数组中的对数，使得乘积大于总和

给定一个非负整数数组 a[]。计算数组中对 (i, j) 的数量，使得 a[i] + a[j] < a[i]*a[j]。 （这对 (i, j) 和 (j, i) 被认为是相同的，并且 i 不应该等于 j）
例子：

Input : a[] = {3, 4, 5}
Output : 3
Pairs are (3, 4) , (4, 5) and (3,5)

Input  : a[] = {1, 1, 1}
Output : 0

天真的方法
对于每个值 a[i] 计算 a[j] (i > j) 的数量，使得 a[i]*a[j] > a[i] + a[j]

C++

// Naive C++ program to count number of pairs
// such that their sum is more than product.
#include
using namespace std;
 
// Returns the number of valid pairs
int countPairs (int arr[], int n)
{   
    int ans = 0;  // initializing answer
 
    // Traversing the array. For each array
    // element, checking its predecessors that
    // follow the condition
    for (int i = 0; i= 0; j--)
            if (arr[i]*arr[j] > arr[i] + arr[j])
                ans++;
    return ans;
}
 
// Driver function
int main()
{
    int arr[] = {3, 4, 5};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << countPairs(arr, n);
    return 0;
}

Java

// Naive java program to count number of pairs
// such that their sum is more than product.
import java.*;
 
public class GFG
{
     
    // Returns the number of valid pairs
    static int countPairs (int arr[], int n)
    {
        int ans = 0; // initializing answer
     
        // Traversing the array. For each array
        // element, checking its predecessors that
        // follow the condition
        for (int i = 0; i= 0; j--)
                if (arr[i]*arr[j] > arr[i] + arr[j])
                    ans++;
        return ans;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int arr[] = {3, 4, 5};
        int n = arr.length;
        System.out.println(countPairs(arr, n));
    }
}
 
// This code is contributed by Sam007

Python3

# Naive Python program to count number
# of pairs such that their sum is more
# than product.
 
# Returns the number of valid pairs
def countPairs(arr, n):
     
    # initializing answer
    ans = 0
     
    # Traversing the array. For each
    # array element, checking its
    # predecessors that follow the
    # condition
    for i in range(0, n):
        j = i-1
        while(j >= 0):
            if (arr[i] * arr[j] >
                     arr[i] + arr[j]):
                ans = ans + 1
            j = j - 1
    return ans
     
# Driver program to test above function.
arr = [3, 4, 5]
n = len(arr)
k = countPairs(arr, n)
print(k)
     
# This code is contributed by Sam007.

C#

// Naive C# program to count number of pairs
// such that their sum is more than product.
using System;
         
public class GFG
{
    // Returns the number of valid pairs
    static int countPairs (int []arr, int n)
    {
        int ans = 0; // initializing answer
     
        // Traversing the array. For each array
        // element, checking its predecessors that
        // follow the condition
        for (int i = 0; i= 0; j--)
                if (arr[i]*arr[j] > arr[i] + arr[j])
                    ans++;
        return ans;
    }
     
    // driver program
    public static void Main()
    {
        int []arr = {3, 4, 5};
        int n = arr.Length;
        Console.Write( countPairs(arr, n));
    }
}
 
// This code is contributed by Sam007

PHP

= 0; $j--)
            if ($arr[$i] * $arr[$j] >
                $arr[$i] + $arr[$j])
                $ans++;
    return $ans;
}
 
// Driver Code
$arr = array(3, 4, 5);
$n = sizeof($arr);
echo(countPairs($arr, $n));
 
// This code is contributed by Ajit.
?>

Javascript

    // Naive Javascript program to count number of pairs
    // such that their sum is more than product.
     
    // Returns the number of valid pairs
    function countPairs(arr, n)
    {
        let ans = 0; // initializing answer
       
        // Traversing the array. For each array
        // element, checking its predecessors that
        // follow the condition
        for (let i = 0; i<n; i++)
            for (let j = i-1; j>= 0; j--)
                if (arr[i]*arr[j] > (arr[i] + arr[j]))
                    ans++;
        return ans;
    }
     
    let arr = [3, 4, 5];
    let n = arr.length;
    document.write( countPairs(arr, n));
 

输出：

3

高效的方法
当 a[i] = 0 : a[i]*a[j] = 0 且 a[i] + a[j] >= 0 所以如果 a[i] = 0 则找不到对。  
当 a[i] = 1 : a[i]*a[j] = a[j] and a[i] + a[j] = 1 + a[j], 所以当 a[i ] = 1  
当 a[i] = 2 且 a[j] = 2 时： a[i]*a[j] = a[i] + a[j] = 4  
当 a[i] = 2 且 a[j] > 2 或 a[i] > 2 且 a[j] >= 2 时：所有此类对都是有效的。
要解决这个问题，请计算数组中 2 的数量，例如 twoCount。计算数组中大于 2 的数字，比如 twoGreaterCount。答案是 twoCount * twoGreaterCount + twoGreaterCount * (twoGreaterCount-1)/2

C++

// C++ implementation of efficient approach
// to count valid pairs.
#include
using namespace std;
 
// returns the number of valid pairs
int CountPairs (int arr[], int n)
{
    // traversing the array, counting the
    // number of 2s and greater than 2
    // in array
    int twoCount = 0, twoGrCount = 0;
    for (int i = 0; i 2)
            twoGrCount++;
    }
    return twoCount*twoGrCount +
          (twoGrCount*(twoGrCount-1))/2;
}
 
// Driver function
int main()
{
    int arr[] = {3, 4, 5};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << CountPairs(arr, n);
    return 0;
}

Java

// Java implementation of efficient approach
// to count valid pairs.
import java.*;
 
public class GFG
{
    // Returns the number of valid pairs
    static int countPairs (int arr[], int n)
    {
        // traversing the array, counting the
        // number of 2s and greater than 2
        // in array
        int twoCount = 0, twoGrCount = 0;
        for (int i = 0; i 2)
            twoGrCount++;
        }
        return twoCount*twoGrCount +
        (twoGrCount*(twoGrCount-1))/2;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int arr[] = {3, 4, 5};
        int n = arr.length;
        System.out.println(countPairs(arr, n));
    }
}
 
// This code is contributed by Sam007

Python3

# python implementation of efficient approach
# to count valid pairs.
 
# returns the number of valid pairs
def CountPairs (arr,n):
     
    # traversing the array, counting the
    # number of 2s and greater than 2
    # in array
    twoCount = 0
    twoGrCount = 0
    for i in range(0, n):
         
        if (arr[i] == 2):
            twoCount += 1
        else if (arr[i] > 2):
            twoGrCount += 1
     
    return ((twoCount * twoGrCount)
      + (twoGrCount * (twoGrCount - 1)) / 2)
 
# Driver function
arr = [3, 4, 5]
n = len(arr)
print( CountPairs(arr, n))
 
# This code is contributed by Sam007.

C＃

// C# implementation of efficient approach
// to count valid pairs.
using System;
         
public class GFG
{
    // Returns the number of valid pairs
    static int countPairs (int []arr, int n) {
         
    // traversing the array, counting the
    // number of 2s and greater than 2
    // in array
    int twoCount = 0, twoGrCount = 0;
    for (int i = 0; i 2)
            twoGrCount++;
    }
    return twoCount*twoGrCount +
           (twoGrCount*(twoGrCount-1))/2;
    }
     
    // driver program
    public static void Main()
    {
        int []arr = {3, 4, 5};
        int n = arr.Length;
        Console.Write( countPairs(arr, n));
    }
}
 
// This code is contributed by Sam007

PHP

 2)
            $twoGrCount++;
    }
    return $twoCount * $twoGrCount +
          ($twoGrCount * ($twoGrCount -
                               1)) / 2;
}
 
// Driver Code
$arr = array(3, 4, 5);
$n = sizeof($arr);
echo(CountPairs($arr, $n));
 
// This code is contributed by Ajit.
?>

Javascript

    // Javascript implementation of efficient approach to count valid pairs.
     
    // returns the number of valid pairs
    function CountPairs(arr, n)
    {
        // traversing the array, counting the
        // number of 2s and greater than 2
        // in array
        let twoCount = 0, twoGrCount = 0;
        for (let i = 0; i<n; i++)
        {
            if (arr[i] == 2)
                twoCount++;
            else if (arr[i] > 2)
                twoGrCount++;
        }
        return twoCount*twoGrCount + parseInt((twoGrCount*(twoGrCount-1))/2, 10);
    }
     
    let arr = [3, 4, 5];
    let n = arr.length;
    document.write(CountPairs(arr, n));
 

输出：

3

时间复杂度： O(n)
辅助空间： O(1)