数 N 的第 k 个最大因子
给定两个正整数 N 和 K ，任务是打印 N 的第 K 个最大因子。
输入: N = 12，K = 3 输出: 4 说明:12 的因子为{1，2，3，4，6，12}。最大因子为 12，第 3 第 3最大因子为 4。
输入: N = 30，K = 2 输出: 15 说明:30 的因子为{1，2，3，5，6，10，15，30}，2 nd 最大的因子为 15。
逼近:思路是检查[N，1]范围内的每个数字，并打印将 N 完全除的第 k 个数字。
迭代从 N 到 0 的循环。现在，对于这个循环中的每个数字:
- 检查它是否将划分为。
- 如果 N 可被当前数整除，则将 K 的值减 1。
- 当 K 变为 0 时，这意味着当前数是的第 K 个最大因子 n
- 根据以上观察打印答案。
下面是上述方法的实现:
C
// C program for the above approach
#include <stdio.h>
// Function to print Kth largest
// factor of N
int KthLargestFactor(int N, int K)
{
// Check for numbers
// in the range [N, 1]
for (int i = N; i > 0; i--) {
// Check if i is a factor of N
if (N % i == 0)
// If Yes, reduce K by 1
K--;
// If K is 0, it means
// i is the required
// Kth factor of N
if (K == 0) {
return i;
}
}
// When K is more
// than the factors of N
return -1;
}
// Driver Code
int main()
{
int N = 12, K = 3;
printf("%d", KthLargestFactor(N, K));
return 0;
}
C++
// C++ program for the above approach
#include <iostream>
using namespace std;
// Function to print Kth largest
// factor of N
int KthLargestFactor(int N, int K)
{
// Check for numbers
// in the range [N, 1]
for (int i = N; i > 0; i--) {
// Check if i is a factor of N
if (N % i == 0)
// If Yes, reduce K by 1
K--;
// If K is 0, it means
// i is the required
// Kth factor of N
if (K == 0) {
return i;
}
}
// When K is more
// than the factors of N
return -1;
}
// Driver Code
int main()
{
int N = 12, K = 3;
cout << KthLargestFactor(N, K);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
class GFG {
// Function to print Kth largest
// factor of N
static int KthLargestFactor(int N, int K)
{
// Check for numbers
// in the range [N, 1]
for (int i = N; i > 0; i--) {
// Check if i is a factor of N
if (N % i == 0)
// If Yes, reduce K by 1
K--;
// If K is 0, it means
// i is the required
// Kth factor of N
if (K == 0) {
return i;
}
}
// When K is more
// than the factors of N
return -1;
}
// Driver Code
public static void main(String[] args)
{
int N = 12, K = 3;
System.out.println(KthLargestFactor(N, K));
}
}
计算机编程语言
# Python program for the above approach
# Function to print Kth largest
# factor of N
def KthLargestFactor(N, K):
for i in range(N, 0, -1):
if N % i == 0:
K -= 1
if K == 0:
return i
return -1
# Driver Code
N = 12
K = 3
print(KthLargestFactor(N, K))
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print Kth largest
// factor of N
static int KthLargestFactor(int N, int K)
{
// Check for numbers
// in the range [N, 1]
for (int i = N; i > 0; i--) {
// Check if i is a factor of N
if (N % i == 0)
// If Yes, reduce K by 1
K--;
// If K is 0, it means
// i is the required
// Kth factor of N
if (K == 0) {
return i;
}
}
// When K is more
// than the factors of N
return -1;
}
// Driver Code
public static void Main()
{
int N = 12, K = 3;
Console.Write(KthLargestFactor(N, K));
}
}
// This code is contributed by ipg2016107.
java 描述语言
<script>
// JavaScript program for the above approach
// Function to print Kth largest
// factor of N
function KthLargestFactor(N, K)
{
// Check for numbers
// in the range [N, 1]
for (let i = N; i > 0; i--) {
// Check if i is a factor of N
if (N % i == 0)
// If Yes, reduce K by 1
K--;
// If K is 0, it means
// i is the required
// Kth factor of N
if (K == 0) {
return i;
}
}
// When K is more
// than the factors of N
return -1;
}
// Driver Code
let N = 12, K = 3;
document.write(KthLargestFactor(N, K));
// This code is contributed by shivanisinghss2110
</script>
Output:
4
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