3 seconds
256 megabytes
standard input
standard output
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, ..., ak}, where a1 < a2 < a3 < ... < ak, 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.
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.
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).
6
add 4
add 5
add 1
add 2
add 3
sum
3
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
5
11
13
#include<bits/stdc++.h>
using namespace std;
#define ll unsigned long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<endl;
const int N=2e5+,M=1e6+,inf=1e9+;
const ll INF=1e18+,mod=;
int n,tree[N];
int lowbit(int x)
{
return x&-x;
}
void update(int x,int c)
{
while(x<1e5+)
{
tree[x]+=c;
x+=lowbit(x);
}
}
int getsum(int x)
{
int sum=;
while(x>)
{
sum+=tree[x];
x-=lowbit(x);
}
return sum;
}
struct is
{
int lazy;
ll ans[];
}a[N<<];
ll temp[];
void pushup(int pos)
{
for(int i=;i<;i++)
a[pos].ans[i]=a[pos<<].ans[i]+a[pos<<|].ans[i];
}
void change(int pos,int x)
{
x=(x%+)%;
int ji=;
for(int i=;i<;i++)
temp[i]=a[pos].ans[i];
for(int i=x;i<;i++)
a[pos].ans[i]=temp[ji++];
for(int i=;i<x;i++)
a[pos].ans[i]=temp[ji++];
}
void pushdown(int pos)
{
if(a[pos].lazy)
{
a[pos<<].lazy+=a[pos].lazy;
a[pos<<|].lazy+=a[pos].lazy;
change(pos<<,a[pos].lazy);
change(pos<<|,a[pos].lazy);
a[pos].lazy=;
}
}
void build(int l,int r,int pos)
{
a[pos].lazy=;
memset(a[pos].ans,,sizeof(a[pos].ans));
if(l==r)return;
int mid=(l+r)>>;
build(l,mid,pos<<);
build(mid+,r,pos<<|);
}
void update(int L,int R,int c,int l,int r,int pos)
{
if(L<=l&&r<=R)
{
a[pos].lazy+=c;
change(pos,c);
return;
}
pushdown(pos);
int mid=(l+r)>>;
if(L<=mid)
update(L,R,c,l,mid,pos<<);
if(R>mid)
update(L,R,c,mid+,r,pos<<|);
pushup(pos);
}
void point(int p,int k,int c,int l,int r,int pos)
{
if(l==r)
{
a[pos].ans[k]+=c;
return;
}
pushdown(pos);
int mid=(l+r)>>;
if(p<=mid)
point(p,k,c,l,mid,pos<<);
else
point(p,k,c,mid+,r,pos<<|);
pushup(pos);
}
char str[N][];
int b[N];
int s[N],cnt;
int getpos(int x)
{
int pos=lower_bound(s+,s++cnt,x)-s;
return pos;
}
int main()
{
int n;
scanf("%d",&n);
for(int i=;i<=n;i++)
{
scanf("%s",str[i]);
if(str[i][]=='a'||str[i][]=='d')
{
scanf("%d",&b[i]);
s[++cnt]=b[i];
}
}
sort(s+,s++cnt);
cnt=max(,cnt);
build(,cnt,);
for(int i=;i<=n;i++)
{
//cout<<str[i]<<endl;
if(str[i][]=='a')
{
int x=getpos(b[i]);
int now=getsum(x-);
now%=;
//cout<<x<<" "<<now<<" "<<b[i]<<endl;
update(x,);
update(x+,cnt,,,cnt,);
point(x,now,b[i],,cnt,);
}
else if(str[i][]=='d')
{
int x=getpos(b[i]);
int now=getsum(x-);
now%=;
update(x,-);
point(x,now,-b[i],,cnt,);
update(x+,cnt,-,,cnt,);
}
else
printf("%lld\n",a[].ans[]);
//printf("%lld\n",a[1].ans[2]);
}
return ;
}