可被 C 整除且不在[A，B]
范围内的最小正整数
给定三个正整数 A 、 B 和 C 。任务是找到最小整数 X > 0 ，这样:
- X % C = 0 和
- X 一定不属于【A，B】范围
例:
输入: A = 2，B = 4，C = 2 输出: 6 输入: A = 5，B = 10，C = 4 输出: 4
进场:
- 如果 C 不属于【A，B】即 C < A 或 C > B ，那么 C 就是需要的数字。
- 否则取 C 大于 B 的第一倍数，这是必选项。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the required number
int getMinNum(int a, int b, int c)
{
// If doesn't belong to the range
// then c is the required number
if (c < a || c > b)
return c;
// Else get the next multiple of c
// starting from b + 1
int x = ((b / c) * c) + c;
return x;
}
// Driver code
int main()
{
int a = 2, b = 4, c = 4;
cout << getMinNum(a, b, c);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.io.*;
import java.math.*;
public class GFG
{
// Function to return the required number
int getMinNum(int a, int b, int c)
{
// If doesn't belong to the range
// then c is the required number
if (c < a || c > b)
{
return c;
}
// Else get the next multiple of c
// starting from b + 1
int x = ((b / c) * c) + c;
return x;
}
// Driver code
public static void main(String args[])
{
int a = 2;
int b = 4;
int c = 4;
GFG g = new GFG();
System.out.println(g.getMinNum(a, b, c));
}
}
// This code is contributed by Shivi_Aggarwal
Python 3
# Python3 implementation of the approach
# Function to return the required number
def getMinNum(a, b, c):
# If doesn't belong to the range
# then c is the required number
if (c < a or c > b):
return c
# Else get the next multiple of c
# starting from b + 1
x = ((b // c) * c) + c
return x
# Driver code
a, b, c = 2, 4, 4
print(getMinNum(a, b, c))
# This code is contributed by
# Mohit kumar 29
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the required number
static int getMinNum(int a, int b, int c)
{
// If doesn't belong to the range
// then c is the required number
if (c < a || c > b)
{
return c;
}
// Else get the next multiple of c
// starting from b + 1
int x = ((b / c) * c) + c;
return x;
}
// Driver code
static public void Main ()
{
int a = 2, b = 4, c = 4;
Console.WriteLine( getMinNum(a, b, c));
}
}
// This Code is contributed by ajit..
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the above approach
// Function to return the required number
function getMinNum($a, $b, $c)
{
// If doesn't belong to the range
// then c is the required number
if ($c < $a || $c > $b)
return $c;
// Else get the next multiple of c
// starting from b + 1
$x = (floor(($b / $c)) * $c) + $c;
return $x;
}
// Driver code
$a = 2;
$b = 4;
$c = 4;
echo getMinNum($a, $b, $c);
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the required number
function getMinNum(a, b, c)
{
// If doesn't belong to the range
// then c is the required number
if (c < a || c > b)
return c;
// Else get the next multiple of c
// starting from b + 1
let x = (parseInt(b / c) * c) + c;
return x;
}
// Driver code
let a = 2, b = 4, c = 4;
document.write(getMinNum(a, b, c));
// This code is contributed by souravmahato348
</script>
Output:
8
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