分治思想 - 数1的个数

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.

Example

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.

 

题目分析 ： 给你一个 n ,将 n 分成 0 和 1，询问你给定区间内的 1 的个数

思路分析 ： 是一个分治的思想，每次只是查询所给区间的 1 的个数

代码示例 ：

#define ll long long
const int maxn = 1e6+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
ll n, l, r;

ll fun(ll x){
    if (x <= 1) return 1;
    else return fun(x/2)*2+1;
}

ll dfs(ll n, ll l, ll r, ll a, ll b){ 
    if (l > b || r < a) return 0;
    if (n == 0) return 0;
    if (n == 1) return 1;
    ll sum = 0;
    ll mid = (a + b) >> 1;
    sum += dfs(n/2, l, r, a, mid-1);
    sum += dfs(n%2, l, r, mid, mid);
    sum += dfs(n/2, l, r, mid+1, b);
    return sum;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    cin >> n >> l >> r;
    ll len = fun(n);
    printf("%lld\n", dfs(n, l, r, 1ll, len));
    return 0;
}