给定总和的四元组计数|套装3

给定四个包含整数元素和整数和的数组，任务是对四元组进行计数，以使每个元素都从不同的数组中选择，并且所有四个元素的和等于给定的和。

例子：

Input: P[] = {0, 2}, Q[] = {-1, -2}, R[] = {2, 1}, S[] = {2, -1}, sum = 0
Output: 2
(0, -1, 2, -1) and (2, -2, 1, -1) are the required quadruplets.

Input: P[] = {1, -1, 2, 3, 4}, Q[] = {3, 2, 4}, R[] = {-2, -1, 2, 1}, S[] = {4, -1}, sum = 3
Output: 10

方法：本文的Set 1和Set 2中讨论了两种解决此问题的方法。在此，将讨论使用二进制搜索的方法。
选择任意两个数组并计算所有可能的对和，并将它们存储在向量中。现在，选择其他两个数组并计算所有可能的总和，并使用tempSum为每个总和检查使用二进制搜索在先前创建的矢量(排序后)中是否存在sum – temp。

下面是上述方法的实现：

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count
// of the required quadruplets
int countQuadruplets(int arr1[], int n1, int arr2[],
                     int n2, int arr3[], int n3,
                     int arr4[], int n4, int value)
  
{
    vector sum1;
    vector::iterator it;
    vector::iterator it2;
    int cnt = 0;
  
    // Take every possible pair sum
    // from the two arrays
    for (int i = 0; i < n1; i++) {
        for (int j = 0; j < n2; j++) {
  
            // Push the sum to a vector
            sum1.push_back(arr1[i] + arr2[j]);
        }
    }
  
    // Sort the sum vector
    sort(sum1.begin(), sum1.end());
  
    // Calculate the pair sums from
    // the other two arrays
    for (int i = 0; i < n3; i++) {
        for (int j = 0; j < n4; j++) {
  
            // Calculate the sum
            int temp = arr3[i] + arr4[j];
  
            // Check whether temp can be added to any
            // sum stored in the sum1 vector such that
            // the result is the required sum
            if (binary_search(sum1.begin(), sum1.end(), value - temp)) {
  
                // Add the count of such values from the sum1 vector
                it = lower_bound(sum1.begin(), sum1.end(), value - temp);
                it2 = upper_bound(sum1.begin(), sum1.end(), value - temp);
                cnt = cnt + ((it2 - sum1.begin()) - (it - sum1.begin()));
            }
        }
    }
  
    return cnt;
}
  
// Driver code
int main()
{
    int arr1[] = { 0, 2 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
  
    int arr2[] = { -1, -2 };
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
  
    int arr3[] = { 2, 1 };
    int n3 = sizeof(arr3) / sizeof(arr3[0]);
  
    int arr4[] = { 2, -1 };
    int n4 = sizeof(arr4) / sizeof(arr4[0]);
  
    int sum = 0;
  
    cout << countQuadruplets(arr1, n1, arr2, n2,
                             arr3, n3, arr4, n4, sum);
  
    return 0;
}

输出：

2