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
7 7
output
5
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
这题注意的是long long要用%lld输入输出。
#include <stdio.h>
#include <stdlib.h>
int main()
{
long long a,b,result;
while(scanf("%lld%lld",&a,&b)!=EOF)
{
if(a%2!=0)
a=a/2+1;
else
a=a/2;
if(b<=a)
result=b*2-1;
else
result=(b-a)*2;
printf("%lldn",result);
}
//
system("pause");
return 0;
}
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