用给定的一组因子生成 n 个数字
原文:https://www . geesforgeks . org/generation-n-numbers-given-set-factors/
给定一个由 k 个数字组成的数组因子[] ，任务是打印前 n 个数字(按升序排列)，其因子来自给定的数组。
示例:
Input : factor[] = {2, 3, 4, 7}
n = 8
Output : 2 3 4 6 7 8 9 10
Input : factor[] = {3, 5, 7}
n = 10
Output : 3 5 6 7 9 10 12 14 15 18
这个问题主要是丑数问题的延伸。我们创建一个辅助数组下一个[] ，其大小与因子[]相同，以跟踪因子的下一个倍数。在每次迭代中，我们从下一个输出最小元素，然后用相应的因子递增它。如果最小值等于前一个输出值，则递增它(以避免重复)。否则我们打印最小值。
C++
// C++ program to generate n numbers with
// given factors
#include<bits/stdc++.h>
using namespace std;
// Generate n numbers with factors in factor[]
void generateNumbers(int factor[], int n, int k)
{
// array of k to store next multiples of
// given factors
int next[k] = {0};
// Prints n numbers
int output = 0; // Next number to print as output
for (int i=0; i<n; )
{
// Find the next smallest multiple
int toincrement = 0;
for (int j=0; j<k; j++)
if (next[j] < next[toincrement])
toincrement = j;
// Printing minimum in each iteration
// print the value if output is not equal to
// current value(to avoid the duplicates)
if (output != next[toincrement])
{
output = next[toincrement];
printf("%d ", next[toincrement]);
i++;
}
// incrementing the current value by the
// respective factor
next[toincrement] += factor[toincrement];
}
}
// Driver code
int main()
{
int factor[] = {3, 5, 7};
int n = 10;
int k = sizeof(factor)/sizeof(factor[0]);
generateNumbers(factor, n, k);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to generate
// n numbers with given factors
import java.io.*;
class GFG
{
// Generate n numbers with
// factors in factor[]
static void generateNumbers(int factor[],
int n, int k)
{
// array of k to store
// next multiples of
// given factors
int next[] = new int[k];
// Prints n numbers
int output = 0; // Next number to
// print as output
for (int i = 0; i < n;)
{
// Find the next
// smallest multiple
int toincrement = 0;
for (int j = 0; j < k; j++)
if (next[j] < next[toincrement])
toincrement = j;
// Printing minimum in
// each iteration print
// the value if output
// is not equal to current
// value(to avoid the
// duplicates)
if (output != next[toincrement])
{
output = next[toincrement];
System.out.print(
next[toincrement] + " ");
i++;
}
// incrementing the
// current value by the
// respective factor
next[toincrement] +=
factor[toincrement];
}
}
// Driver code
public static void main (String[] args)
{
int factor[] = {3, 5, 7};
int n = 10;
int k = factor.length;
generateNumbers(factor, n, k);
}
}
// This code is contributed
// by ajit
Python 3
# Python3 program to generate n
# numbers with given factors
# Generate n numbers with
# factors in factor[]
def generateNumbers(factor, n, k):
# array of k to store next
# multiples of given factors
next = [0] * k;
# Prints n numbers
output = 0; # Next number to
# print as output
i = 0;
while(i < n):
# Find the next smallest
# multiple
toincrement = 0;
for j in range(k):
if(next[j] < next[toincrement]):
toincrement = j;
# Printing minimum in each
# iteration print the value
# if output is not equal to
# current value(to avoid the
# duplicates)
if(output != next[toincrement]):
output = next[toincrement];
print(next[toincrement], end = " ");
i += 1;
# incrementing the current
# value by the respective factor
next[toincrement] += factor[toincrement];
# Driver code
factor = [3, 5, 7];
n = 10;
k = len(factor);
generateNumbers(factor, n, k);
# This code is contributed by mits
C
// C# program to generate
// n numbers with given factors
using System;
class GFG
{
// Generate n numbers with
// factors in factor[]
static void generateNumbers(int []factor,
int n, int k)
{
// array of k to store
// next multiples of
// given factors
int []next = new int[k];
// Prints n numbers
int output = 0; // Next number to
// print as output
for (int i = 0; i < n;)
{
// Find the next
// smallest multiple
int toincrement = 0;
for (int j = 0; j < k; j++)
if (next[j] < next[toincrement])
toincrement = j;
// Printing minimum in
// each iteration print
// the value if output
// is not equal to current
// value(to avoid the
// duplicates)
if (output != next[toincrement])
{
output = next[toincrement];
Console.Write(
next[toincrement] + " ");
i++;
}
// incrementing the
// current value by the
// respective factor
next[toincrement] +=
factor[toincrement];
}
}
// Driver code
static public void Main ()
{
int []factor = {3, 5, 7};
int n = 10;
int k = factor.Length;
generateNumbers(factor, n, k);
}
}
// This code is contributed
// by akt_mit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to generate n numbers
// with given factors
// Generate n numbers with factors
// in factor[]
function generateNumbers($factor, $n, $k)
{
// array of k to store next
// multiples of given factors
$next = array_fill(0, $k, 0);
// Prints n numbers
$output = 0; // Next number to print
// as output
for ($i = 0; $i < $n; )
{
// Find the next smallest multiple
$toincrement = 0;
for ($j = 0; $j < $k; $j++)
if ($next[$j] < $next[$toincrement])
$toincrement = $j;
// Printing minimum in each iteration
// print the value if output is not
// equal to current value(to avoid
// the duplicates)
if ($output != $next[$toincrement])
{
$output = $next[$toincrement];
echo $next[$toincrement] . " ";
$i++;
}
// incrementing the current value
// by the respective factor
$next[$toincrement] += $factor[$toincrement];
}
}
// Driver code
$factor = array(3, 5, 7);
$n = 10;
$k = count($factor);
generateNumbers($factor, $n, $k);
// This code is contributed by mits
?>
java 描述语言
<script>
// Javascript program to generate
// n numbers with given factors
// Generate n numbers with
// factors in factor[]
function generateNumbers(factor, n, k)
{
// Array of k to store
// next multiples of
// given factors
let next = new Array(k);
next.fill(0);
// Prints n numbers
let output = 0; // Next number to
// print as output
for(let i = 0; i < n;)
{
// Find the next
// smallest multiple
let toincrement = 0;
for(let j = 0; j < k; j++)
if (next[j] < next[toincrement])
toincrement = j;
// Printing minimum in
// each iteration print
// the value if output
// is not equal to current
// value(to avoid the
// duplicates)
if (output != next[toincrement])
{
output = next[toincrement];
document.write(next[toincrement] + " ");
i++;
}
// Incrementing the
// current value by the
// respective factor
next[toincrement] += factor[toincrement];
}
}
// Driver code
let factor = [ 3, 5, 7 ];
let n = 10;
let k = factor.length;
generateNumbers(factor, n, k);
// This code is contributed by divyesh072019
</script>
输出:
3 5 6 7 9 10 12 14 15 18
时间复杂度–O(n * k) T3】辅助空间–O(k)
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