奇、偶数（规律）A. Even Odds-优快云博客

本文链接：https://blog.youkuaiyun.com/tjracjhd/article/details/109605933

该问题涉及Volodya重新排列1到n的自然数，先列出所有奇数，再列出所有偶数。根据输入的n和k，找出排列后位于k位置的数字。当n为奇数且k小于(n+1)/2时，位置k的数字是2k-1；当n为偶数或k大于等于(n+1)/2时，位置k的数字是n+1。

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				    	**A. Even Odds
				time limit per test1 second
			memory limit per test256 megabytes
					inputstandard input
				   outputstandard output**

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.

思路：
正常做法会爆，这里给的是找规律。
奇数排前面，偶数排后面；
如果 n 是奇数，那么奇数的个数就会比偶数的个数多 “一个”；
如果 n 是偶数，则奇数和偶数的个数是一样多的。
当 k < (n+1) / 2 时，第 k个数的值就是 2*k-1 ;
当 k >= （n+1）/ 2 时，值就为 2 * ( n+1 ) / 2 ;
这里解释一下为什么是（n+1）而不是 n , 就我是把奇偶一起判的，当然你也可以分开判。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;

int main()
{
    ll n,k,ans,j;
    cin>>n>>k;
    j=(n+1)/2;
    if(k>j)
    {
        ans = 2*(k-j);
    }
    else
    {
        ans = 2*k-1;
    }
    cout<<ans<<endl;
    return 0;
}