1,1,n);    char op[10];    while(t--)    {        scanf("%s",op);        if(op[0]=='s')        {            printf("%d\n",sum[1]);        }        else if(op[0]=='d')        {            scanf("%d",&l);            Update(1,1,n,l,-l);        }        else        {            cnt++;            //printf("fafa\n");            scanf("%d",&l);            //printf("%d\n",l);            Update(1,1,n,l,l);            //printf("fafa\n");            //cout<<t<<endl;        }    }    return 0;}

接着用STL秒杀

#include<bits/stdc++.h>using namespace std;typedef long long ll;int main() {vector<int> v;int n; cin>>n;for (int i=0; i<n; i++) {string s; cin>>s;if(s[0]=='s') {ll ret=0;for (int i=2; i<v.size(); i+=5)ret+=v[i];cout << ret << endl;} else {int x; cin>>x;if(s[0]=='a')v.insert(lower_bound(v.begin(),v.end(),x),x);elsev.erase(lower_bound(v.begin(),v.end(),x));}}return 0;}