将给定的数乘以 2，使其可被 10 整除
给定一个数字，唯一允许的操作是将该数字乘以 2。计算最小运算数，使该数可被 10 整除。 注:如果无法转换，则打印-1。 示例:
输入: 10 输出: 0 由于给定的数本身可以被 10 整除， 的答案是 0。 输入: 1 输出: -1 由于通过乘以 2，给定的编号不能被 转换成可被 10 整除的数字， 因此答案是-1。
方法:任何给定的数只有在最后一位数字为 0 时才能被 10 整除。对于这个问题，提取输入数字的最后一位数字，并通过以下方式进行检查: 1)如果最后一位数字是 0，那么它已经可以被 10 整除，因此最小步数是 0。 2)如果最后一位数字是 5，那么将其乘以 2 一次将使其可被 10 整除，因此最小步数是 1。 3)如果最后一个数字是偶数或奇数(除了 0 和 5)，那么将其乘以 2 的任何次数都只会产生偶数，因此我们永远无法使其被 10 整除。因此步数为-1。
C++
// C++ code for finding
// number of operations
#include <bits/stdc++.h>
using namespace std;
int multiplyBy2(int n)
{
int rem, value;
// Find the last digit or remainder
rem = n % 10;
switch (rem) {
// If the last digit is 0
case 0:
value = 0;
break;
// If the last digit is 5
case 5:
value = 1;
break;
// If last digit is other
// than 0 and 5.
default:
value = -1;
}
return value;
}
// Driver code
int main()
{
int n = 28;
cout << multiplyBy2(n) << endl;
n = 255;
cout << multiplyBy2(n) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// JAVA code for finding
// number of operations
import java.io.*;
class GFG
{
static int multiplyBy2(int n)
{
int rem, value;
// Find the last digit
// or remainder
rem = n % 10;
switch (rem)
{
// If the last digit is 0
case 0:
value = 0;
break;
// If the last digit is 5
case 5:
value = 1;
break;
// If last digit is other
// than 0 and 5.
default:
value = -1;
}
return value;
}
// Driver code
public static void main (String[] args)
{
int n = 28;
System.out.println(multiplyBy2(n));
n = 255;
System.out.println(multiplyBy2(n));
}
}
// This code is contributed
// by shiv_bhakt.
Python 3
# Python3 code for finding number
# of operations
def dig(argu):
switcher = {
0: 0,
5: 1,
}
return switcher.get(argu, -1)
def multiplyBy2(n):
# Find the last digit or remainder
rem = n % 10;
return dig(rem);
# Driver code
n = 28;
print(multiplyBy2(n));
n = 255;
print(multiplyBy2(n));
# This code is contributed by mits
C
// C# code for finding
// number of operations
using System;
class GFG
{
static int multiplyBy2(int n)
{
int rem, value;
// Find the last digit
// or remainder
rem = n % 10;
switch (rem)
{
// If the last
// digit is 0
case 0:
value = 0;
break;
// If the last
// digit is 5
case 5:
value = 1;
break;
// If last digit is
// other than 0 and 5.
default:
value = -1;
break;
}
return value;
}
// Driver code
public static void Main ()
{
int n = 28;
Console.WriteLine(multiplyBy2(n));
n = 255;
Console.WriteLine(multiplyBy2(n));
}
}
// This code is contributed
// by shiv_bhakt.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP code for finding
// number of operations
function multiplyBy2($n)
{
$rem;
$value;
// Find the last digit or remainder
$rem = $n % 10;
switch ($rem) {
// If the last digit is 0
case 0:
$value = 0;
break;
// If the last digit is 5
case 5:
$value = 1;
break;
// If last digit is other
// than 0 and 5.
default:
$value = -1;
}
return $value;
}
// Driver code
$n = 28;
echo multiplyBy2($n),"\n";
$n = 255;
echo multiplyBy2($n),"\n";
// This code is contributed by aj_36
?>
java 描述语言
<script>
// Javascript code for finding
// number of operations
function multiplyBy2(n)
{
var rem, value;
// Find the last digit
// or remainder
rem = n % 10;
switch (rem) {
// If the last digit is 0
case 0:
value = 0;
break;
// If the last digit is 5
case 5:
value = 1;
break;
// If last digit is other
// than 0 and 5.
default:
value = -1;
}
return value;
}
// Driver code
var n = 28;
document.write(multiplyBy2(n)+"<br/>");
n = 255;
document.write(multiplyBy2(n));
// This code is contributed by todaysgaurav
</script>
Output:
-1
1
版权属于：月萌API www.moonapi.com，转载请注明出处
