找到给定长度的复合数范围

给定一个整数n,我们需要找到一个正整数的范围,使得该范围内的所有数字都是复合的,且该范围的长度为n。如果有多个答案,您可以打印任何范围。复合数是一个正整数,它至少有一个除1之外的除数和自身(来源: 维基 )

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例如:

Input : 3Output : [122, 124]Explanation 122, 123, 124 are all composite numbers

解决方案有点棘手。因为有很多可能的答案,我们在这里讨论一个广义解。

Let the length of range be n and range starts from a then, a, a+1, a+2, ...., a+n-1 all should be composite. So the problem boils down to findingsuch 'a'.If we closely observe p! (where p is a positive integers) then we will find that, p! has factors of2, 3, 4, ..., p-1,Hence if we add i to p! such that 1 < i < p,then p! + i has a factor i, so p! + i must be composite. So we end up finding p! + 2, p! + 3, .... p! + p-1 are all composite and continuous integers forming a range [p! + 2, p! + p-1]The above range consists of p-2 elements.For a range of n elements we need to consider (n+2)!If we take a = (n+2)! + 2, Then, a + 1 = (n+2)! + 3Then, a + 2 = (n+2)! + 4...Then, a + n-1 = (n+2)! + n+1Hence,a = (n+2)! + 2 = 2*3*....*(n+2) + 2a has 2 as its divisor because (n+2)! and 2 both divides 2a + 1 = 2*3*....*(n+2) + 3a + 1 has 3 as its divisor because (n+2)! and 3 both divides 3...a + n-1 = 2*3*....*(n+2) + n+1a + n-1 has n+1 as its divisor because (n+2)! and n+1 both divides n+1Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]

上述算法的示例

n = 3Then a = (n+2)! + 2a = 5! + 2a + 1 = 5! + 3a + 2 = 5! + 4Here a is divisible by 2Here a + 1 is divisible by 3Here a + 2 is divisible by 4Hence a, a+1, a+2 are all composites

C++

// C++ program to find a range of
// composite numbers of given length
#include <bits/stdc++.h>
using namespace std;
// method to find factorial
// of given number
int factorial ( int n)
{
if (n == 0)
return 1;
return n * factorial(n-1);
}
// to print range of length n
// having all composite integers
int printRange( int n)
{
int a = factorial(n + 2) + 2;
int b = a + n - 1;
cout << "[" << a << ", " << b << "]" ;
return 0;
}
// Driver method
int main()
{
int n = 3 ;
printRange(n);
return 0;
}
// This code is contributed by Anshika Goyal


JAVA

// Java program to find a range of composite
// numbers of given length
class Test
{
// method to find factorial of given number
static int factorial( int n)
{
if (n == 0 )
return 1 ;
return n*factorial(n- 1 );
}
// to print range of length n
//  having all composite integers
static void printRange( int n)
{
int a = factorial(n + 2 ) + 2 ;
int b = a + n - 1 ;
System.out.println( "[" + a + ", " + b + "]" );
}
// Driver method
public static void main(String args[]) throws Exception
{
int n = 3 ;
printRange(n);
}
}


Python3

# Python program to find a range of composite
# numbers of given length
# function to calculate factorial
def factorial(n):
a = 1
for i in range ( 2 , n + 1 ):
a * = i
return a
# to print range of length n
# having all composite integers
def printRange(n):
a = factorial(n + 2 ) + 2
b = a + n - 1
print ( "[" + str (a) + ", " + str (b) + "]" )
# driver code to test above functions
n = 3
printRange(n)


C#

// C# program to find a range of
// composite numbers of given
// length
using System;
class GFG {
// method to find factorial
// of given number
static int factorial( int n)
{
if (n == 0)
return 1;
return n*factorial(n-1);
}
// to print range of length n
// having all composite integers
static void printRange( int n)
{
int a = factorial(n + 2) + 2;
int b = a + n - 1;
Console.WriteLine( "[" + a +
", " + b + "]" );
}
// Driver method
public static void Main()
{
int n = 3 ;
printRange(n);
}
}
// This code is contributed by anuj_67.


PHP

<?php
// PHP program to find a range of
// composite numbers of given length
// method to find factorial
// of given number
function factorial ( $n )
{
if ( $n == 0)
return 1;
return $n * factorial( $n - 1);
}
// to print range of length n
// having all composite integers
function printRange( $n )
{
$a = factorial( $n + 2) + 2;
$b = $a + $n - 1;
echo "[" , $a , ", " , $b , "]" ;
return 0;
}
// Driver Code
$n = 3 ;
printRange( $n );
// This code is contributed by anuj_67.
?>


Javascript

<script>
// Javascript program to find a range of
// composite numbers of given length
// Method to find factorial
// of given number
function factorial(n)
{
if (n == 0)
return 1;
return n * factorial(n - 1);
}
// To print range of length n
// having all composite integers
function printRange(n)
{
let a = factorial(n + 2) + 2;
let b = a + n - 1;
document.write(`[${a}, ${b}]`);
return 0;
}
// Driver Code
let n = 3;
printRange(n);
// This code is contributed by _saurabh_jaiswal.
</script>


输出:

[122, 124]

对上述算法的分析 时间复杂度:O(n) 辅助空间:O(1)

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