Codeforces Round #188 (Div. 2) A. Even Odds 水题

A. Even Odds

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/318/problem/A

Description

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.

Sample Input

10 3

Sample Output

5

HINT

题意

给你n，然后把这些数按照先奇数和后偶数排序之后，输出第k个

题解：

很明显这是一道不是输出k*2+1就是输出k*2的答案

我们只要自己出几个无聊的数据测测就好了

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

int main()
{
    ll n=read(),k=read();
    if(n%2==1)
    {
        if(k<=n/2+1)
            printf("%lld",k*2-1);
        else
            printf("%lld",(k-n/2-1)*2);
    }
    else
    {
        if(k<=n/2)
            printf("%lld",k*2-1);
        else
            printf("%lld",(k-n/2)*2);
    }
}