Codeforces 768B B. 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 $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ 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:

　　 $\text{[math]}$

　　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:

　　 $\text{[math]}$

　　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

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 ll n, l, r, s = 1, ans;
 5 void solve(ll a, ll b, ll l, ll r, ll d){//二分的思想
 6     if ( a > b || l > r )    return;
 7     else{
 8         ll mid = (a+b)/2;
 9         if ( r < mid )solve(a,mid-1,l,r,d/2);
10         else if ( mid < l )solve(mid+1,b,l,r,d/2);
11         else {
12             ans += d%2;
13             solve(a,mid-1,l,mid-1,d/2);
14             solve(mid+1,b,mid+1,r,d/2);
15         }
16     }
17 }
18 int main(){
19     cin >> n >> l >> r;
20     ll p = n;
21     while ( p >= 2 ){
22         p /= 2;
23         s = s*2+1;
24     }
25     solve(1,s,l,r,n);
26     cout << ans << endl;
27     return 0;
28 }

Codeforces 768B B. Code For 1

B. Code For 1

思路：

解法： 分治的思想，二分法

解法：分治的思想，二分法