A. Even Odds

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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 %I64dspecifier.

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.

解題說明：簡單的數學題，先是排列從1到不超過n的奇數，然後排列從2到不超過n的偶數，然後找出第k個數是多少。注意此題輸入需要用long long類型，輸出用double，否則越界。

#include<iostream>#include<map>#include<string>#include<algorithm>#include<cstdio>#include<cmath>using namespace std;int main(){long long int n,k;double ans;scanf("%lld %lld",&n,&k);if(n%2==0){if(k<=n/2){ans=2*k-1;}else{k=k-n/2;ans=2*k;}}else{if(k<=n/2+1){ans=2*k-1;}else{k=k-n/2-1;ans=2*k;}}printf("%.0lf\n",ans);return 0;}