Codeforces_768_B_(二分)

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 lto 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.

题意：将一个数列中大于1的数x，经过一个操作，在其位置上用  , ,  代替x，直至将整个数列化作01序列。

也是悲剧，比赛时把n的规模看成了10^50，卡了好久。。。

 直接用dfs会超时，因为n的规模是2^50。看题解，用了类似线段树的区间查询，二分区间查询。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define LL long long

LL  getLen(LL num)
{
    if(num==0)
        return 0;
    if(num==1)
        return 1;
    return getLen(num/2)*2+1;
}

int query(LL nu,LL L,LL R,LL l,LL r)
{
    if(R<l||L>r||nu==0)
        return 0;
    if(nu==1)
        return 1;
    LL mid=l+getLen(nu/2);
    return query(nu/2,L,R,l,mid-1)+query(nu%2,L,R,mid,mid)+query(nu/2,L,R,mid+1,r);
}

int main()
{
    //cout<<getLen(7)<<endl;
    LL n,l,r;
    scanf("%I64d%I64d%I64d",&n,&l,&r);
    LL len=getLen(n);
    //cout<<len<<endl;
    int res=query(n,l,r,1,len);
    printf("%d
",res);
    return 0;
}