给定三个整数A ， B ， C ，任务是查找
Σ甲I = 1Σ乙J = 1Σc ^ k = 1时d(IJK)，其中d(x)是x的约数的数量。答案可能非常大，因此，以10 9 +7为模数打印答案。
例子：
Input: A = 2, B = 2, c = 2
Output: 20
Explanation: d(1.1.1) = d(1) = 1;
d(1·1·2) = d(2) = 2;
d(1·2·1) = d(2) = 2;
d(1·2·2) = d(4) = 3;
d(2·1·1) = d(2) = 2;
d(2·1·2) = d(4) = 3;
d(2·2·1) = d(4) = 3;
d(2·2·2) = d(8) = 4.
Input: A = 5, B = 6, C = 7
Output: 1520
方法：
- 查找范围为[1，N]的所有数字的除数
- 使用迭代器i，j，k从1到N开始运行三个嵌套循环
- 然后使用预先计算的除数来找到d(ijk)。
下面是上述方法的实现：
C++
#include
using namespace std;
#define N 100005
#define mod 1000000007
// To store the number of divisors
int cnt[N];
// Function to find the number of divisors
// of all numbers in the range 1 to n
void Divisors()
{
memset(cnt, 0, sizeof cnt);
// For every number 1 to n
for (int i = 1; i < N; i++) {
// Increase divisors count for every number
for (int j = 1; j * i < N; j++)
cnt[i * j]++;
}
}
// Function to find the sum of divisors
int Sumofdivisors(int A, int B, int C)
{
// To store sum
int sum = 0;
Divisors();
for (int i = 1; i <= A; i++) {
for (int j = 1; j <= B; j++) {
for (int k = 1; k <= C; k++) {
int x = i * j * k;
// Count the diviosrs
sum += cnt[x];
if (sum >= mod)
sum -= mod;
}
}
}
return sum;
}
// Driver code
int main()
{
int A = 5, B = 6, C = 7;
// Function call
cout << Sumofdivisors(A, B, C);
return 0;
} Java
// Java code for above given approach
class GFG
{
static int N = 100005;
static int mod = 1000000007;
// To store the number of divisors
static int cnt[] = new int[N];
// Function to find the number of divisors
// of all numbers in the range 1 to n
static void Divisors()
{
// For every number 1 to n
for (int i = 1; i < N; i++)
{
// Increase divisors count for every number
for (int j = 1; j * i < N; j++)
{
cnt[i * j]++;
}
}
}
// Function to find the sum of divisors
static int Sumofdivisors(int A, int B, int C)
{
// To store sum
int sum = 0;
Divisors();
for (int i = 1; i <= A; i++)
{
for (int j = 1; j <= B; j++)
{
for (int k = 1; k <= C; k++)
{
int x = i * j * k;
// Count the diviosrs
sum += cnt[x];
if (sum >= mod)
{
sum -= mod;
}
}
}
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int A = 5, B = 6, C = 7;
// Function call
System.out.println(Sumofdivisors(A, B, C));
}
}
/* This code contributed by PrinciRaj1992 */Python3
# Python3 code for above given approach
N = 100005
mod = 1000000007
# To store the number of divisors
cnt = [0] * N;
# Function to find the number of divisors
# of all numbers in the range 1 to n
def Divisors() :
# For every number 1 to n
for i in range(1, N) :
# Increase divisors count
# for every number
for j in range(1, N // i) :
cnt[i * j] += 1;
# Function to find the sum of divisors
def Sumofdivisors(A, B, C) :
# To store sum
sum = 0;
Divisors();
for i in range(1,A + 1) :
for j in range(1, B + 1) :
for k in range(1, C + 1) :
x = i * j * k;
# Count the diviosrs
sum += cnt[x];
if (sum >= mod) :
sum -= mod;
return sum;
# Driver code
if __name__ == "__main__" :
A = 5; B = 6; C = 7;
# Function call
print(Sumofdivisors(A, B, C));
# This code is contributed by RyugaC#
// C# code for above given approach
using System;
class GFG
{
static int N = 100005;
static int mod = 1000000007;
// To store the number of divisors
static int []cnt = new int[N];
// Function to find the number of divisors
// of all numbers in the range 1 to n
static void Divisors()
{
// For every number 1 to n
for (int i = 1; i < N; i++)
{
// Increase divisors count for every number
for (int j = 1; j * i < N; j++)
{
cnt[i * j]++;
}
}
}
// Function to find the sum of divisors
static int Sumofdivisors(int A, int B, int C)
{
// To store sum
int sum = 0;
Divisors();
for (int i = 1; i <= A; i++)
{
for (int j = 1; j <= B; j++)
{
for (int k = 1; k <= C; k++)
{
int x = i * j * k;
// Count the diviosrs
sum += cnt[x];
if (sum >= mod)
{
sum -= mod;
}
}
}
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
int A = 5, B = 6, C = 7;
// Function call
Console.WriteLine(Sumofdivisors(A, B, C));
}
}
// This code contributed by Rajput-Ji输出：
1520