找到唯一的对，使得每个元素小于或等于 N
原文:https://www . geesforgeks . org/find-unique-pairs-so-每个元素小于或等于 n/
给定一个整数 N，找到并显示满足以下条件的对的数量:
注意:只应显示同时符合上述两个条件的配对，这些数字必须小于或等于 n
示例:
Input: 10
Output: No. of pairs = 1
Pair no. 1 --> (2, 4)
Input: 500
Output: No. of pairs = 7
Pair no. 1 --> (2, 4)
Pair no. 2 --> (12, 18)
Pair no. 3 --> (36, 48)
Pair no. 4 --> (80, 100)
Pair no. 5 --> (150, 180)
Pair no. 6 --> (252, 294)
Pair no. 7 --> (392, 448)
解释: 下面的表格将给出一个清晰的视图:
上表显示了由两个连续数字及其对应倍数的乘积形成的 GCD，其中每个值对应一个唯一配对。每行中的绿色条目为相应的 GCD 形成一个唯一的对。 注:在上表中，
- 对于第一个条目，GCD=2，2 的第一个和第二个倍数形成唯一对(2，4)
- 类似地，对于第二个条目，GCD=6，6 的第二个和第三个倍数形成唯一对(12，18)
- 类似地，继续，对于 Zth 条目，即对于 GCD = Z(Z+1)，很明显唯一对将包括 Zth 和 GCD = Z(Z+1)的(Z+1)倍。现在，GCD 的 Zth 倍数为 Z * (Z(Z+1))，GCD 的(Z+1)倍将为(Z + 1) * (Z(Z+1))。
- 由于限制是 N，因此唯一对中的第二个数字必须小于或等于 N。因此，(Z+1)(Z (Z+1))<= N. Simplifying it further, the desired relation is derived Z3+(2 * Z2)+Z<= N
这形成了一种模式，并且从数学计算中，可以得出，对于给定的 N，这种唯一对的总数(比如 Z)将遵循如下所示的数学关系:
Z3 + (2*Z2) + Z <= N
以下是所需的实现:
C
// C program for finding the required pairs
#include <stdio.h>
#include <stdlib.h>
// Finding the number of unique pairs
int No_Of_Pairs(int N)
{
int i = 1;
// Using the derived formula
while ((i * i * i) + (2 * i * i) + i <= N)
i++;
return (i - 1);
}
// Printing the unique pairs
void print_pairs(int pairs)
{
int i = 1, mul;
for (i = 1; i <= pairs; i++) {
mul = i * (i + 1);
printf("Pair no. %d --> (%d, %d)\n",
i, (mul * i), mul * (i + 1));
}
}
// Driver program to test above functions
int main()
{
int N = 500, pairs, mul, i = 1;
pairs = No_Of_Pairs(N);
printf("No. of pairs = %d \n", pairs);
print_pairs(pairs);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for finding
// the required pairs
import java.io.*;
class GFG
{
// Finding the number
// of unique pairs
static int No_Of_Pairs(int N)
{
int i = 1;
// Using the derived formula
while ((i * i * i) +
(2 * i * i) + i <= N)
i++;
return (i - 1);
}
// Printing the unique pairs
static void print_pairs(int pairs)
{
int i = 1, mul;
for (i = 1; i <= pairs; i++)
{
mul = i * (i + 1);
System.out.println("Pair no. " + i + " --> (" +
(mul * i) + ", " +
mul * (i + 1) + ")");
}
}
// Driver code
public static void main (String[] args)
{
int N = 500, pairs, mul, i = 1;
pairs = No_Of_Pairs(N);
System.out.println("No. of pairs = " + pairs);
print_pairs(pairs);
}
}
// This code is contributed by Mahadev.
Python 3
# Python3 program for finding the required pairs
# Finding the number of unique pairs
def No_Of_Pairs(N):
i = 1;
# Using the derived formula
while ((i * i * i) + (2 * i * i) + i <= N):
i += 1;
return (i - 1);
# Printing the unique pairs
def print_pairs(pairs):
i = 1;
mul = 0;
for i in range(1, pairs + 1):
mul = i * (i + 1);
print("Pair no." , i, " --> (", (mul * i),
", ", mul * (i + 1), ")");
# Driver Code
N = 500;
i = 1;
pairs = No_Of_Pairs(N);
print("No. of pairs = ", pairs);
print_pairs(pairs);
# This code is contributed
# by mits
C
// C# program for finding
// the required pairs
using System;
class GFG
{
// Finding the number
// of unique pairs
static int No_Of_Pairs(int N)
{
int i = 1;
// Using the derived formula
while ((i * i * i) +
(2 * i * i) + i <= N)
i++;
return (i - 1);
}
// Printing the unique pairs
static void print_pairs(int pairs)
{
int i = 1, mul;
for (i = 1; i <= pairs; i++)
{
mul = i * (i + 1);
Console.WriteLine("Pair no. " + i + " --> (" +
(mul * i) + ", " +
mul * (i + 1) + ")");
}
}
// Driver code
static void Main()
{
int N = 500, pairs;
pairs = No_Of_Pairs(N);
Console.WriteLine("No. of pairs = " +
pairs);
print_pairs(pairs);
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program for finding
// the required pairs
// Finding the number
// of unique pairs
function No_Of_Pairs($N)
{
$i = 1;
// Using the
// derived formula
while (($i * $i * $i) +
(2 * $i * $i) +
$i <= $N)
$i++;
return ($i - 1);
}
// Printing the unique pairs
function print_pairs($pairs)
{
$i = 1; $mul;
for ($i = 1;
$i <= $pairs; $i++)
{
$mul = $i * ($i + 1);
echo "Pair no." ,
$i, " --> (" ,
($mul * $i), ", ",
$mul * ($i + 1),") \n";
}
}
// Driver Code
$N = 500; $pairs;
$mul; $i = 1;
$pairs = No_Of_Pairs($N);
echo "No. of pairs = ",
$pairs , " \n";
print_pairs($pairs);
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>
java 描述语言
<script>
// Javascript program for finding the
// required pairs
// Finding the number of unique pairs
function No_Of_Pairs(N)
{
let i = 1;
// Using the derived formula
while ((i * i * i) +
(2 * i * i) + i <= N)
i++;
return (i - 1);
}
// Printing the unique pairs
function print_pairs(pairs)
{
let i = 1, mul;
for(i = 1; i <= pairs; i++)
{
mul = i * (i + 1);
document.write("Pair no. " + i +
" --> (" + (mul * i) +
", " + mul * (i + 1) +
")<br>");
}
}
// Driver code
let N = 500, pairs, mul, i = 1;
pairs = No_Of_Pairs(N);
document.write("No. of pairs = " +
pairs + "<br>");
print_pairs(pairs);
// This code is contributed by mohit kumar 29
</script>
Output:
No. of pairs = 7
Pair no. 1 --> (2, 4)
Pair no. 2 --> (12, 18)
Pair no. 3 --> (36, 48)
Pair no. 4 --> (80, 100)
Pair no. 5 --> (150, 180)
Pair no. 6 --> (252, 294)
Pair no. 7 --> (392, 448)
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