原文:
给定一个由 n 个整数组成的数组 arr[] ，任务是在选择一个随机对时，找出从数组中获得最大和对 (arr[i]，arr[j]) 的概率。 举例:
输入: arr[] = {3，3，3，3} 输出: 1 所有对将给出最大和，即 6。 输入: arr[] = {1，1，1，2，2，2} 输出: 0.2 只有对(2，2)、(2，2)和(2，2)会给出 15 对中的最大和。 3 / 15 = 0.2
方法:运行两个嵌套循环以获得每一对的总和，保留任何对的最大总和及其计数(即给出该总和的对的数量)。现在，得到这个总数的概率将是(计数/总计对)，其中总计对=(n *(n–1))/2。 以下是上述方法的实现:
c
// c implementation of the approach
#include
using namespace std;
// function to return the probability
// of getting the maximum pair sum
// when a random pair is chosen
// from the given array
float findprob(int arr[], int n)
{
// initialize the maximum sum, its count
// and the count of total pairs
long maxsum = int_min, maxcount = 0, totalpairs = 0;
// for every single pair
for (int i = 0; i < n - 1; i ) {
for (int j = i 1; j < n; j ) {
// get the sum of the current pair
int sum = arr[i] arr[j];
// if the sum is equal to the current
// maximum sum so far
if (sum == maxsum) {
// increment its count
maxcount ;
}
// if the sum is greater than
// the current maximum
else if (sum > maxsum) {
// update the current maximum and
// re-initialize the count to 1
maxsum = sum;
maxcount = 1;
}
totalpairs ;
}
}
// find the required probability
float prob = (float)maxcount / (float)totalpairs;
return prob;
}
// driver code
int main()
{
int arr[] = { 1, 1, 1, 2, 2, 2 };
int n = sizeof(arr) / sizeof(int);
cout << findprob(arr, n);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java implementation of the approach
import java.util.*;
class gfg
{
// function to return the probability
// of getting the maximum pair sum
// when a random pair is chosen
// from the given array
static float findprob(int arr[], int n)
{
// initialize the maximum sum, its count
// and the count of total pairs
long maxsum = integer.min_value,
maxcount = 0, totalpairs = 0;
// for every single pair
for (int i = 0; i < n - 1; i )
{
for (int j = i 1; j < n; j )
{
// get the sum of the current pair
int sum = arr[i] arr[j];
// if the sum is equal to the current
// maximum sum so far
if (sum == maxsum)
{
// increment its count
maxcount ;
}
// if the sum is greater than
// the current maximum
else if (sum > maxsum)
{
// update the current maximum and
// re-initialize the count to 1
maxsum = sum;
maxcount = 1;
}
totalpairs ;
}
}
// find the required probability
float prob = (float)maxcount /
(float)totalpairs;
return prob;
}
// driver code
public static void main(string args[])
{
int arr[] = { 1, 1, 1, 2, 2, 2 };
int n = arr.length;
system.out.println(findprob(arr, n));
}
}
// this code is contributed by rajput-ji
python 3
# python3 implementation of the approach
import sys
# function to return the probability
# of getting the maximum pair sum
# when a random pair is chosen
# from the given array
def findprob(arr, n) :
# initialize the maximum sum, its count
# and the count of total pairs
maxsum = -(sys.maxsize - 1);
maxcount = 0;
totalpairs = 0;
# for every single pair
for i in range(n - 1) :
for j in range(i 1, n) :
# get the sum of the current pair
sum = arr[i] arr[j];
# if the sum is equal to the current
# maximum sum so far
if (sum == maxsum) :
# increment its count
maxcount = 1;
# if the sum is greater than
# the current maximum
elif (sum > maxsum) :
# update the current maximum and
# re-initialize the count to 1
maxsum = sum;
maxcount = 1;
totalpairs = 1;
# find the required probability
prob = maxcount / totalpairs;
return prob;
# driver code
if __name__ == "__main__" :
arr = [ 1, 1, 1, 2, 2, 2 ];
n = len(arr);
print(findprob(arr, n));
# this code is contributed by ankitrai01
c
// c# implementation of above approach
using system;
class gfg
{
// function to return the probability
// of getting the maximum pair sum
// when a random pair is chosen
// from the given array
static float findprob(int []arr, int n)
{
// initialize the maximum sum, its count
// and the count of total pairs
long maxsum = int.minvalue,
maxcount = 0, totalpairs = 0;
// for every single pair
for (int i = 0; i < n - 1; i )
{
for (int j = i 1; j < n; j )
{
// get the sum of the current pair
int sum = arr[i] arr[j];
// if the sum is equal to the current
// maximum sum so far
if (sum == maxsum)
{
// increment its count
maxcount ;
}
// if the sum is greater than
// the current maximum
else if (sum > maxsum)
{
// update the current maximum and
// re-initialize the count to 1
maxsum = sum;
maxcount = 1;
}
totalpairs ;
}
}
// find the required probability
float prob = (float)maxcount /
(float)totalpairs;
return prob;
}
// driver code
public static void main(string []args)
{
int []arr = { 1, 1, 1, 2, 2, 2 };
int n = arr.length;
console.writeline(findprob(arr, n));
}
}
// this code is contributed by 29ajaykumar
java 描述语言
output:
0.2
时间复杂度: o(n 2 )
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