CF768B Code For 1

Jon fought bravely to rescue the wildlings who were attacked by the white-walkers at Hardhome. On his arrival, Sam tells him that he wants to go to Oldtown to train at the Citadel to become a maester, so he can return and take the deceased Aemon's place as maester of Castle Black. Jon agrees to Sam's proposal and Sam sets off his journey to the Citadel. However becoming a trainee at the Citadel is not a cakewalk and hence the maesters at the Citadel gave Sam a problem to test his eligibility.

Initially Sam has a list with a single element n . Then he has to perform certain operations on this list. In each operation Sam must remove any element x , such that x>1 , from the list and insert at the same position 

, 

, 

sequentially. He must continue with these operations until all the elements in the list are either 0 or 1 .

Now the masters want the total number of 1 s in the range l to r ( 1 -indexed). Sam wants to become a maester but unfortunately he cannot solve this problem. Can you help Sam to pass the eligibility test?

输入格式

The first line contains three integers n , l , r ( 0<=n&lt;2^{50} , 0<=r−l<=105 , r>=1 , l>=1 ) – initial element and the range l to r .

It is guaranteed that r is not greater than the length of the final list.

输出格式

Output the total number of 1 s in the range l to r in the final sequence.

Examples

Inputcopy	Outputcopy
7 2 5	4

Inputcopy	Outputcopy
10 3 10	5

Note

Consider first example:

Elements on positions from 2-nd to 5-th in list is [1, 1, 1, 1]. The number of ones is 4.

For the second example:

Elements on positions from 3-rd to 10-th in list is [1, 1, 1, 0, 1, 0, 1, 0]. The number of ones is 5.

题意翻译

• 有一个序列，初始时只有一个数 n(0≤n≤250)。
• 对于序列中每一个 >1 的数，拆分成三个数 ⌊2n​⌋,nmod2,⌊2n​⌋ 并替换原数，直到序列中没有>1的数为止。
• 查询最终序列中 [l,r](1≤l,r) 中有多少 1。
• 同时 l,r 满足 0≤r−l≤105。

#include<bits/stdc++.h>
#define int long long
#define N 1000005
#define INF 0x3f3f3f3f
using namespace std;
int f(int n,int p)
{
   int mid=1; //初始化mid为1，保证后续能够计算整个序列的中间位置
   int y=n; 
   while(y>>1) //每次对y/2，mid*2,当y=0时，mid正好为中间位置
   {
      y>>=1;
      mid<<=1; }
      if(p<mid) return f(n>>1,p); //二分，如果所查询的位置在mid之前，那递归前半段,并继续查找p的位置
      else if(p>mid) return f(n>>1,p-mid); //如果查询的位置在mid之后，那递归后半段
      else return n&1; //如果找到了，那就对n取模，因为这个位置只能由n%2产生，就算之前是n/2的，那也是其中序列里n%2产生的
  
}
signed main()
 {
    ios::sync_with_stdio(false);
    int n,l,r;
    cin>>n>>l>>r;
    int ans=0;
    for(int i=l;i<=r;i++)
   ans+=f(n,i);   //递归要查询的每一个值
   cout<<ans<<endl;
    return 0; 
 }