GZHU18级寒假训练:Leo's Trial B

本文探讨了一个有趣的问题:如何重新排列前n个自然数,首先按升序列出所有奇数,然后按升序列出所有偶数,并找出特定位置k上的数字。通过分析,提供了一种高效算法来解决这一挑战。

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Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first n. He writes down the following sequence of numbers: firstly all odd integers from 1 to n (in ascending order), then all even integers from 1 to n (also in ascending order). Help our hero to find out which number will stand at the position number k.
Input
The only line of input contains integers n and k (1 ≤ k ≤ n ≤ 1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print the number that will stand at the position number k after Volodya’s manipulations.
Examples
Input
10 3
Output
5
Input
7 7
Output
6
Note
In the first sample Volodya’s sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.

#include <iostream>
using namespace std;
int main()
{
	_int64 n, k;

	while (cin >> n)
	{
		cin >> k;
		if (n % 2 == 0)
		{
			cout << ((n / 2) >= k ? 2 * (k - 1) + 1 : 2 * (k - n / 2)) << endl;
		}
		else if (n == 1)
			cout << 1 << endl;
		else
		{
			if (k == n)
				cout << n-1 << endl;
			else cout << (((n +1)/ 2) >= k ? 2 * (k - 1) + 1 : 2 * (k - (n+1 )/ 2)) << endl;
		}
	}
}

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