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 nlr (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-;
const double pi=acos(-1.0);
const int K=1e6+;
const int mod=1e9+; LL n,l,r,ans;
LL a[],p[],sum; int sc(LL x,int t)
{
while(p[t]!=x)
{
if(x>p[t]) x=p[t]*-x;
t--;
}
return a[sum-t];
} int main(void)
{
p[]=;
cin>>n>>l>>r;
while(n>)
a[sum++]=n%,n/=,p[sum]=p[sum-]*;
if(n!=)
a[sum]=;
for(LL i=l;i<=r;i++)
ans+=sc(i,sum);
cout<<ans<<endl;
return ;
}

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