Codeforces 768B B. Code For 1

参考自：https://www.cnblogs.com/ECJTUACM-873284962/p/6423483.html

B. Code For 1

time limit per test：2 seconds

memory limit per test：256 megabytes

input：standard input

output：standard output

　　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.

思路：

　　给你一个数n和区间(l，r)，每次都能把任意数拆成 n/2，n%2，n/2 三个数，直到变成0和1，问区间l，r里有多少个1？

　　如 7 2 5

　　　　7 → 3  1  3；

　　　　3 → 1  1  1；

　　所以能拆成 7个 1，所以在2--5之间数字1的个数为4。

　　同理 10  3  10

　　　　10 → 5  0  5；

　　　　  5 → 2  1  2；

　　　　  2 → 1  0  1；

　　故拆成 → [ 1  0  1      1      1  0  1     0     1  0  1      1      1  0  1  ]

　　3--10之间数字1的个数为5.

解法：  分治的思想，二分法

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 ll n, l, r, s = , ans;
 void solve(ll a, ll b, ll l, ll r, ll d){//二分的思想
     if ( a > b || l > r )    return;
     else{
         ll mid = (a+b)/;
         if ( r < mid )solve(a,mid-,l,r,d/);
         else if ( mid < l )solve(mid+,b,l,r,d/);
         else {
             ans += d%;
             solve(a,mid-,l,mid-,d/);
             solve(mid+,b,mid+,r,d/);
         }
     }
 }
 int main(){
     cin >> n >> l >> r;
     ll p = n;
     while ( p >=  ){
         p /= ;
         s = s*+;
     }
     solve(,s,l,r,n);
     cout << ans << endl;
     return ;
 }