HDU-4973-A simple simulation problem. (Binary + tree array)

Problem descriptionthere 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 3 4 4 5 }. after some operations this queue cocould become V Ery long, and I can't figure out maximum count of cells of same type. Could you help me?
Inputthe 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)
Outputfor 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

15 5D 5 5Q 5 6D 2 3D 1 2Q 1 7

 
Sample output

Case #1:23

 
Source2014 multi-university training contest 10
Idea: Perform binary search directly when finding a location. The two arrays Save the interval and the single point count respectively.

#include <stdio.h>#define max(A,B)(A>B?A:B)long long node[50005],cnt[50005];int n,m;long long sum(int x){    long long res=0;    while(x>=1)    {        res+=node[x];        x-=x&-x;    }    return res;}void add(int x,long long val){    while(x<=n)    {        node[x]+=val;        x+=x&-x;    }}int getpos(long long val){    int l=1,r=n,mid,res;    while(l<=r)    {        mid=(l+r)>>1;        if(sum(mid)>=val)        {            res=mid;            r=mid-1;        }        else l=mid+1;    }    return res;}int main(){    int T,i,l,r,cases=1;    long long a,b,tempa,tempb,ans;    char s[5];    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        printf("Case #%d:\n",cases++);        for(i=1;i<=n;i++) node[i]=0,cnt[i]=1;        for(i=1;i<=n;i++) add(i,1);        while(m--)        {            scanf("%s%I64d%I64d",s,&a,&b);            if(s[0]=='D')            {                l=getpos(a);                r=getpos(b);                if(l==r)                {                    cnt[l]+=b-a+1;                    add(l,b-a+1);                }                else                {                    tempa=sum(l)-a+1;                    tempb=b-sum(r-1);                    cnt[l]+=tempa;                    add(l,tempa);                    cnt[r]+=tempb;                    add(r,tempb);                    for(i=l+1;i<=r-1;i++)                    {                        add(i,cnt[i]);                        cnt[i]+=cnt[i];                    }                }            }            else            {                l=getpos(a);                r=getpos(b);                if(l==r) printf("%I64d\n",b-a+1);                else                {                    ans=max(sum(l)-a+1,b-sum(r-1));                    for(i=l+1;i<=r-1;i++) ans=max(ans,cnt[i]);                    printf("%I64d\n",ans);                }            }        }    }}

HDU-4973-A simple simulation problem. (Binary + tree array)