一个数组中乘积大于总和的对的数量
原文:https://www . geesforgeks . org/number-pairs-array-product-greater-sum/
给定非负整数数组 a[。计算数组中的对(I，j)的数量，使得 a[i] + a[j] < a[i]*a[j]. (the pair (i, j) and (j, i) are considered same and i should not be equal to j) 示例:
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)的个数，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<bits/stdc++.h>
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<n; i++)
for (int j = i-1; j>= 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 语言(一种计算机语言，尤用于创建网站)
// 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<n; i++)
for (int j = i-1; j>= 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
Python 3
# 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<n; i++)
for (int j = i-1; j>= 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Naive PHP program to
// count number of pairs
// such that their sum
// is more than product.
// Returns the number
// of valid pairs
function countPairs ($arr, $n)
{
// initializing answer
$ans = 0;
// Traversing the array.
// For each array
// element, checking
// its predecessors that
// follow the condition
for ($i = 0; $i < $n; $i++)
for ($j = $i - 1; $j >= 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.
?>
java 描述语言
<script>
// 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));
</script>
输出:
3
高效进场 当 a[i] = 0 时:a[i]a[j] = 0 且 a[i] + a[j] > = 0 所以如果 a[i] = 0 则找不到配对。 当 a[I]= 1:a[I] a[j]= a[j]和 a[i] + a[j] = 1 + a[j]时，因此当 a[i] = 1 当 a[i] = 2 且 a[j]= 2:a[I] a[j]= a[I]时，找不到配对 要解决这个问题，统计数组中 2s 的个数，比如 twoCount。计算数组中大于 2 的数字，比如 twoGreaterCount。答案是两个计数两个更大的计数+两个更大的计数*(两个更大的计数-1)/2
C++
// C++ implementation of efficient approach
// to count valid pairs.
#include<bits/stdc++.h>
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<n; i++)
{
if (arr[i] == 2)
twoCount++;
else if (arr[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 语言(一种计算机语言，尤用于创建网站)
// 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<n; i++)
{
if (arr[i] == 2)
twoCount++;
else if (arr[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
Python 3
# 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
elif (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<n; i++)
{
if (arr[i] == 2)
twoCount++;
else if (arr[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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP 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
$twoCount = 0; $twoGrCount = 0;
for ($i = 0; $i < $n; $i++)
{
if ($arr[$i] == 2)
$twoCount++;
else if ($arr[$i] > 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.
?>
java 描述语言
<script>
// 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));
</script>
输出:
3
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