原文：
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。
示例：
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.
暴力法：以n ^ 2的时间复杂度生成所有可能的对，并计算最大加权对。
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(count1 * count2) / 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(count1 * count2) / 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 shrikant13
c
// 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
?>
输出：
0.222222
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