103361: [Atcoder]ABC336 B - CTZ
Description
Score: $200$ points
Problem Statement
For a positive integer $X$, let $\text{ctz}(X)$ be the (maximal) number of consecutive zeros at the end of the binary notation of $X$.
If the binary notation of $X$ ends with a $1$, then $\text{ctz}(X)=0$.
You are given a positive integer $N$. Print $\text{ctz}(N)$.
Constraints
- $1\leq N\leq 10^9$
- $N$ is an integer.
Input
The input is given from Standard Input in the following format:
$N$
Output
Print $\text{ctz}(N)$.
Sample Input 1
2024
Sample Output 1
3
$2024$ is 11111101000
in binary, with three consecutive 0
s from the end, so $\text{ctz}(2024)=3$.
Thus, print $3$.
Sample Input 2
18
Sample Output 2
1
$18$ is 10010
in binary, so $\text{ctz}(18)=1$.
Note that we count the trailing zeros.
Sample Input 3
5
Sample Output 3
0
Input
Output
问题描述
对于一个正整数$X$,令$\text{ctz}(X)$为$X$的二进制表示中结尾的(最大)连续零的个数。
如果$X$的二进制表示以$1$结尾,则$\text{ctz}(X)=0$。
约束条件
- $1\leq N\leq 10^9$
- $N$是一个整数。
输入
输入通过标准输入给出以下格式:
$N$
输出
打印$\text{ctz}(N)$。
样例输入1
2024
样例输出1
3
$2024$的二进制表示为11111101000
,从结尾开始有三个连续的0
,因此$\text{ctz}(2024)=3$。
因此,打印$3$。
样例输入2
18
样例输出2
1
$18$的二进制表示为10010
,因此$\text{ctz}(18)=1$。
注意我们计算的是尾随的零。
样例输入3
5
样例输出3
0