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?
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 the total number of 1s in the range l to r in the final sequence.
7 2 5
4
10 3 10
5
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.
题目链接：http://codeforces.com/contest/768/problem/B
分析：二分的模版题！来围观看看！
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 { 7 if ( a > b || l > r ) return; 8 else 9 { 10 ll mid = (a+b)/2; 11 if ( r < mid )solve(a,mid-1,l,r,d/2); 12 else if ( mid < l )solve(mid+1,b,l,r,d/2); 13 else { 14 ans += d%2; 15 solve(a,mid-1,l,mid-1,d/2); 16 solve(mid+1,b,mid+1,r,d/2); 17 } 18 } 19 } 20 int main() 21 { 22 cin >> n >> l >> r; 23 long long p = n; 24 while ( p >= 2 ) 25 { 26 p /= 2; 27 s = s*2+1; 28 } 29 solve(1,s,l,r,n); 30 cout << ans << endl; 31 return 0; 32 }
