Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) B. Code For 1

地址：http://codeforces.com/contest/768/problem/B

题目：

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 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.

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.

 

思路：因为r-l<=1e5，所以可以求出l,r中每个数的值，然后求和。

　　求值的时候，可以通过不断对称求出，就像一张纸可以不断对折。（画出那棵树你就能看出规律了）。

　　O((r-l)logn)

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;

LL n,l,r,ans;
LL a[100],p[100],sum;

int sc(LL x,int t)
{
    while(p[t]!=x)
    {
        if(x>p[t])  x=p[t]*2-x;
        t--;
    }
    return a[sum-t];
}

int main(void)
{
    p[0]=1;
    cin>>n>>l>>r;
    while(n>1)
        a[sum++]=n%2,n/=2,p[sum]=p[sum-1]*2;
    if(n!=0)
        a[sum]=1;
    for(LL i=l;i<=r;i++)
        ans+=sc(i,sum);
    cout<<ans<<endl;
    return 0;
}