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 = 1, 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)/2;
        if ( r < mid )solve(a,mid-1,l,r,d/2);
        else if ( mid < l )solve(mid+1,b,l,r,d/2);
        else {
            ans += d%2;
            solve(a,mid-1,l,mid-1,d/2);
            solve(mid+1,b,mid+1,r,d/2);
        }
    }
}
int main(){
    cin >> n >> l >> r;
    ll p = n;
    while ( p >= 2 ){
        p /= 2;
        s = s*2+1;
    }
    solve(1,s,l,r,n);
    cout << ans << endl;
    return 0;
}