cf85D Sum of Medians

In one well-known algorithm of finding the k-th order statistics we should divide all elements into groups of five consecutive elements and find the median of each five. A median is called the middle element of a sorted array (it's the third largest element for a group of five). To increase the algorithm's performance speed on a modern video card, you should be able to find a sum of medians in each five of the array.

A sum of medians of a sorted k-element set S = {a₁, a₂, ..., a_k}, where a₁ < a₂ < a₃ < ... < a_k, will be understood by as

The operator stands for taking the remainder, that is stands for the remainder of dividing x by y.

To organize exercise testing quickly calculating the sum of medians for a changing set was needed.

Input

The first line contains number n (1 ≤ n ≤ 10⁵), the number of operations performed.

Then each of n lines contains the description of one of the three operations:

• add x — add the element x to the set;
• del x — delete the element x from the set;
• sum — find the sum of medians of the set.

For any add x operation it is true that the element x is not included in the set directly before the operation.

For any del x operation it is true that the element x is included in the set directly before the operation.

All the numbers in the input are positive integers, not exceeding 10⁹.

Output

For each operation sum print on the single line the sum of medians of the current set. If the set is empty, print 0.

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams (also you may use the %I64d specificator).

Examples

Input

6
add 4
add 5
add 1
add 2
add 3
sum

Output

3

Input

14
add 1
add 7
add 2
add 5
sum
add 6
add 8
add 9
add 3
add 4
add 10
sum
del 1
sum

Output

5
11
13

 

要有一个支持自动排序的数据结构，然后支持询问所有rnk=k*5+3的数字的和

离散完实际上只有10w个不同的数

按照rnk建个线段树，每个节点存5个点，sum[i]是当前rnk%5==i的数字之和

每次加一个数，只要在rnk[a[i]]+1到n上面所有sum[k]往右移一位

 

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<cstdlib>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<deque>
  9 #include<set>
 10 #include<map>
 11 #include<ctime>
 12 #define LL long long
 13 #define inf 0x7ffffff
 14 #define pa pair<int,int>
 15 #define mkp(a,b) make_pair(a,b)
 16 #define pi 3.1415926535897932384626433832795028841971
 17 using namespace std;
 18 inline LL read()
 19 {
 20     LL x=0,f=1;char ch=getchar();
 21     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 22     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 23     return x*f;
 24 }
 25 char s[10];
 26 int n,m,x;
 27 int d[100010];
 28 map<int,int>mp;
 29 struct opr{int op,x;}q[100010];
 30 struct segtree{
 31     LL sum[5];
 32     int l,r,tag,tot;
 33 }t[100010*8];
 34 LL swp[5];
 35 inline void gao(int k,int tt)
 36 {
 37     int t2=0;t[k].tot+=tt;
 38     for (int i=0;i<5;i++)
 39     {
 40         t2=i+tt;if (t2>=5)t2-=5;if (t2<0)t2+=5;
 41         swp[t2]=t[k].sum[i];
 42     }
 43     for (int i=0;i<5;i++)t[k].sum[i]=swp[i];
 44 }
 45 inline void pushdown(int k)
 46 {
 47     int tt=t[k].tag;t[k].tag=0;
 48     t[k<<1].tag+=tt;t[k<<1|1].tag+=tt;
 49     if (t[k<<1].tag>=5)t[k<<1].tag-=5;if (t[k<<1].tag<0)t[k<<1].tag+=5;
 50     if (t[k<<1|1].tag>=5)t[k<<1|1].tag-=5;if (t[k<<1|1].tag<0)t[k<<1|1].tag+=5;
 51     gao(k<<1,tt);gao(k<<1|1,tt);
 52 }
 53 inline void update(int k)
 54 {
 55     for (int i=0;i<5;i++)t[k].sum[i]=t[k<<1].sum[i]+t[k<<1|1].sum[i];
 56 }
 57 inline void buildtree(int now,int l,int r)
 58 {
 59     t[now].l=l;t[now].r=r;
 60     if (l==r)return;
 61     int mid=(l+r)>>1;
 62     buildtree(now<<1,l,mid);
 63     buildtree(now<<1|1,mid+1,r);
 64 }
 65 inline void rotate(int now,int x,int y,int d)
 66 {
 67     pushdown(now);
 68     int l=t[now].l,r=t[now].r;
 69     if(l==x&&r==y)
 70     {
 71         t[now].tag+=d;
 72         gao(now,d);
 73         return;
 74     }
 75     int mid=(l+r)>>1;
 76     if (y<=mid)rotate(now<<1,x,y,d);
 77     else if (x>mid)rotate(now<<1|1,x,y,d);
 78     else rotate(now<<1,x,mid,d),rotate(now<<1|1,mid+1,y,d);
 79     update(now);
 80 }
 81 inline void insert(int now,int x,int z)
 82 {
 83     pushdown(now);
 84     int l=t[now].l,r=t[now].r;
 85     if (l==x&&r==x)
 86     {
 87         if (z==1)t[now].sum[t[now].tot%5]+=d[x];
 88         else if (z==-1)t[now].sum[t[now].tot%5]-=d[x];
 89         return;
 90     }
 91     int mid=(l+r)>>1;
 92     if (x<=mid)insert(now<<1,x,z);
 93     else insert(now<<1|1,x,z);
 94     update(now);
 95 }
 96 int main()
 97 {
 98     m=read();
 99     for (int i=1;i<=m;i++)
100     {
101         scanf("%s",s+1);
102         if (s[1]=='a')
103         {
104             x=read();
105             q[i].op=1;q[i].x=x;
106             d[++n]=x;
107         }else if (s[1]=='d')
108         {
109             x=read();
110             q[i].op=2;q[i].x=x;
111         }else q[i].op=3;
112     }
113     if (!n)
114     {
115         for (int i=1;i<=m;i++)puts("0");
116         return 0;
117     }
118     sort(d+1,d+n+1);
119     for (int i=1;i<=n;i++)mp[d[i]]=i;
120     for (int i=1;i<=m;i++)if (q[i].x)q[i].x=mp[q[i].x];
121     buildtree(1,1,n);
122     for (int i=1;i<=m;i++)
123     {
124         if (q[i].op==1)
125         {
126             rotate(1,q[i].x,n,1);
127             insert(1,q[i].x,1);
128         }else if (q[i].op==2)
129         {
130             rotate(1,q[i].x,n,-1);
131             insert(1,q[i].x,-1);
132         }else if (q[i].op==3)
133         {
134             printf("%lld\n",t[1].sum[3]);
135         }
136     }
137 }

cf85D

 

转载于:https://www.cnblogs.com/zhber/p/7284725.html