CodeForces - 768B Code For 1（找规律）

最新推荐文章于 2019-08-17 11:29:52 发布

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最新推荐文章于 2019-08-17 11:29:52 发布

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分类专栏：思维题／模拟／乱搞／贪心

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本文介绍了一个关于序列操作的问题，初始元素通过特定规则转换成由0和1组成的序列，目标是计算指定范围内1的数量。文章提供了算法思路及实现代码。

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题目链接：http://codeforces.com/problemset/problem/768/B点击打开链接

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 $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ 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:

$\text{[math]}$

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:

$\text{[math]}$

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.

仔细观察可以发现每个数被拆分成的最终序列是有规律的这里以10举例

先将该数用二进制表示 1010

观察序列发现二进制中的第一位的1位于1 3 5 7 9 11 13 15的位置

第二位的0位于2 6 10 14

第三位的1位于4 12

第四位的0位于8

将1010这个二进制数的每一位附上权值每一位对应为1 2 4 8 从左到右位置为 0 1 2 3 （其实就是2的次方对应关系）

于是可以发现对于序列中任意一个位置的值都具有以下规律

设当前位置为position

position的值对应于 position的因子中最大的2的N次方数即为该数二进制的第N位的值

有点难懂还是举10的例子

对于序列的第3位他的因子中最大的2的N次方数为1 即N=0 所以为二进制1010的第0位也就是第一个1

对于序列的第6位她的因子中最大的2的N次方数为2 即N=1 所以对应1010的第一个0

。。。。。。12。。。。。。。。。。。。。。。4 即N=2 所以对应第三位也就是第二个1

找出规律之后代码就很简单了

因为范围不大因此枚举他需要的区间的每个位置根据规律计算当前的数是不是1

#include <iostream>
#include <stdio.h>
#include <limits.h>
#include <stack>
#include <algorithm>
#include <queue>
#include <string.h>
#include <set>
using namespace std;
vector<long long int > s;
int main()
{
   long long int n,l,r;
   cin >> n >> l >> r;
   while(n!=0)
   {
       s.push_back(n%2);
       n/=2;
   }
   long long int cnt=0;
   reverse(s.begin(),s.end());
   s.push_back(0);
   for(long long int i=l;i<=r;i++)
   {
        long long int j=0;
        long long int mid=i;
        while(mid%2==0)
            j++,mid/=2;
        if(s[j]==1)
            cnt++;
   }
   cout << cnt;
}