给定整数N。任务是找到N的所有因子，并打印N的四个因子的乘积，使得：
- 四个因子的总和等于N。
- 这四个因素的乘积最大。
如果找不到4个这样的因素，则打印“不可能”。
注意：为了使乘积最大化，这四个因素可以彼此相等，并且可能存在大量查询。
例子：
Input: 24
Output: Product -> 1296
All factors are -> 1 2 3 4 6 8 12 24
Choose the factor 6 four times,
Therefore, 6+6+6+6 = 24 and product is maximum.
Input: 100
Output: Product -> 390625
All the factors are -> 1 2 4 5 10 10 20 25 50 100
Choose the factor 25 four times.
这个想法是找到从1到N(这是n的最大值)的所有数字的因数。
- 如果给出给定的答案将是不可能的
是首要的。 - 如果给定的n可被4整除，则答案将是pow(q，4)，其中n是被n除以4的商。
- 如果有可能找到答案，则答案必须包含两次倒数第二个因素。并针对其他两个因素运行嵌套循环。
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
using namespace std;
// Function to find primes
bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function to find factors
void factors(int N, vector& v[])
{
for (int i = 2; i < N; i++) {
// run a loop upto square root of that number
for (int j = 1; j * j <= i; j++) {
if (i % j == 0) {
// if the n is perfect square
if (i / j == j)
v[i].push_back(j);
// otherwise push it's two divisors
else {
v[i].push_back(j);
v[i].push_back(i / j);
}
}
}
// sort the divisors
sort(v[i].begin(), v[i].end());
}
}
// Function to find max product
int product(int n)
{
// To store factors of 'n'
vector v[n + 100];
// find factors
factors(n + 100, v);
// if it is divisible by 4.
if (n % 4 == 0) {
int x = n / 4;
x *= x;
return x * x;
}
else {
// if it is prime
if (isPrime[n])
return -1;
// otherwise answer will be possible
else {
int ans = -1;
if (v[n].size() > 2) {
// include last third factor
int fac = v[n][v[n].size() - 3];
// nested loop to find other two factors
for (int i = v[n].size() - 1; i >= 0; i--) {
for (int j = v[n].size() - 1; j >= 0; j--) {
if ((fac * 2) + (v[n][j] + v[n][i]) == n)
ans = max(ans, fac * fac * v[n][j] * v[n][i]);
}
}
return ans;
}
}
}
}
// Driver code
int main()
{
int n = 24;
// function call
cout << product(n);
return 0;
} Java
// Java implementation of above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
// Function to find primes
static boolean isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static Vector > v = new Vector >();
// Function to find factors
static void factors(int N )
{
for (int i = 2; i < N; i++)
{
// run a loop upto square root of that number
for (int j = 1; j * j <= i; j++)
{
if (i % j == 0)
{
// if the n is perfect square
if (i / j == j)
v.get(i).add(j);
// otherwise push it's two divisors
else
{
v.get(i).add(j);
v.get(i).add(i / j);
}
}
}
// sort the divisors
Collections.sort(v.get(i));
}
}
// Function to find max product
static int product(int n)
{
// To store factors of 'n'
v.clear();
for(int i = 0; i < n + 100; i++)
v.add(new Vector());
// find factors
factors(n + 100);
// if it is divisible by 4.
if (n % 4 == 0)
{
int x = n / 4;
x *= x;
return x * x;
}
else
{
// if it is prime
if (isPrime(n))
return -1;
// otherwise answer will be possible
else
{
int ans = -1;
if (v.get(n).size() > 2)
{
// include last third factor
int fac = v.get(n).get(v.get(n).size() - 3);
// nested loop to find other two factors
for (int i = v.get(n).size() - 1; i >= 0; i--)
{
for (int j = v.get(n).size() - 1; j >= 0; j--)
{
if ((fac * 2) + (v.get(n).get(j) +
v.get(n).get(i)) == n)
ans = Math.max(ans, fac * fac *
v.get(n).get(j) *
v.get(n).get(i));
}
}
return ans;
}
}
}
return 0;
}
// Driver code
public static void main(String args[])
{
int n = 24;
// function call
System.out.println( product(n));
}
}
// This code is contributed by Arnab Kundu Python3
# Python3 implementation of above approach
from math import sqrt, ceil, floor
# Function to find primes
def isPrime(n):
# Corner cases
if (n <= 1):
return False
if (n <= 3):
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0):
return False
for i in range(5, ceil(sqrt(n)), 6):
if (n % i == 0 or n % (i + 2) == 0):
return False
return True
# Function to find factors
def factors(N, v):
for i in range(2, N):
# run a loop upto square root of that number
for j in range(1,ceil(sqrt(i)) + 1):
if (i % j == 0):
# if the n is perfect square
if (i // j == j):
v[i].append(j)
# otherwise push it's two divisors
else:
v[i].append(j)
v[i].append(i // j)
# sort the divisors
v = sorted(v)
# Function to find max product
def product(n):
# To store factors of 'n'
v = [[]] * (n + 100)
# find factors
factors(n + 100, v)
# if it is divisible by 4.
if (n % 4 == 0):
x = n // 4
x *= x
return x * x
else :
# if it is prime
if (isPrime[n]):
return -1
# otherwise answer will be possible
else :
ans = -1
if (len(v[n]) > 2):
# include last third factor
fac = v[n][len(v[n]) - 3]
# nested loop to find other two factors
for i in range(len(v[n] - 1), -1, -1):
for j in range(len(v[n] - 1), -1, -1):
if ((fac * 2) + (v[n][j] + v[n][i]) == n):
ans = max(ans, fac * fac * v[n][j] * v[n][i])
return ans
# Driver code
n = 24
# function call
print(product(n))
# This code is contributed by mohit kumar 29C#
// C# implementation of above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find primes
static bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static List > v = new List >();
// Function to find factors
static void factors(int N )
{
for (int i = 2; i < N; i++)
{
// run a loop upto square root of that number
for (int j = 1; j * j <= i; j++)
{
if (i % j == 0)
{
// if the n is perfect square
if (i / j == j)
v[i].Add(j);
// otherwise push it's two divisors
else
{
v[i].Add(j);
v[i].Add(i / j);
}
}
}
// sort the divisors
v[i].Sort();
}
}
// Function to find max product
static int product(int n)
{
// To store factors of 'n'
v.Clear();
for(int i = 0; i < n + 100; i++)
v.Add(new List());
// find factors
factors(n + 100);
// if it is divisible by 4.
if (n % 4 == 0)
{
int x = n / 4;
x *= x;
return x * x;
}
else
{
// if it is prime
if (isPrime(n))
return -1;
// otherwise answer will be possible
else
{
int ans = -1;
if (v[n].Count > 2)
{
// include last third factor
int fac = v[n][v[n].Count - 3];
// nested loop to find other two factors
for (int i = v[n].Count - 1; i >= 0; i--)
{
for (int j = v[n].Count - 1; j >= 0; j--)
{
if ((fac * 2) + (v[n][j] +
v[n][i]) == n)
ans = Math.Max(ans, fac * fac *
v[n][j] *
v[n][i]);
}
}
return ans;
}
}
}
return 0;
}
// Driver code
public static void Main(String []args)
{
int n = 24;
// function call
Console.WriteLine( product(n));
}
}
// This code is contributed by 29AjayKumar 输出：
1296