生成 n 个连续合成数字的列表(一个有趣的方法)
原文:https://www . geesforgeks . org/generate-list-n-continuous-composite-numbers-interest-method/
给定一个数字 n，生成 n 个复合数字的列表。 例:
Input : 5
Output : 122, 123, 124, 125
Input : 10
Output : 3628802, 3628803, 3628804, 3628805, 3628806,
3628807, 3628808, 3628809, 3628810
这里的想法是利用
的属性。既然
，那么数字
，都分
。因此
可以被 2 整除，
可以被 3 整除…..
可被 n 整除。通过上述模式，它们是连续的复合物。
我们发现(n+1)！，那么我们打印数字(n+1)！+ 2，(n+1)！+ 3, ….(n+1)！+ (n + 1)。
以下是上述方法的实施:
C++
// CPP program to print n consecutive composite
// numbers.
#include <iostream>
using namespace std;
// function to find factorial of given
// number
unsigned long long int factorial(unsigned int n)
{
unsigned long long int res = 1;
for (int i=2; i<=n; i++)
res *= i;
return res;
}
// Prints n consecutive numbers.
void printNComposite(int n)
{
unsigned long long int fact = factorial(n+1);
for (int i = 2; i <= n+1; ++i)
cout << fact + i << " ";
}
// Driver program to test above function
int main()
{
int n = 4;
printNComposite(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print n consecutive composite
// numbers
class GFG {
// function to find factorial of given
// number
static long factorial(int n) {
long res = 1;
for (int i = 2; i <= n; i++) {
res *= i;
}
return res;
}
// Prints n consecutive numbers.
static void printNComposite(int n) {
long fact = factorial(n + 1);
for (int i = 2; i <= n + 1; ++i) {
System.out.print(fact + i + " ");
}
}
// Driver program to test above function
public static void main(String[] args) {
int n = 4;
printNComposite(n);
}
}
Python 3
# Python3 program to print n consecutive
# composite numbers.
# function to find factorial
# of given number
def factorial( n):
res = 1;
for i in range(2, n + 1):
res *= i;
return res;
# Prints n consecutive numbers.
def printNComposite(n):
fact = factorial(n + 1);
for i in range(2, n + 2):
print(fact + i, end = " ");
# Driver Code
n = 4;
printNComposite(n);
# This code is contributed by mits
C
// C# program to print n consecutive composite
// numbers
using System;
public class Program{
// function to find factorial of given
// number
static long factorial(int n) {
long res = 1;
for (int i = 2; i <= n; i++) {
res *= i;
}
return res;
}
// Prints n consecutive numbers.
static void printNComposite(int n) {
long fact = factorial(n + 1);
for (int i = 2; i <= n + 1; ++i) {
Console.Write(fact + i + " ");
}
}
// Driver program to test above function
public static void Main() {
int n = 4;
printNComposite(n);
}
}
// This code is contributed by Rajput-Ji
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to print n consecutive
// composite numbers.
// function to find factorial of given
// number
function factorial( $n)
{
$res = 1;
for ($i = 2; $i <= $n; $i++)
$res *= $i;
return $res;
}
// Prints n consecutive numbers.
function printNComposite(int $n)
{
$fact = factorial($n + 1);
for($i = 2; $i <= $n + 1; ++$i)
echo $fact + $i ," ";
}
// Driver Code
$n = 4;
printNComposite($n);
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// JavaScript program to print n consecutive composite
// numbers
// function to find factorial of given
// number
function factorial(n) {
let res = 1;
for (let i = 2; i <= n; i++) {
res *= i;
}
return res;
}
// Prints n consecutive numbers.
function printNComposite(n) {
let fact = factorial(n + 1);
for (let i = 2; i <= n + 1; ++i) {
document.write(fact + i + " ");
}
}
// Driver code
let n = 4;
printNComposite(n);
// This code is contributed by code_hunt.
</script>
Output:
122 123 124 125
上述解决方案很快就会导致溢出(对于小的 n 值)。我们可以用技巧求大数的阶乘，避免溢出。
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