Code For 1线段树与区间更新

H - Code For 1 Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status

Description

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.

Sample Input

Input

7 2 5

Output

4

Input

10 3 10

Output

5

Hint

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.

#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long LL;

int query(LL n,LL left,LL right,LL qleft,LL qright)
{
    if(right<qleft||left>qright||n==0){
        return 0;
    }
    
    if(n==1){
     return 1;
    }
       
    LL mid=qleft+qright>>1;
    
    return query(n>>1,left,right,qleft,mid-1)+query(n%2,left,right,mid,mid)+query(n>>1,left,right,mid+1,qright);
}

int main()
{
    LL n,l,r;
    cin>>n>>l>>r;
    LL len=1,x=n;
    
    while(x>1){
        len=len*2+1;
        x=x>>1;
    }
    
    printf("%d\n",query(n,l,r,1,len));
    retur

结合二分查找

标签：Code,Sam,线段,list,mid,更新,query,LL,Citadel
来源： https://www.cnblogs.com/killjoyskr/p/16535158.html