给定两个数组A和B，从数组A中选择一个元素，数组B中另一个元素，从中选取一个随机对。输出该对被最大加权的概率。
例子：
Input : A[] = 1 2 3
B[] = 1 3 3
Output : 0.222
Explanation : Possible pairs are : {1, 1},
{1, 3}, {1, 3}, {2, 1}, {2, 3}, {2, 3},
{3, 1}, {3, 3}, {3, 3} i.e. 9.
The pair with maximum weight is {3, 3} with
frequency 2. So, the probability of random
pair being maximum is 2/9 = 0.2222.蛮力法：以N ^ 2的时间复杂度和计数生成所有可能的对
最大加权对。
更好的方法：对两个数组都进行排序，并对A和B中的最后一个(最大)元素进行计数。最大加权对的数量将是两个计数的乘积。概率将是
(计数的乘积)/ sizeof(A)* sizeof(B)
最佳方法最佳方法将是遍历两个数组并计算最大元素。最大加权对的数量将是两个计数的乘积。概率将是(计数的乘积)/ sizeof(A)* sizeof(B)
下面是实现：
C++
#include
using namespace std;
// Function to return probability
double probability(int a[], int b[], int size1,
int size2)
{
// Count occurrences of maximum element
// in A[]
int max1 = INT_MIN, count1 = 0;
for (int i = 0; i < size1; i++) {
if (a[i] > max1) {
max1 = a[i];
count1 = 1;
}
else if (a[i] == max1) {
count1++;
}
}
// Count occurrences of maximum element
// in B[]
int max2 = INT_MIN, count2 = 0;
for (int i = 0; i < size2; i++) {
if (b[i] > max2) {
max2 = b[i];
count2 = 1;
}
else if (b[i] == max2) {
count2++;
}
}
// Returning probability
return (double)(count1 * count2) /
(size1 * size2);
}
// Driver code
int main()
{
int a[] = { 1, 2, 3 };
int b[] = { 1, 3, 3 };
int size1 = sizeof(a) / sizeof(a[0]);
int size2 = sizeof(b) / sizeof(b[0]);
cout << probability(a, b, size1, size2);
return 0;
} Java
// Java program to find Probability
// of a random pair being the maximum
// weighted pair
import java.io.*;
class GFG {
// Function to return probability
static double probability(int a[], int b[],
int size1,int size2)
{
// Count occurrences of maximum
// element in A[]
int max1 = Integer.MIN_VALUE, count1 = 0;
for (int i = 0; i < size1; i++) {
if (a[i] > max1) {
max1 = a[i];
count1 = 1;
}
else if (a[i] == max1) {
count1++;
}
}
// Count occurrences of maximum
// element in B[]
int max2 = Integer.MIN_VALUE, count2 = 0;
for (int i = 0; i < size2; i++) {
if (b[i] > max2) {
max2 = b[i];
count2 = 1;
}
else if (b[i] == max2) {
count2++;
}
}
// Returning probability
return (double)(count1 * count2) / (size1 * size2);
}
// Driver code
public static void main(String args[])
{
int a[] = { 1, 2, 3 };
int b[] = { 1, 3, 3 };
int size1 = a.length;
int size2 = b.length;
System.out.println(probability(a, b,
size1, size2));
}
}
/*This code is contributed by Nikita Tiwari.*/Python3
import sys
# Function to return probability
def probability(a, b, size1, size2):
# Count occurrences of maximum
# element in A[]
max1 = -(sys.maxsize - 1)
count1 = 0
for i in range(size1):
if a[i] > max1:
count1 = 1
elif a[i] == max1:
count1 += 1
# Count occurrences of maximum
# element in B[]
max2 = -(sys.maxsize - 1)
count2 = 0
for i in range(size2):
if b[i] > max2:
max2 = b[i]
count2 = 1
elif b[i] == max2:
count2 += 1
# Returning probability
return round((count1 * count2) /
(size1 * size2), 6)
# Driver code
a = [1, 2, 3]
b = [1, 3, 3]
size1 = len(a)
size2 = len(b)
print(probability(a, b, size1, size2))
# This code is contributed
# by Shrikant13C#
// C# program to find Probability of a random
// pair being the maximum weighted pair
using System;
class GFG {
// Function to return probability
static float probability(int []a, int []b,
int size1,int size2)
{
// Count occurrences of maximum
// element in A[]
int max1 = int.MinValue, count1 = 0;
for (int i = 0; i < size1; i++) {
if (a[i] > max1) {
max1 = a[i];
count1 = 1;
}
else if (a[i] == max1) {
count1++;
}
}
// Count occurrences of maximum
// element in B[]
int max2 = int.MinValue, count2 = 0;
for (int i = 0; i < size2; i++) {
if (b[i] > max2) {
max2 = b[i];
count2 = 1;
}
else if (b[i] == max2) {
count2++;
}
}
// Returning probability
return (float)(count1 * count2) /
(size1 * size2);
}
// Driver code
public static void Main()
{
int []a = { 1, 2, 3 };
int []b = { 1, 3, 3 };
int size1 = a.Length;
int size2 = b.Length;
Console.WriteLine(probability(a, b,
size1, size2));
}
}
/* This code is contributed by vt_m.*/PHP
$max1)
{
$max1 = $a[$i];
$count1 = 1;
}
else if ($a[$i] == $max1)
{
$count1++;
}
}
// Count occurrences of maximum
// element in B[]
$max2 = PHP_INT_MIN; $count2 = 0;
for ($i = 0; $i < $size2; $i++)
{
if ($b[$i] > $max2)
{
$max2 = $b[$i];
$count2 = 1;
}
else if ($b[$i] == $max2)
{
$count2++;
}
}
// Returning probability
return (double)($count1 * $count2) /
($size1 * $size2);
}
// Driver code
$a = array(1, 2, 3);
$b = array(1, 3, 3);
$size1 = sizeof($a);
$size2 = sizeof($b);
echo probability($a, $b,
$size1, $size2);
// This code is contributed by ajit
?>Javascript
输出：
0.222222