codeforces-768B Code for 1

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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 1s 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?
Input
The first line contains three integers n, l, r (0 ≤ n < 250, 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.
Output
Output the total number of 1s in the range l to r in the final sequence.
Examples
input
7 2 5
output
4
input
10 3 10
output
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.

          7

    2     1     2

1   1  1  1  1  1  1
①  ②  ③ ④ ⑤ ⑥ ⑦ 
   -->
   -->
         ④1
     ②1      1⑥
    1    1  1    1
    ①   ③ ⑤    ⑦

叶子结点可以变成一个以中序遍历的二叉树
奇数结点一定为1，因为任意的一个数一直除二，最后一定为1，然后停止。
偶数结点即它所在的层数产生的余数；

#include<math.h>
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
typedef long long ll;
vector<int>v;
int main()
{
    ll n,l,r;
    while(~scanf("%I64d",&n)){
        v.clear();
        scanf("%I64d%I64d",&l,&r);
        if(n==0){
            cout<<"0\n";
            continue;
        }
        ll N=n,ceng=0,ans=0;
        while(N!=1&&N!=0){
            ceng++;
            v.push_back(N%2);
            N/=2;
        }
        for(ll i=l;i<=r;++i){
            if(i%2) ans++;
            else{
                ll j=i,s=0;//s为倒数第几层
                while(j%2==0){
                    s++;j/=2;
                }
                ans+=v[ceng-s];
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}