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;
}
