codeforce round 399# B

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本文介绍了一道编程题的解决方法，题目背景基于《权力的游戏》剧情，Sam需解决一个数学问题以成为学士。任务是对一个整数进行递归分解，直至所有数字变为0或1，并统计指定范围内的1的数量。

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      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.

**求：一个数分解为0和1后，区间【l，r】之间1的个数
思路：将一个数分解dfs+dp[x]（表示x分解后的长度）

#include<iostream>
#include<cmath>
#include<map>
#include<string>
#include<cstdio>
using namespace std;
long long n,l,r;
map<long,long> dp;
long long dfs(long long x)      //求出 dp[x]（表示x分解后的长度）
{
    if (x != 0)
        dp[x] += 2 * dfs(x / 2) + 1;     
    return dp[x];
}
int main()
{
        cin >> n >> l >> r;
        dp[1] = 1;
        dp.clear();
        long long m = n;
        dfs(m);
        long long num = 0;
        for (long long i = l;i <= r;i++)       //从l到r逐个判断是否为1

        {
            long long ans = 0, temp = n;
            while (temp)
            {
                if (ans + dp[temp / 2] + 1 == i)
                {
                    if(temp%2==1) 
                        num++;
                    break;
                }
                else
                {
                    if (ans + dp[temp / 2] > i)
                    {
                        temp /= 2;
                    }
                    else
                    {
                        if (ans + dp[temp / 2] == i)
                        {
                            num++;
                            break;
                        }
                        else
                        {
                            ans += dp[temp / 2] + 1;
                        }
                    }
                }
                if (ans == i)
                {
                    num++;
                    break;
                }


            }
        }
        cout << num << endl;
    return 0;
}