Aninteresting game hdu-5975（树状数组原理）

Let’s play a game.We add numbers 1,2...n in increasing order from 1 and put them into some sets. 
When we add i,we must create a new set, and put iinto it.And meanwhile we have to bring i−lowbit(i)+1,i−1i−lowbit(i)+1,i−1 from their original sets, and put them into the new set,too.When we put one integer into a set,it costs us one unit physical strength. But bringing integer from old set does not cost any physical strength. 
After we add 1,2...n,we have q queries now.There are two different kinds of query: 
1 L R:query the cost of strength after we add all of L,RL,R(1≤L≤R≤n) 
2 x:query the units of strength we cost for putting x(1≤x≤n) into some sets. 

InputThere are several cases,process till end of the input. 
For each case,the first line contains two integers n and q.Then q lines follow.Each line contains one query.The form of query has been shown above. 
n≤10^18,q≤10^5 
OutputFor each query, please output one line containing your answer for this querySample Input

10 2
1 8 9
2 6

Sample Output

9
2

        
 

Hint

lowbit(i) =i&(-i).It means the size of the lowest nonzero bits in binary of i. For example, 610=1102, lowbit(6) =102= 210
When we add 8,we should bring [1,7] and 8 into new set.
When we add 9,we should bring [9,8] (empty) and 9 into new set.
So the first answer is 8+1=9.
When we add 6 and 8,we should put 6 into new sets.
So the second answer is 2. 

 题意：有n个集合，第 i 个集合的数是[i-lowbit(i)+1,i-1]+i；有m个询问，询问方式有两种，第一种方式是集合[a,b]共有多少个数？第二种方式是数字x在几个集合里面？

input
先输入n和m，然后输入1或者2，若是1，接着输入a和b；若是2，输入x
Output
对于每个询问输出一个结果

思路：了解了有关于树状数组的知识后能够知道：

集合：1 2 3 4 5 6 7 8
长度：1 2 1 4 1 2 1 8
能够发现每k个 k ∈{1,2,4,8……}会出现1个k的倍数，所以我们能够知道[1,i]中有多少个不同的k以及每个k的个数p。所以能够求得[1,i]的长度和是多少，进而求得[a,b]的长度和。要找到x在几个集合中，可用x=x+lowbit（x）找出有多少满足条件的x，就能得到集合的个数

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <map>
#include <set>
#include <stdlib.h>
#include <stack>
#include <vector>
#include <cmath>
#define ll long long
using namespace std;
ll tmp[100];
ll count_ans(ll x)
{
    ll ans=0,cnt;
    for(int i=0;tmp[i]<=x;i++)
    {
        ans=ans+((x/tmp[i]-x/tmp[i+1])*tmp[i]);
    }
    return ans;
}
int main()
{
    ll n,a,b;
    int m,x;
    tmp[0]=1;
    for(int i=1;i<63;i++)
    {
        tmp[i]=tmp[i-1]*2;
    }
    while(~scanf("%lld%d",&n,&m))
    {
        while(m--)
        {
            scanf("%d",&x);
            if(x==1)
            {
                scanf("%lld%lld",&a,&b);
                printf("%lld
",count_ans(b)-count_ans(a-1));
            }
            if(x==2)
            {
                scanf("%lld",&a);
                int ans=0;
                while(a<=n)
                {
                    ans++;
                    a+=(a&(-a));
                }
                printf("%d
",ans);
            }
        }
    }
    return 0;
}