查找所有有效对的 a[i]%a[j] 的总和

给定一个大小为N的数组arr[] 。任务是找到所有有效对的arr[i] % arr[j]的总和。答案可能很大。所以，输出答案模 1000000007
例子：

Input: arr[] = {1, 2, 3}
Output: 5
(1 % 1) + (1 % 2) + (1 % 3) + (2 % 1) + (2 % 2)
+ (2 % 3) + (3 % 1) + (3 % 2) + (3 % 3) = 5
Input: arr[] = {1, 2, 4, 4, 4}
Output: 10

编程需要懂一点英语

方法：存储每个元素的频率并在频率数组上运行嵌套循环并找到所需的答案。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
#define mod (int)(1e9 + 7)
 
// Function to return the sum of (a[i] % a[j])
// for all valid pairs
int Sum_Modulo(int a[], int n)
{
    int max = *max_element(a, a + n);
 
    // To store the frequency of each element
    int cnt[max + 1] = { 0 };
 
    // Store the frequency of each element
    for (int i = 0; i < n; i++)
        cnt[a[i]]++;
 
    // To store the required answer
    long long ans = 0;
 
    // For all valid pairs
    for (int i = 1; i <= max; i++) {
        for (int j = 1; j <= max; j++) {
 
            // Update the count
            ans = ans + cnt[i] * cnt[j] * (i % j);
            ans = ans % mod;
        }
    }
 
    return (int)(ans);
}
 
// Driver code
int main()
{
    int a[] = { 1, 2, 3 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << Sum_Modulo(a, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
static int mod = (int)(1e9 + 7);
 
// Function to return the sum of (a[i] % a[j])
// for all valid pairs
static int Sum_Modulo(int a[], int n)
{
    int max = Arrays.stream(a).max().getAsInt();
 
    // To store the frequency of each element
    int []cnt=new int[max + 1];
 
    // Store the frequency of each element
    for (int i = 0; i < n; i++)
        cnt[a[i]]++;
 
    // To store the required answer
    long ans = 0;
 
    // For all valid pairs
    for (int i = 1; i <= max; i++)
    {
        for (int j = 1; j <= max; j++)
        {
 
            // Update the count
            ans = ans + cnt[i] *
                        cnt[j] * (i % j);
            ans = ans % mod;
        }
    }
 
    return (int)(ans);
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 1, 2, 3 };
    int n = a.length;
 
    System.out.println(Sum_Modulo(a, n));
}
}
 
// This code is contributed
// by PrinciRaj1992

Python3

# Python3 implementation of the approach
mod = 10**9 + 7
 
# Function to return the sum of
# (a[i] % a[j]) for all valid pairs
def Sum_Modulo(a, n):
 
    Max = max(a)
 
    # To store the frequency of each element
    cnt = [0 for i in range(Max + 1)]
 
    # Store the frequency of each element
    for i in a:
        cnt[i] += 1
 
    # To store the required answer
    ans = 0
 
    # For all valid pairs
    for i in range(1, Max + 1):
        for j in range(1, Max + 1):
 
            # Update the count
            ans = ans + cnt[i] * \
                        cnt[j] * (i % j)
            ans = ans % mod
 
    return ans
 
# Driver code
a = [1, 2, 3]
n = len(a)
 
print(Sum_Modulo(a, n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
using System.Linq;
class GFG
{
     
static int mod = (int)(1e9 + 7);
 
// Function to return the sum of (a[i] % a[j])
// for all valid pairs
static int Sum_Modulo(int []a, int n)
{
    int max = a.Max();
 
    // To store the frequency of each element
    int []cnt = new int[max + 1];
 
    // Store the frequency of each element
    for (int i = 0; i < n; i++)
        cnt[a[i]]++;
 
    // To store the required answer
    long ans = 0;
 
    // For all valid pairs
    for (int i = 1; i <= max; i++)
    {
        for (int j = 1; j <= max; j++)
        {
 
            // Update the count
            ans = ans + cnt[i] *
                        cnt[j] * (i % j);
            ans = ans % mod;
        }
    }
    return (int)(ans);
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = { 1, 2, 3 };
    int n = a.Length;
 
    Console.WriteLine(Sum_Modulo(a, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

输出：