hdu 4973 A simple simulation problem.（2014 Multi-University Training Contest 10）

A simple simulation problem.

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 392    Accepted Submission(s): 155

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

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

 

Sample Output

Case #1:23

 

树状数组做的

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int maxn=50010;long long sum[maxn];long long coun[maxn];int n,m;int low(int k){    return k&(-k);}void add(int i,long long v){    while(i<=n)    {        sum[i]+=v;        i+=low(i);    }}long long getsum(int k){    long long ans=0;    while(k>0)    {        ans+=sum[k];        k-=low(k);    }    return ans;}int find(long long k){  int mid,l=1,r=n;  while(l<r)  {      mid=(l+r)>>1;      long long temp=getsum(mid);      if(k>temp)      {          l=mid+1;      }      else      r=mid;  }  return l;}void update(long long l,long long r){    int L=find(l);    int R=find(r);    if(L==R)    {        add(L,r-l+1);        coun[L]+=(r-l+1);        return;    }    long long x=getsum(L);    long long y=getsum(R-1);    add(L,x-l+1);    coun[L]+=(x-l+1);    add(R,r-y);    coun[R]+=(r-y);    for(int i=L+1;i<R;i++)    {        add(i,coun[i]);        coun[i]+=coun[i];    }    return;}void quarry(long long l,long long r){    int L=find(l);    int R=find(r);    if(L==R)    {        printf("%I64d\n",r-l+1);        return;    }    long long maxn;    maxn=max(getsum(L)-l+1,r-getsum(R-1));    for(int i=L+1;i<R;i++)    {        if(maxn<coun[i])        maxn=coun[i];    }    printf("%I64d\n",maxn);    return;}int main(){   int t,ca=1;   scanf("%d",&t);   while(t--)   {       char s;       long long x,y;       scanf("%d%d",&n,&m);       memset(sum,0,sizeof(sum));       for(int i=1;i<=n;i++)       {           add(i,1);           coun[i]=1;       }       printf("Case #%d:\n",ca++);       while(m--)       {           scanf(" %c%I64d%I64d",&s,&x,&y);           if(s=='D')           {           update(x,y);           }           else           quarry(x,y);       }   }}