除范围[L，R]
内所有自然数的最大除数
给定两个整数 L 和 R ，任务是找到在【L，R】范围内除所有自然数的最大除数。 举例:
输入: L = 3，R = 12 输出: 1 输入: L = 24，R = 24 输出: 24
方法:对于一系列连续的整数元素，有两种情况:
- 如果 L = R 那么答案将是 L 。
- 如果 L < R 那么这个范围内所有连续的自然数都是同素的。因此， 1 是唯一能够划分范围内所有元素的数字。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the greatest divisor that
// divides all the natural numbers in the range [l, r]
int find_greatest_divisor(int l, int r)
{
if (l == r)
return l;
return 1;
}
// Driver Code
int main()
{
int l = 2, r = 12;
cout << find_greatest_divisor(l, r);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function to return the greatest divisor that
// divides all the natural numbers in the range [l, r]
static int find_greatest_divisor(int l, int r) {
if (l == r) {
return l;
}
return 1;
}
// Driver Code
public static void main(String[] args) {
int l = 2, r = 12;
System.out.println(find_greatest_divisor(l, r));
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 implementation of the approach
# Function to return the greatest divisor that
# divides all the natural numbers in the range [l, r]
def find_greatest_divisor(l, r):
if (l == r):
return l;
return 1;
# Driver Code
l = 2;
r = 12;
print(find_greatest_divisor(l, r));
#This code is contributed by Shivi_Aggarwal
C
// C# implementation of the approach
using System;
class GFG {
// Function to return the greatest divisor that
// divides all the natural numbers in the range [l, r]
static int find_greatest_divisor(int l, int r) {
if (l == r) {
return l;
}
return 1;
}
// Driver Code
public static void Main() {
int l = 2, r = 12 ;
Console.WriteLine(find_greatest_divisor(l, r));
}
// This code is contributed by Ryuga
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the greatest
// divisor that divides all the natural
// numbers in the range [l, r]
function find_greatest_divisor($l, $r)
{
if ($l == $r)
return $l;
return 1;
}
// Driver Code
$l = 2;
$r = 12;
echo find_greatest_divisor($l, $r);
// This code is contributed by jit_t
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the greatest
// divisor that divides all the natural
// numbers in the range [l, r]
function find_greatest_divisor(l, r)
{
if (l == r)
return l;
return 1;
}
// Driver Code
let l = 2;
let r = 12;
document.write( find_greatest_divisor(l, r));
// This code is contributed
// by bobby
</script>
Output:
1
时间复杂度: O(1)
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