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

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我是靠谱客的博主传统饼干，最近开发中收集的这篇文章主要介绍Codeforces Round #188 (Div. 2) A.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

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.

这是一个超级大水题，之所以写这个题，原因非常惭愧，因为这是我第一次用__int64这个东西，感谢永乐一语点醒我，让我能很快的搞定这个题

题目的大致意思就是给你1-n的自然数，前面的数全部都是奇数，按照升序排列，后面的都是偶数，同样还是按照升序排列。然后找到第k个位置上的数是多少，其实非常简单，把这对数从中间切开，找出n在奇偶情况下，奇序列和偶序列的通项公式。下面是代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
using namespace std;
int main(){
__int64 n;
__int64 k;
scanf("%I64d",&n);
scanf("%I64d",&k);
if(n%2==0){
if(k<=n/2){
__int64 count=2*k-1;
k=count;
}
else{
__int64 count=2*(k-n/2);
k=count;
}
}
if(n%2!=0){
if(k<=(n/2+1)){
__int64 count=2*k-1;
k=count;
}
else{
__int64 q=k-(n/2+1);
__int64 count=2*q;
k=count;
}
}
printf("%I64d",k);
return 0;
}