给定大小为n的整数的数组arr [] 。对于每个元素，必须打印使用该元素的除数形成的每个三元组的乘积之和。
例子：
Input: arr[] = {4}
Output: 8
4 has three divisors 1, 2 and 4.
1 * 2 * 4 = 8
Input: arr[] = {9, 5, 6}
Output: 27 0 72
9 has three divisors 1, 3 and 9. 1 * 3 * 9 = 27
5 has two divisors 1 and 5. So, no triplet is formed .
Similarly, 6 has four divisors 1, 2, 3 and 6. (1 * 2 * 3) + (1 * 3 * 6) + (2 * 3 * 6) + (1 * 6 * 2) = 72
天真的方法：存储数字的所有除数并应用蛮力方法，对于每个三元组，将这三个元素的乘积添加到答案中。
高效的方法：此方法类似于Eratosthenes的筛网，我们用于查找范围内的所有素数。
- 首先，我们创建一个数组sum1，该数组在sum1 [x]处存储数字x的所有除数之和。因此，首先迭代所有小于maximum_Element的数字，然后将此数字加到该数字的倍数的和中。因此，我们将能够在该数字位置存储任意数目的所有除数的总和。
- 填充sum1数组后，需要填充第二个数组sum2的时间，第二个数组sum2会将数字x的每个对数的乘积之和存储在sum2 [x]处。为此，我们对与step1类似的每个数字进行求和并用达到更高的价值。
- 为了填充sum3数组，我们将对小于max_Element的所有数字进行相似的处理，并将j的除数之和相加，以使i为j的除数。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define ll long long
const ll max_Element = 1e6 + 5;
// Global array declaration
int sum1[max_Element], sum2[max_Element], sum3[max_Element];
// Function to find the sum of multiplication of
// every triplet in the divisors of a number
void precomputation(int arr[], int n)
{
// sum1[x] represents the sum of all the divisors of x
for (int i = 1; i < max_Element; i++)
for (int j = i; j < max_Element; j += i)
// Adding i to sum1[j] because i
// is a divisor of j
sum1[j] += i;
// sum2[x] represents the sum of all the divisors of x
for (int i = 1; i < max_Element; i++)
for (int j = i; j < max_Element; j += i)
// Here i is divisor of j and sum1[j] - i
// represents sum of all divisors of
// j which do not include i so we add
// i * (sum1[j] - i) to sum2[j]
sum2[j] += (sum1[j] - i) * i;
// In the above implementation we have considered
// every pair two times so we have to divide
// every sum2 array element by 2
for (int i = 1; i < max_Element; i++)
sum2[i] /= 2;
// Here i is the divisor of j and we are trying to
// add the sum of multiplication of all triplets of
// divisors of j such that one of the divisors is i
for (int i = 1; i < max_Element; i++)
for (int j = i; j < max_Element; j += i)
sum3[j] += i * (sum2[j] - i * (sum1[j] - i));
// In the above implementation we have considered
// every triplet three times so we have to divide
// every sum3 array element by 3
for (int i = 1; i < max_Element; i++)
sum3[i] /= 3;
// Print the results
for (int i = 0; i < n; i++)
cout << sum3[arr[i]] << " ";
}
// Driver code
int main()
{
int arr[] = { 9, 5, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
// Precomputing
precomputation(arr, n);
return 0;
} Java
// Java implementation of the approach
class GFGq
{
static int max_Element = (int) (1e6 + 5);
// Global array declaration
static int sum1[] = new int[max_Element], sum2[] =
new int[max_Element], sum3[] = new int[max_Element];
// Function to find the sum of multiplication of
// every triplet in the divisors of a number
static void precomputation(int arr[], int n)
{
// sum1[x] represents the sum of all the divisors of x
for (int i = 1; i < max_Element; i++)
for (int j = i; j < max_Element; j += i)
// Adding i to sum1[j] because i
// is a divisor of j
sum1[j] += i;
// sum2[x] represents the sum of all the divisors of x
for (int i = 1; i < max_Element; i++)
for (int j = i; j < max_Element; j += i)
// Here i is divisor of j and sum1[j] - i
// represents sum of all divisors of
// j which do not include i so we add
// i * (sum1[j] - i) to sum2[j]
sum2[j] += (sum1[j] - i) * i;
// In the above implementation we have considered
// every pair two times so we have to divide
// every sum2 array element by 2
for (int i = 1; i < max_Element; i++)
sum2[i] /= 2;
// Here i is the divisor of j and we are trying to
// add the sum of multiplication of all triplets of
// divisors of j such that one of the divisors is i
for (int i = 1; i < max_Element; i++)
for (int j = i; j < max_Element; j += i)
sum3[j] += i * (sum2[j] - i * (sum1[j] - i));
// In the above implementation we have considered
// every triplet three times so we have to divide
// every sum3 array element by 3
for (int i = 1; i < max_Element; i++)
sum3[i] /= 3;
// Print the results
for (int i = 0; i < n; i++)
System.out.print(sum3[arr[i]] + " ");
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 9, 5, 6 };
int n = arr.length;
// Precomputing
precomputation(arr, n);
}
}
// This code has been contributed by 29AjayKumarPython3
# Python 3 implementation of the approach
# Global array declaration
#global max_Element
max_Element = 100005
sum1 = [0 for i in range(max_Element)]
sum2 = [0 for i in range(max_Element)]
sum3 = [0 for i in range(max_Element)]
# Function to find the sum of multiplication of
# every triplet in the divisors of a number
def precomputation(arr, n):
# global max_Element
# sum1[x] represents the sum of
# all the divisors of x
for i in range(1, max_Element, 1):
for j in range(i, max_Element, i):
# Adding i to sum1[j] because i
# is a divisor of j
sum1[j] += i
# sum2[x] represents the sum of
# all the divisors of x
for i in range(1, max_Element, 1):
for j in range(i, max_Element, i):
# Here i is divisor of j and sum1[j] - i
# represents sum of all divisors of
# j which do not include i so we add
# i * (sum1[j] - i) to sum2[j]
sum2[j] += (sum1[j] - i) * i
# In the above implementation we have considered
# every pair two times so we have to divide
# every sum2 array element by 2
for i in range(1, max_Element, 1):
sum2[i] = int(sum2[i] / 2)
# Here i is the divisor of j and we are trying to
# add the sum of multiplication of all triplets of
# divisors of j such that one of the divisors is i
for i in range(1, max_Element, 1):
for j in range(i, max_Element, i):
sum3[j] += i * (sum2[j] - i *
(sum1[j] - i))
# In the above implementation we have considered
# every triplet three times so we have to divide
# every sum3 array element by 3
for i in range(1, max_Element, 1):
sum3[i] = int(sum3[i] / 3)
# Print the results
for i in range(n):
print(sum3[arr[i]], end = " ")
# Driver code
if __name__ == '__main__':
arr = [9, 5, 6]
n = len(arr)
# Precomputing
precomputation(arr, n)
# This code is contributed by
# Surendra_GangwarC#
// C# implementation of the approach
using System;
class GFG
{
static int max_Element = (int) (1e6 + 5);
// Global array declaration
static int []sum1 = new int[max_Element];
static int []sum2 = new int[max_Element];
static int []sum3 = new int[max_Element];
// Function to find the sum of multiplication of
// every triplet in the divisors of a number
static void precomputation(int []arr, int n)
{
// sum1[x] represents the sum of all the divisors of x
for (int i = 1; i < max_Element; i++)
for (int j = i; j < max_Element; j += i)
// Adding i to sum1[j] because i
// is a divisor of j
sum1[j] += i;
// sum2[x] represents the sum of all the divisors of x
for (int i = 1; i < max_Element; i++)
for (int j = i; j < max_Element; j += i)
// Here i is divisor of j and sum1[j] - i
// represents sum of all divisors of
// j which do not include i so we add
// i * (sum1[j] - i) to sum2[j]
sum2[j] += (sum1[j] - i) * i;
// In the above implementation we have considered
// every pair two times so we have to divide
// every sum2 array element by 2
for (int i = 1; i < max_Element; i++)
sum2[i] /= 2;
// Here i is the divisor of j and we are trying to
// add the sum of multiplication of all triplets of
// divisors of j such that one of the divisors is i
for (int i = 1; i < max_Element; i++)
for (int j = i; j < max_Element; j += i)
sum3[j] += i * (sum2[j] - i * (sum1[j] - i));
// In the above implementation we have considered
// every triplet three times so we have to divide
// every sum3 array element by 3
for (int i = 1; i < max_Element; i++)
sum3[i] /= 3;
// Print the results
for (int i = 0; i < n; i++)
Console.Write(sum3[arr[i]] + " ");
}
// Driver code
public static void Main()
{
int []arr = { 9, 5, 6 };
int n = arr.Length;
// Precomputing
precomputation(arr, n);
}
}
// This code has been contributed by RyugaPHP
输出：
27 0 72