数字和为 1 的复合数字
原文:https://www . geesforgeks . org/composite-numbers-with-digital-sum-1/
给定一个范围【L，R】，任务是从该范围中找出所有复合的数字，并且它们的数字的最终总和是 1 。 例:
输入: L = 10，R = 100 输出:10 28 46 55 64 82 91 100 10 = 1+0 = 1 28 = 2+8 = 10 = 1+0 = 1 … 91 = 9+1 = 10 = 1+0 = 1 100 = 1+0+0 = 1 输入: L
方法:对于范围内的每个数字，检查该数字是否为复合，即它的除数不是 1，而是数字本身。如果当前数字是一个复合数字，那么继续计算它的数字总和，直到数字减少到一个数字，如果这个数字是 1，那么选择的数字是一个有效的数字。 以下是上述方法的实施:
C++
// C++ implementation of the above approach
#include <iostream>
using namespace std;
// Function that returns true if number n
// is a composite number
bool isComposite(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return false;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return true;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function that returns true if the eventual
// digit sum of number nm is 1
bool isDigitSumOne(int nm)
{
// Loop till the sum is not single digit number
while (nm > 9) {
// Initialize the sum as zero
int sum_digit = 0;
// Find the sum of digits
while (nm > 0) {
int digit = nm % 10;
sum_digit = sum_digit + digit;
nm = nm / 10;
}
nm = sum_digit;
}
// If sum is eventually 1
if (nm == 1)
return true;
else
return false;
}
// Function to print the required numbers
// from the given range
void printValidNums(int l, int r)
{
for (int i = l; i <= r; i++) {
// If i is one of the required numbers
if (isComposite(i) && isDigitSumOne(i))
cout << i << " ";
}
}
// Driver code
int main(void)
{
int l = 10, r = 100;
printValidNums(l, r);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
public class GFG {
// Function that returns true if number n
// is a composite number
static boolean isComposite(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return false;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return true;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function that returns true if the eventual
// digit sum of number nm is 1
static boolean isDigitSumOne(int nm)
{
// Loop till the sum is not single
// digit number
while (nm > 9) {
// Initialize the sum as zero
int sum_digit = 0;
// Find the sum of digits
while (nm > 0) {
int digit = nm % 10;
sum_digit = sum_digit + digit;
nm = nm / 10;
}
nm = sum_digit;
}
// If sum is eventually 1
if (nm == 1)
return true;
else
return false;
}
// Function to print the required numbers
// from the given range
static void printValidNums(int l, int r)
{
for (int i = l; i <= r; i++) {
// If i is one of the required numbers
if (isComposite(i) && isDigitSumOne(i))
System.out.print(i + " ");
}
}
// Driver code
public static void main(String arg[])
{
int l = 10, r = 100;
printValidNums(l, r);
}
}
Python 3
# Python3 implementation of the approach
# Function that returns true if number n
# is a composite number
def isComposite(n):
# Corner cases
if (n <= 1):
return False
if (n <= 3):
return False
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0):
return True
i = 5
while(i * i <= n):
if (n % i == 0 or n % (i + 2) == 0):
return True
i = i + 6
return False
# Function that returns true if the eventual
# digit sum of number nm is 1
def isDigitSumOne(nm) :
# Loop till the sum is not single
# digit number
while(nm>9) :
# Initialize the sum as zero
sum_digit = 0
# Find the sum of digits
while(nm != 0) :
digit = nm % 10
sum_digit = sum_digit + digit
nm = nm // 10
nm = sum_digit
# If sum is eventually 1
if(nm == 1):
return True
else:
return False
# Function to print the required numbers
# from the given range
def printValidNums(m, n ):
for i in range(m, n + 1):
# If i is one of the required numbers
if(isComposite(i) and isDigitSumOne(i)) :
print(i, end =" ")
# Driver code
l = 10
r = 100
printValidNums(m, n)
C
// C# implementation of the above approach
using System;
class GFG
{
// Function that returns true if number n
// is a composite number
static bool isComposite(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return false;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return true;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function that returns true if the
// eventual digit sum of number nm is 1
static bool isDigitSumOne(int nm)
{
// Loop till the sum is not single
// digit number
while (nm > 9)
{
// Initialize the sum as zero
int sum_digit = 0;
// Find the sum of digits
while (nm > 0)
{
int digit = nm % 10;
sum_digit = sum_digit + digit;
nm = nm / 10;
}
nm = sum_digit;
}
// If sum is eventually 1
if (nm == 1)
return true;
else
return false;
}
// Function to print the required numbers
// from the given range
static void printValidNums(int l, int r)
{
for (int i = l; i <= r; i++)
{
// If i is one of the required numbers
if (isComposite(i) && isDigitSumOne(i))
Console.Write(i + " ");
}
}
// Driver code
static public void Main ()
{
int l = 10, r = 100;
printValidNums(l, r);
}
}
// This code is contributed by jit_t
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the above approach
// Function that returns true if number n
// is a composite number
function isComposite($n)
{
// Corner cases
if ($n <= 1)
return false;
if ($n <= 3)
return false;
// This is checked so that we can skip
// middle five numbers in below loop
if ($n % 2 == 0 || $n % 3 == 0)
return true;
for ($i = 5; $i * $i <= $n; $i = $i + 6)
if ($n % $i == 0 || $n % ($i + 2) == 0)
return true;
return false;
}
// Function that returns true if the eventual
// digit sum of number nm is 1
function isDigitSumOne($nm)
{
// Loop till the sum is not single
// digit number
while ($nm > 9)
{
// Initialize the sum as zero
$sum_digit = 0;
// Find the sum of digits
while ($nm > 0)
{
$digit = $nm % 10;
$sum_digit = $sum_digit + $digit;
$nm = floor($nm / 10);
}
$nm = $sum_digit;
}
// If sum is eventually 1
if ($nm == 1)
return true;
else
return false;
}
// Function to print the required numbers
// from the given range
function printValidNums($l, $r)
{
for ($i = $l; $i <= $r; $i++)
{
// If i is one of the required numbers
if (isComposite($i) && isDigitSumOne($i))
echo $i, " ";
}
}
// Driver code
$l = 10; $r = 100;
printValidNums($l, $r);
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// Javascript implementation of the above approach
// Function that returns true if number n
// is a composite number
function isComposite(n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return false;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return true;
for (let i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return true;
return false;
}
// Function that returns true if the eventual
// digit sum of number nm is 1
function isDigitSumOne(nm)
{
// Loop till the sum is not single digit number
while (nm > 9) {
// Initialize the sum as zero
let sum_digit = 0;
// Find the sum of digits
while (nm > 0) {
let digit = nm % 10;
sum_digit = sum_digit + digit;
nm = Math.floor(nm / 10);
}
nm = sum_digit;
}
// If sum is eventually 1
if (nm == 1)
return true;
else
return false;
}
// Function to print the required numbers
// from the given range
function printValidNums(l, r)
{
for (let i = l; i <= r; i++) {
// If i is one of the required numbers
if (isComposite(i) && isDigitSumOne(i))
document.write(i + " ");
}
}
// Driver code
let l = 10, r = 100;
printValidNums(l, r);
// This code is contributed by Mayank Tyagi
</script>
Output:
10 28 46 55 64 82 91 100
优化:我们可以使用筛选算法预计算复合数。
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