*【CodeForces - 768B】Code For 1 （分治策略，模拟二分思想，模拟线段树思想）

题干：

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

Input

10 3 10

Output

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.

题目大意：

给一个数n，和一个区间[l,r] （r-l<1e5，n<2^50），每次需要把序列中大于1的数字分成（n/2，n%2，n/2）（其中n/2是向下取整），直到所有数变成0或1，问[l,r]区间内有多少个1？

解题报告：

想着先把数字分好，然后再o(n)区间查询一下？有点麻烦，，其实分数字的过程中，答案就可以带出来了。

AC代码：

#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
ll l,r;
ll dfs(ll n,ll curl,ll curr) {if(curl > r || curr < l) return 0 ;if(n == 1) return 1;ll res = 0;ll mid = (curl+curr)/2;if(n % 2 == 1 && (mid<=r && mid>=l)) res++;return res + dfs(n>>1,curl,mid-1) + dfs(n>>1,mid+1,curr);
}int main()
{ll n,x,len = 1;cin>>n>>l>>r;x = n;while(x > 1) {len = len*2+1;x>>=1;}printf("%lld\n",dfs(n,1,len));return 0 ;}

总结：

刚开始写代码的时候两个地方写萎了

第一处：两个出口写颠倒了，想想也是不对啊，肯定是首先要在区间内部啊！！这是毋庸置疑的啊！！

第二处：if(n % 2 == 1) res++;

至于为什么没有像线段树一样写上如果两者都在所求区间内部的话单独返回一个值，这是因为我们提前把len算好了，所以最后分出的答案一定是0或者1，可以直接通过那个n==1来防止无限递归。然后就是对于0的处理，只需要那个res那里处理一下就可以了，因为只有中间的值才会出现等于0的情况，两边的值是不会出现为0的情况的，因为是除2操作嘛！！枚举几个终点可能值，就可以简单证明了，3/2=1，，,2/2=1，，，,1的话就是出口了，所以拆数的时候不会在两边有0出现，只可能是中间会有0出现。

所以其实这题我们也可以直接模拟一下过程，找找会有几个mid值是偶数，那么才会有0出现，假设求出0的数量是cnt0，那么结果就是r-l-cnt0。但是其实还是要模拟线段树build的过程的、、、、因为还要找和l，r的大小关系。

另附一个网络的二分思路的代码：（还未看）

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n, l, r, s = 1, ans;
void solve(ll a, ll b, ll l, ll r, ll d) { //二分的思想if ( a > b || l > r ) return;ll mid = (a+b)/2;if ( r < mid )solve(a,mid-1,l,r,d/2);else if( mid < l )solve(mid+1,b,l,r,d/2);else {ans += d%2;solve(a,mid-1,l,mid-1,d/2);solve(mid+1,b,mid+1,r,d/2);}
}
int main() {cin >> n >> l >> r;long long p = n;while ( p >= 2 ) {p /= 2;s = s*2+1;}solve(1,s,l,r,n);cout << ans << endl;return 0;
}

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*【CodeForces - 768B】Code For 1 （分治策略，模拟二分思想，模拟线段树思想）

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