求最接近 n 且能被 m 整除的数
原文:https://www . geesforgeks . org/find-number-close-n-除尽-m/
给定两个整数 n 和 m 。问题是找到最接近 n 且能被 m 整除的数。如果有一个以上这样的数字，那么输出具有最大绝对值的那个。如果 n 完全被 m 整除，则只输出 n 。要求 O(1)的时间复杂度。 约束: m！= 0 示例:
Input : n = 13, m = 4
Output : 12
Input : n = -15, m = 6
Output : -18
Both -12 and -18 are closest to -15, but
-18 has the maximum absolute value.
来源: 微软面试心得|第 125 集。
我们找到 n/m 的值，让这个值为 q，然后我们找到两种可能性中最接近的一种。一个是 q * m，另一个是(m * (q + 1))或(m (q–1))，这取决于给定的两个数字中是否有一个是负数。 算法:*
closestNumber(n, m)
Declare q, n1, n2
q = n / m
n1 = m * q
if (n * m) > 0
n2 = m * (q + 1)
else
n2 = m * (q - 1)
if abs(n-n1) < abs(n-n2)
return n1
return n2
C++
// C++ implementation to find the number closest to n
// and divisible by m
#include <bits/stdc++.h>
using namespace std;
// function to find the number closest to n
// and divisible by m
int closestNumber(int n, int m)
{
// find the quotient
int q = n / m;
// 1st possible closest number
int n1 = m * q;
// 2nd possible closest number
int n2 = (n * m) > 0 ? (m * (q + 1)) : (m * (q - 1));
// if true, then n1 is the required closest number
if (abs(n - n1) < abs(n - n2))
return n1;
// else n2 is the required closest number
return n2;
}
// Driver program to test above
int main()
{
int n = 13, m = 4;
cout << closestNumber(n, m) << endl;
n = -15; m = 6;
cout << closestNumber(n, m) << endl;
n = 0; m = 8;
cout << closestNumber(n, m) << endl;
n = 18; m = -7;
cout << closestNumber(n, m) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to find the number closest to n
// and divisible by m
public class close_to_n_divisible_m {
// function to find the number closest to n
// and divisible by m
static int closestNumber(int n, int m)
{
// find the quotient
int q = n / m;
// 1st possible closest number
int n1 = m * q;
// 2nd possible closest number
int n2 = (n * m) > 0 ? (m * (q + 1)) : (m * (q - 1));
// if true, then n1 is the required closest number
if (Math.abs(n - n1) < Math.abs(n - n2))
return n1;
// else n2 is the required closest number
return n2;
}
// Driver program to test above
public static void main(String args[])
{
int n = 13, m = 4;
System.out.println(closestNumber(n, m));
n = -15; m = 6;
System.out.println(closestNumber(n, m));
n = 0; m = 8;
System.out.println(closestNumber(n, m));
n = 18; m = -7;
System.out.println(closestNumber(n, m));
}
}
// This code is contributed by Sumit Ghosh
Python 3
# Python 3 implementation to find
# the number closest to n
# Function to find the number closest
# to n and divisible by m
def closestNumber(n, m) :
# Find the quotient
q = int(n / m)
# 1st possible closest number
n1 = m * q
# 2nd possible closest number
if((n * m) > 0) :
n2 = (m * (q + 1))
else :
n2 = (m * (q - 1))
# if true, then n1 is the required closest number
if (abs(n - n1) < abs(n - n2)) :
return n1
# else n2 is the required closest number
return n2
# Driver program to test above
n = 13; m = 4
print(closestNumber(n, m))
n = -15; m = 6
print(closestNumber(n, m))
n = 0; m = 8
print(closestNumber(n, m))
n = 18; m = -7
print(closestNumber(n, m))
# This code is contributed by Nikita tiwari.
C
// C# implementation to find the
// number closest to n and divisible by m
using System;
class GFG {
// function to find the number closest to n
// and divisible by m
static int closestNumber(int n, int m)
{
// find the quotient
int q = n / m;
// 1st possible closest number
int n1 = m * q;
// 2nd possible closest number
int n2 = (n * m) > 0 ? (m * (q + 1)) : (m * (q - 1));
// if true, then n1 is the required closest number
if (Math.Abs(n - n1) < Math.Abs(n - n2))
return n1;
// else n2 is the required closest number
return n2;
}
// Driver program to test above
public static void Main()
{
int n = 13, m = 4;
Console.WriteLine(closestNumber(n, m));
n = -15;
m = 6;
Console.WriteLine(closestNumber(n, m));
n = 0;
m = 8;
Console.WriteLine(closestNumber(n, m));
n = 18;
m = -7;
Console.WriteLine(closestNumber(n, m));
}
}
// This code is contributed by Sam007
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation to find
// the number closest to n and
// divisible by m
// function to find the number
// closest to n and divisible by m
function closestNumber($n, $m)
{
// find the quotient
$q = (int) ($n / $m);
// 1st possible closest number
$n1 = $m * $q;
// 2nd possible closest number
$n2 = ($n * $m) > 0 ?
($m * ($q + 1)) : ($m * ($q - 1));
// if true, then n1 is the
// required closest number
if (abs($n - $n1) < abs($n - $n2))
return $n1;
// else n2 is the required
// closest number
return $n2;
}
// Driver Code
$n = 13;
$m = 4;
echo closestNumber($n, $m), "\n";
$n = -15;
$m = 6;
echo closestNumber($n, $m), "\n";
$n = 0;
$m = 8;
echo closestNumber($n, $m), "\n";
$n = 18;
$m = -7;
echo closestNumber($n, $m), "\n";
// This code is contributed by jit_t
?>
java 描述语言
<script>
// Javascript implementation to find
// the number closest to n and
// divisible by m
// function to find the number
// closest to n and divisible by m
function closestNumber(n, m)
{
// find the quotient
let q = parseInt(n / m);
// 1st possible closest number
let n1 = m * q;
// 2nd possible closest number
let n2 = (n * m) > 0 ?
(m * (q + 1)) : (m * (q - 1));
// if true, then n1 is the
// required closest number
if (Math.abs(n - n1) < Math.abs(n - n2))
return n1;
// else n2 is the required
// closest number
return n2;
}
// Driver Code
let n = 13;
let m = 4;
document.write(closestNumber(n, m) + "<br>");
n = -15;
m = 6;
document.write(closestNumber(n, m) + "<br>");
n = 0;
m = 8;
document.write(closestNumber(n, m) + "<br>");
n = 18;
m = -7;
document.write(closestNumber(n, m) + "<br>");
// This code is contributed by gfgking
</script>
输出:
12
-18
0
21
时间复杂度: O(1) 本文由阿育什·乔哈里供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用contribute.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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