acm日常CodeForces - 318A

CodeForces - 318A

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.

解题思想：
①开始的错误思想：列原题数组，再进行检测。×
②应该是观察发现，其实就是由两个等差数列组成，找规律，发现数学表达式。√

ac代码

#include<iostream>
using namespace std;
int main()
{
long long n, k;
cin >> n >> k;
if (n % 2)
{
if (k <= n / 2 + 1)//奇数
cout << 2 * k - 1 << endl;
else
cout << 2 * (k-n/2-1)<< endl;
}
else
{
if (k <= n / 2)
cout << 2 * k - 1 << endl;
else
cout << 2 * (k - n / 2)
<< endl;
}
return 0;
}

最后

以上就是诚心苗条为你收集整理的acm日常CodeForces - 318A的全部内容，希望文章能够帮你解决acm日常CodeForces - 318A所遇到的程序开发问题。

如果觉得靠谱客网站的内容还不错，欢迎将靠谱客网站推荐给程序员好友。