给定范围内所有可能的三元组乘积的总和

给定三个整数A ， B和C ，任务是查找表达式的值

$\sum_{i = 1}^{i = A}\sum_{j = 1}^{j = B}\sum_{k = 1}^{k = C} (i * j * k)$

由于答案可能非常大，请以10 ⁹ + 7模输出答案。

例子：

Input: A = 1, B = 1, C = 2 
Output: 3 
Explanation: The value of the given expression is: (1 * 1 * 1 + 1 * 1 * 2) % (10⁹ + 7) = 3 
Therefore, the required output is 3.

Input: A = 10, B =100, C = 1000  
Output: 13874027

天真的方法：解决此问题的最简单方法是生成所有可能的三元组(i，j，k)并打印所有可能乘积(i * j * k)mod(10 ⁹ + 7)的总和。

下面是上述方法的实现。

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
#define M 1000000007
 
// Function to find the sum of all
// possible triplet products (i * j * k)
long long findTripleSum(long long A,
                        long long B,
                        long long C)
{
 
    // Stores sum required sum
    long long sum = 0;
 
    // Iterate over all
    // possible values of i
    for (long long i = 1; i <= A;
         i++) {
 
        // Iterate over all
        // possible values of j
        for (long long j = 1; j <= B;
             j++) {
 
            // Iterate over all
            // possible values of k
            for (long long k = 1; k <= C;
                 k++) {
 
                // Stores the product
                // of (i * j *k)
                long long prod = (((i % M)
 
                                   * (j % M))
                                  % M
                                  * (k % M))
                                 % M;
 
                // Update sum
                sum = (sum + prod) % M;
            }
        }
    }
 
    return sum;
}
 
// Driver Code
int main()
{
 
    long long A = 10;
    long long B = 100;
    long long C = 1000;
    cout << findTripleSum(A, B, C);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
static int M = 1000000007;
 
// Function to find the sum of all
// possible triplet products (i * j * k)
static int findTripleSum(int A, int B,
                         int C)
{
     
    // Stores sum required sum
    int sum = 0;
 
    // Iterate over all
    // possible values of i
    for(int i = 1; i <= A; i++)
    {
         
        // Iterate over all
        // possible values of j
        for(int j = 1; j <= B; j++)
        {
             
            // Iterate over all
            // possible values of k
            for(int k = 1; k <= C; k++)
            {
                 
                // Stores the product
                // of (i * j *k)
                int prod = (((i % M) * (j % M)) %
                                   M * (k % M)) % M;
 
                // Update sum
                sum = (sum + prod) % M;
            }
        }
    }
    return sum;
}
 
// Driver Code
public static void main(String args[])
{
    int A = 10;
    int B = 100;
    int C = 1000;
     
    System.out.println(findTripleSum(A, B, C));
}
}
 
// This code is contributed by bgangwar59

Python3

# Python3 program to implement
# the above approach
M = 1000000007
 
# Function to find the sum
# of all possible triplet
# products (i * j * k)
def findTripleSum(A, B, C):
     
    # Stores sum required sum
    sum = 0
 
    # Iterate over all
    # possible values of i
    for i in range(1, A + 1):
         
        # Iterate over all
        # possible values of j
        for j in range(1, B + 1):
             
            # Iterate over all
            # possible values of k
            for k in range(1, C + 1):
                 
                # Stores the product
                # of (i * j *k)
                prod = (((i % M) * (j % M)) %
                          M * (k % M)) % M
 
                # Update sum
                sum = (sum + prod) % M
 
    return sum
 
# Driver Code
if __name__ == '__main__':
 
    A = 10
    B = 100
    C = 1000
     
    print(findTripleSum(A, B, C))
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach 
using System;
 
class GFG{
  
static int M = 1000000007;
  
// Function to find the sum of all
// possible triplet products (i * j * k)
static int findTripleSum(int A, int B,
                         int C)
{
     
    // Stores sum required sum
    int sum = 0;
  
    // Iterate over all
    // possible values of i
    for(int i = 1; i <= A; i++)
    {
         
        // Iterate over all
        // possible values of j
        for(int j = 1; j <= B; j++)
        {
             
            // Iterate over all
            // possible values of k
            for(int k = 1; k <= C; k++)
            {
                 
                // Stores the product
                // of (i * j *k)
                int prod = (((i % M) * (j % M)) %
                                   M * (k % M)) % M;
  
                // Update sum
                sum = (sum + prod) % M;
            }
        }
    }
    return sum;
}
  
// Driver Code
public static void Main()
{
    int A = 10;
    int B = 100;
    int C = 1000;
      
    Console.WriteLine(findTripleSum(A, B, C));
}
}
 
// This code is contributed by code_hunt

Javascript

 
// JavaScript program to implement the above approach
 
let M = 1000000007;
 
// Function to find the sum of all
// possible triplet products (i * j * k)
function findTripleSum(A, B, C)
{
     
    // Stores sum required sum
    let sum = 0;
 
    // Iterate over all
    // possible values of i
    for(let i = 1; i <= A; i++)
    {
         
        // Iterate over all
        // possible values of j
        for(let j = 1; j <= B; j++)
        {
             
            // Iterate over all
            // possible values of k
            for(let k = 1; k <= C; k++)
            {
                 
                // Stores the product
                // of (i * j *k)
                let prod = (((i % M) * (j % M)) %
                                   M * (k % M)) % M;
 
                // Update sum
                sum = (sum + prod) % M;
            }
        }
    }
    return sum;
}
 
// Driver Code
    let A = 10;
    let B = 100;
    let C = 1000;
     
    document.write(findTripleSum(A, B, C));
     
    // This code is cotributed by susmitakunndugoaldanga.

输出：

13874027

时间复杂度： O(A * B * C)
辅助空间： O(1)

高效的方法：可以基于以下观察来优化上述方法：

Single summation: 
 

$\sum_{i = 1}^{i = A} i = (A * (A + 1)) / 2$
 

Double summation: 
 

$\sum_{i = 1}^{i = A}\sum_{j = 1}^{j = B} (i * j) = ((B * (B + 1)) / 2) * \sum_{i = 1}^{i = A} i$
 

= ((A * (A + 1) / 2) * (B * (B + 1) / 2))

 

Similarly, for triple summation: 
 

$\sum_{i = 1}^{i = A}\sum_{j = 1}^{j = B}\sum_{k = 1}^{k = C} (i * j * k)$
 

= ((A * (A + 1) / 2) * (B * (B + 1) / 2) * (C * (C + 1) / 2)) 
 

 

请按照以下步骤解决问题：

• 初始化一个变量，例如说MMI，以存储8的模乘逆数。
• 初始化一个变量M来存储10 ⁹ + 7的值。

Finally, print the value of ((A * (A + 1) % M) * (B * (B + 1) % M) * (C * (C + 1) % M) * MMI) % M.

C++

// C++ implementation to implement
// the above approach
 
#include 
using namespace std;
#define M 1000000007
 
// Function to find the value
// of power(X, N) % M
long long power(long long x,
                long long N)
{
 
    // Stores the value
    // of (X ^ N) % M
    long long res = 1;
 
    // Calculate the value of
    // power(x, N) % M
    while (N > 0) {
 
        // If N is odd
        if (N & 1) {
 
            // Update res
            res = (res * x) % M;
        }
 
        // Update x
        x = (x * x) % M;
 
        // Update N
        N = N >> 1;
    }
    return res;
}
 
// Function to find modulo multiplicative
// inverse of X under modulo M
long long modinv(long long X)
{
    return power(X, M - 2);
}
 
// Function to find the sum of all
// possible triplet products (i * j * k)
int findTripleSum(long long A, long long B,
                  long long C)
{
 
    // Stores modulo multiplicative
    // inverse of 8
    long long MMI = modinv(8);
 
    // Stores the sum of all
    // possible values of (i * j * k)
    long long res = 0;
 
    // Update res
    res = ((((A % M * (A + 1) % M)
             % M
             * (B % M * (B + 1) % M)
             % M)
            % M
            * (C % M * (C + 1) % M)
            % M)
           % M
           * MMI)
          % M;
 
    return res;
}
 
// Driver Code
int main()
{
 
    long long A = 10;
    long long B = 100;
    long long C = 1000;
    cout << findTripleSum(A, B, C);
 
    return 0;
}

Java

// Java implementation to implement
// the above approach
import java.util.*;
 
class GFG{
   
static final int M = 1000000007;
 
// Function to find the value
// of power(X, N) % M
static long power(long x, long N)
{
   
    // Stores the value
    // of (X ^ N) % M
    long res = 1;
 
    // Calculate the value of
    // power(x, N) % M
    while (N > 0)
    {
       
        // If N is odd
        if (N % 2 == 1)
        {
           
            // Update res
            res = (res * x) % M;
        }
 
        // Update x
        x = (x * x) % M;
 
        // Update N
        N = N >> 1;
    }
    return res;
}
 
// Function to find modulo multiplicative
// inverse of X under modulo M
static long modinv(long X)
{
    return power(X, M - 2);
}
 
// Function to find the sum of all
// possible triplet products (i * j * k)
static long findTripleSum(long A, long B,
                          long C)
{
   
    // Stores modulo multiplicative
    // inverse of 8
    long MMI = modinv(8);
 
    // Stores the sum of all
    // possible values of (i * j * k)
    long res = 0;
 
    // Update res
    res = ((((A % M * (A + 1) % M) % M *
             (B % M * (B + 1) % M) % M) % M *
             (C % M * (C + 1) % M) % M) % M *
               MMI) % M;
   
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    long A = 10;
    long B = 100;
    long C = 1000;
   
    System.out.print(findTripleSum(A, B, C));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to
# implement the above approach
M = 1000000007
 
# Function to find the value
# of power(X, N) % M
def power(x,N):
   
    global M
     
    # Stores the value
    # of (X ^ N) % M
    res = 1
 
    # Calculate the value of
    # power(x, N) % M
    while (N > 0):
       
        # If N is odd
        if (N & 1):
           
            # Update res
            res = (res * x) % M
 
        # Update x
        x = (x * x) % M
 
        # Update N
        N = N >> 1
    return res
 
# Function to find modulo
# multiplicative inverse
# of X under modulo M
def modinv(X):
   
    return power(X, M - 2)
 
# Function to find the
# sum of all possible
# triplet products (i * j * k)
def findTripleSum(A, B, C):
   
    global M
     
    # Stores modulo multiplicative
    # inverse of 8
    MMI = modinv(8)
    # Stores the sum of all
    # possible values of (i * j * k)
    res = 0
 
    # Update res
    res = ((((A % M * (A + 1) % M) % M *
             (B % M * (B + 1) % M) % M) % M *
             (C % M * (C + 1) % M) % M) % M *
              MMI)% M
    return res
 
# Driver Code
if __name__ == '__main__':
   
    A = 10
    B = 100
    C = 1000
    print(findTripleSum(A, B, C))
 
# This code is contributed by SURENDRA_GANGWAR

C＃

// C# implementation to implement
// the above approach
using System;
 
class GFG{
   
static readonly int M = 1000000007;
 
// Function to find the value
// of power(X, N) % M
static long power(long x, long N)
{
     
    // Stores the value
    // of (X ^ N) % M
    long res = 1;
 
    // Calculate the value of
    // power(x, N) % M
    while (N > 0)
    {
         
        // If N is odd
        if (N % 2 == 1)
        {
           
            // Update res
            res = (res * x) % M;
        }
 
        // Update x
        x = (x * x) % M;
 
        // Update N
        N = N >> 1;
    }
    return res;
}
 
// Function to find modulo multiplicative
// inverse of X under modulo M
static long modinv(long X)
{
    return power(X, M - 2);
}
 
// Function to find the sum of all
// possible triplet products (i * j * k)
static long findTripleSum(long A, long B,
                          long C)
{
   
    // Stores modulo multiplicative
    // inverse of 8
    long MMI = modinv(8);
 
    // Stores the sum of all
    // possible values of (i * j * k)
    long res = 0;
 
    // Update res
    res = ((((A % M * (A + 1) % M) % M *
             (B % M * (B + 1) % M) % M) % M *
             (C % M * (C + 1) % M) % M) % M *
               MMI) % M;
   
    return res;
}
 
// Driver Code
public static void Main(String[] args)
{
    long A = 10;
    long B = 100;
    long C = 1000;
   
    Console.Write(findTripleSum(A, B, C));
}
}
 
// This code is contributed by Amit Katiyar

输出：

13874027

时间复杂度： O(log ₂ N)，其中N =(A * B * C)
辅助空间： O(1)