Even Odds

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我是靠谱客的博主殷勤雨，最近开发中收集的这篇文章主要介绍Even Odds，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

A. Even Odds
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≤1e14).

Output
Print the number that will stand at the position number k after Volodya's manipulations.

Sample test(s)
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.
Hint
注意数据的类型
use __int64 to read data and use %I64d to output data

题目分析：

这道题是入队选拔赛的第一题，也是最水的一道题，意思就是将输入数字n，先将从1到n的奇数排列出来，再将从2到n的偶数排列出来，找出第m个数是几。这道题直接for循环可能会超时，所以找规律即可。比较简单，我直接贴代码。

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#include<iostream>
#define ll __int64
using namespace std;
int main()
{
    long long num,ans,pos;
    while(scanf("%I64d %I64d",&num,&pos)!=EOF){
          if(num%2==0){
              if(pos<=num/2)printf("%I64dn",2*pos-1);
              else printf("%I64dn",2*pos-num);
          }
          else {
               if( pos<=(num/2+1) )printf("%I64dn",2*pos-1);
               else printf("%I64dn",2*pos-num-1);
          }                         
    
    }
    return 0;    
}