Problem Description
There are n types of cells in the lab, numbered from 1 to n. These cells are put in a queue, the i-th cell belongs to type i. Each time I can use mitogen to double the cells in the interval [l, r]. For instance, the original queue is {1 2 3 3 4 5}, after using
a mitogen in the interval [2, 5] the queue will be {1 2 2 3 3 3 3 4 4 5}. After some operations this queue could become very long, and I can’t figure out maximum count of cells of same type. Could you help me?
Input
The first line contains a single integer t (1 <= t <= 20), the number of test cases.
For each case, the first line contains 2 integers (1 <= n,m<= 50000) indicating the number of cell types and the number of operations.
For the following m lines, each line represents an operation. There are only two kinds of operations: Q and D. And the format is:
“Q l r”, query the maximum number of cells of same type in the interval [l, r];
“D l r”, double the cells in the interval [l, r];
(0 <= r – l <= 10^8, 1 <= l, r <= the number of all the cells)
Output
For each case, output the case number as shown. Then for each query "Q l r", print the maximum number of cells of same type in the interval [l, r].
Take the sample output for more details.
Sample Input
1 5 5 D 5 5 Q 5 6 D 2 3 D 1 2 Q 1 7
Sample Output
Case #1:
2
3
#include<stdio.h>
#define L 50005
#define ll __int64
#define ls 2*i
#define rs 2*i+1
#define w(x) while(x)
#define max(a,b) (a>b?a:b)
int t,n,m,cas=1;
ll x,y,ans,res;
char str[10];
struct node
{
int l,r,flag;
ll maxn,sum;
} a[L<<2];
void PushUp(int i)
{
a[i].sum=a[2*i].sum+a[2*i+1].sum;
a[i].maxn=max(a[2*i].maxn,a[2*i+1].maxn);
}
void PushDown(int i)
{
if(a[i].flag)
{
a[ls].flag += a[i].flag;
a[rs].flag += a[i].flag;
w(a[i].flag--)
{
a[ls].maxn *= 2;
a[rs].maxn *= 2;
a[ls].sum *= 2;
a[rs].sum *= 2;
}
a[i].flag=0;
}
}
void init(int l,int r,int i)
{
a[i].l=l;
a[i].r=r;
a[i].flag=0;
a[i].maxn=1;
if(l==r)
{
a[i].sum=1;
return;
}
int mid=(l+r)/2;
init(l,mid,ls);
init(mid+1,r,rs);
PushUp(i);
}
int find(int i,ll x,ll &k,ll &tem)
{
if(a[i].l==a[i].r)
{
k=x;
tem=a[i].sum;
return a[i].l;
}
PushDown(i);
if(x<=a[ls].sum) return find(ls,x,k,tem);
else return find(rs,x-a[ls].sum,k,tem);
}
void insert1(int i,int p,ll v)
{
if(a[i].l==a[i].r)
{
a[i].maxn+=v;
a[i].sum+=v;
return;
}
PushDown(i);
int mid=(a[i].l+a[i].r)/2;
if(p<=mid) insert1(ls,p,v);
else insert1(rs,p,v);
PushUp(i);
}
void insert2(int l,int r,int i)
{
if(l<=a[i].l&&a[i].r<=r)
{
PushDown(i);
a[i].flag++;
a[i].sum*=2;
a[i].maxn*=2;
return;
}
PushDown(i);
int mid=(a[i].l+a[i].r)/2;
if(l<=mid) insert2(l,r,ls);
if(r>mid) insert2(l,r,rs);
PushUp(i);
}
void query(int l,int r,int i)
{
if(l<=a[i].l&&a[i].r<=r)
{
res=max(res,a[i].maxn);
return;
}
PushDown(i);
int mid=(a[i].l+a[i].r)/2;
if(l<=mid) query(l,r,ls);
if(r>mid) query(l,r,rs);
}
int main()
{
scanf("%d",&t);
w(t--)
{
scanf("%d%d",&n,&m);
init(1,n,1);
printf("Case #%d:\n",cas++);
w(m--)
{
scanf("%s%I64d%I64d",str,&x,&y);
ll lk,rk,tem;
int r=find(1,y,rk,tem);
int l=find(1,x,lk,tem);
if(str[0]=='D')
{
if(l==r)
{
insert1(1,l,rk-lk+1);
continue;
}
insert1(1,r,rk);
insert1(1,l,tem-lk+1);
if(r-l>1)
insert2(l+1,r-1,1);
}
else
{
ans=0;
if(l==r)
{
printf("%I64d\n",rk-lk+1);
continue;
}
ans=max(ans,max(rk,tem-lk+1));
if(r-l>1)
{
res=0;
query(l+1,r-1,1);
ans=max(ans,res);
}
printf("%I64d\n",ans);
}
}
}
return 0;
}