最小绝对差配对，乘积为 N+1 或 N+2
原文:https://www . geesforgeks . org/pair-with-min-绝对差-and-谁的产品是-n1-或-n2/
给定一个整数 N ，任务是找到一对乘积为 N + 1 或 N + 2 且对的绝对差最小的对。
示例:
输入: N = 8 输出: 3，3 解释: 3 * 3 = 8 + 1 输入: N = 123 输出: 5，25 解释: 5 * 25 = 123 + 2
方法:思想是用一个循环变量 I 从 sqrt(N+2)到 1 迭代一个循环，并检查以下条件:
- 如果 (n + 1) % i = 0 ，那么我们将打印该对 (i，(n + 1) / i) 。
- 如果 (n + 2) % i = 0 ，那么我们将打印该对 (i，(n + 2) / i) 。
- 打印的第一对将是绝对差异最小的一对。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print pair (a, b)
// such that a*b=N+1 or N+2
void closestDivisors(int n)
{
// Loop to iterate over the
// desired possible values
for (int i = sqrt(n + 2);
i > 0; i--) {
// Check for condition 1
if ((n + 1) % i == 0) {
cout << i << ", "
<< (n + 1) / i;
break;
}
// Check for condition 2
if ((n + 2) % i == 0) {
cout << i << ", "
<< (n + 2) / i;
break;
}
}
}
// Driver Code
int main()
{
// Given Number
int N = 123;
// Function Call
closestDivisors(N);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function to print pair (a, b)
// such that a*b=N+1 or N+2
static void closestDivisors(int n)
{
// Loop to iterate over the
// desired possible values
for (int i = (int)Math.sqrt(n + 2); i > 0; i--)
{
// Check for condition 1
if ((n + 1) % i == 0)
{
System.out.print(i + ", " +
(n + 1) / i);
break;
}
// Check for condition 2
if ((n + 2) % i == 0)
{
System.out.print(i + ", " +
(n + 2) / i);
break;
}
}
}
// Driver Code
public static void main(String[] args)
{
// Given Number
int N = 123;
// Function Call
closestDivisors(N);
}
}
// This code is contributed by rock_cool
计算机编程语言
# Python3 program for the above approach
from math import sqrt, ceil, floor
# Function to prpair (a, b)
# such that a*b=N+1 or N+2
def closestDivisors(n):
# Loop to iterate over the
# desired possible values
for i in range(ceil(sqrt(n + 2)), -1, -1):
# Check for condition 1
if ((n + 1) % i == 0):
print(i, ",", (n + 1) // i)
break
# Check for condition 2
if ((n + 2) % i == 0):
print(i, ",", (n + 2) // i)
break
# Driver Code
if __name__ == '__main__':
# Given Number
N = 123
# Function Call
closestDivisors(N)
# This code is contributed by Mohit Kumar
C
// C# program for the above approach
using System;
class GFG{
// Function to print pair (a, b)
// such that a*b=N+1 or N+2
static void closestDivisors(int n)
{
// Loop to iterate over the
// desired possible values
for (int i = (int)Math.Sqrt(n + 2); i > 0; i--)
{
// Check for condition 1
if ((n + 1) % i == 0)
{
Console.Write(i + ", " +
(n + 1) / i);
break;
}
// Check for condition 2
if ((n + 2) % i == 0)
{
Console.Write(i + ", " +
(n + 2) / i);
break;
}
}
}
// Driver Code
public static void Main(string[] args)
{
// Given Number
int N = 123;
// Function Call
closestDivisors(N);
}
}
// This code is contributed by Ritik Bansal
java 描述语言
<script>
// Javascript program for the above approach
// Function to print pair (a, b)
// such that a*b=N+1 or N+2
function closestDivisors(n)
{
// Loop to iterate over the
// desired possible values
for (var i = parseInt(Math.sqrt(n + 2));
i > 0; i--) {
// Check for condition 1
if ((n + 1) % i == 0) {
document.write(i+", "+parseInt((n + 1) / i));
break;
}
// Check for condition 2
if ((n + 2) % i == 0) {
document.write(i+", "+parseInt((n + 2) / i));
break;
}
}
}
// Driver Code
// Given Number
N = 123;
// Function Call
closestDivisors(N);
</script>
Output:
5, 25
时间复杂度:O(sqrt(N)) T5】辅助空间: O(1)
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