给定非负整数的数组a []。任务是计算数组中对(i,j)的数量,以使LCM(a [i],a [j])= HCF(a [i],a [j])。
注意:(i,j)和(j,i)对被认为是相同的,并且我不应该等于j。
例子:
Input : a[] = {3, 4, 3, 4, 5}
Output : 2
Pairs are (3, 3) and (4, 4)
Input : a[] = {1, 1, 1}
Output : 3
天真的方法:生成所有可能的对,并用相等的HCF和LCM计数对。
C++
// Naive C++ program to count number of pairs
// such that their hcf and lcm are equal
#include
using namespace std;
// Function to return HCF of two numbers
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Function to return LCM of two numbers
int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
// 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 if it
// follow the condition
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j]))
ans++;
return ans;
}
// Driver function
int main()
{
int arr[] = { 1, 1, 1 };
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 hcf and lcm are equal
import java.util.*;
class GFG{
// Function to return HCF of two numbers
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Function to return LCM of two numbers
static int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
// 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 if it
// follow the condition
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j]))
ans++;
return ans;
}
// Driver function
public static void main(String[] args)
{
int arr[] = { 1, 1, 1 };
int n = arr.length;
System.out.print(countPairs(arr, n));
}
}
// This code is contributed by Rajput-Ji
Python3
# Naive Python program to count number of pairs
# such that their hcf and lcm are equal
# Function to return HCF of two numbers
def gcd(a, b):
if (a == 0):
return b;
return gcd(b % a, a);
# Function to return LCM of two numbers
def lcm(a, b):
return (a * b) / gcd(a, b);
# Returns the number of valid pairs
def countPairs(arr, n):
ans = 0; # initializing answer
# Traversing the array. For each array
# element, checking if it
# follow the condition
for i in range(n):
for j in range(i+1,n):
if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j])):
ans+=1;
return ans;
# Driver function
if __name__ == '__main__':
arr = [ 1, 1, 1 ];
n = len(arr);
print(countPairs(arr, n));
# This code is contributed by 29AjayKumar
C#
// Naive C# program to count number of pairs
// such that their hcf and lcm are equal
using System;
class GFG{
// Function to return HCF of two numbers
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Function to return LCM of two numbers
static int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
// 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 if it
// follow the condition
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j]))
ans++;
return ans;
}
// Driver function
public static void Main(String[] args)
{
int []arr = { 1, 1, 1 };
int n = arr.Length;
Console.Write(countPairs(arr, n));
}
}
// This code is contributed by 29AjayKumar
Javascript
C++
// C++ program to count number of pairs
// such that their hcf and lcm are equal
#include
using namespace std;
// Function to count number of pairs
// such that their hcf and lcm are equal
int countPairs(int a[], int n)
{
// Store frequencies of array elements
unordered_map frequency;
for (int i = 0; i < n; i++) {
frequency[a[i]]++;
}
int count = 0;
// Count of pairs (arr[i], arr[j])
// where arr[i] = arr[j]
for (auto x : frequency) {
int f = x.second;
count += f * (f - 1) / 2;
}
// Count of pairs (arr[i], arr[j]) where
// arr[i] = arr[j],
return count;
}
// Driver function
int main()
{
int arr[] = { 1, 1, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << countPairs(arr, n);
return 0;
}
Java
// Java program to count number of pairs
// such that their hcf and lcm are equal
import java.util.*;
class GFG{
// Function to count number of pairs
// such that their hcf and lcm are equal
static int countPairs(int arr[], int n)
{
// Store frequencies of array elements
HashMap frequency =
new HashMap();
for (int i = 0; i < n; i++) {
if(frequency.containsKey(arr[i])){
frequency.put(arr[i], frequency.get(arr[i])+1);
}else{
frequency.put(arr[i], 1);
}
}
int count = 0;
// Count of pairs (arr[i], arr[j])
// where arr[i] = arr[j]
for (Map.Entry x : frequency.entrySet()) {
int f = x.getValue();
count += f * (f - 1) / 2;
}
// Count of pairs (arr[i], arr[j]) where
// arr[i] = arr[j],
return count;
}
// Driver function
public static void main(String[] args)
{
int arr[] = { 1, 1, 1 };
int n = arr.length;
System.out.print(countPairs(arr, n));
}
}
// This code contributed by sapnasingh4991
C#
// C# program to count number of pairs
// such that their hcf and lcm are equal
using System;
using System.Collections.Generic;
class GFG{
// Function to count number of pairs
// such that their hcf and lcm are equal
static int countPairs(int []a, int n)
{
// Store frequencies of array elements
Dictionary frequency = new Dictionary();
for (int i = 0; i < n; i++)
{
if (frequency.ContainsKey(a[i]))
{
var val = frequency[a[i]];
frequency.Remove(a[i]);
frequency.Add(a[i], val + 1);
}
else
{
frequency.Add(a[i], 1);
}
}
int count = 0;
// Count of pairs (arr[i], arr[j])
// where arr[i] = arr[j]
foreach(KeyValuePair entry in frequency) {
int f = entry.Value;
count += f * (f - 1) / 2;
}
// Count of pairs (arr[i], arr[j]) where
// arr[i] = arr[j],
return count;
}
// Driver function
public static void Main(String[] args)
{
int []arr = { 1, 1, 1 };
int n = arr.Length;
Console.Write(countPairs(arr, n));
}
}
// This code is contributed by shivanisinghss2110
Python3
# Python 3 program to count number of pairs
# such that their hcf and lcm are equal
from collections import defaultdict
# Function to count number of pairs
# such that their hcf and lcm are equal
def countPairs(a, n):
# Store frequencies of array elements
frequency = defaultdict(int)
for i in range(n) :
frequency[a[i]] += 1
count = 0
# Count of pairs (arr[i], arr[j])
# where arr[i] = arr[j]
for x in frequency.keys():
f = frequency[x]
count += f * (f - 1) // 2
# Count of pairs (arr[i], arr[j]) where
# arr[i] = arr[j],
return count
# Driver function
if __name__ == "__main__":
arr = [ 1, 1, 1 ]
n = len(arr)
print(countPairs(arr, n))
# This code is contributed by chitranayal
输出:
3
高效的方法:如果仔细观察,可以证明只有当数字也相等时,两个数字的HCF和LCM才能相等。
证明:
Let,
HCF(a[i], a[j]) = LCM(a[i], a[j]) = K
Since HCF(a[i], a[j]) = k,
a[i] = k*n1, a[j] = k*n2, for some natural numbers n1, n2
We know that,
HCF × LCM = Product of the two numbers
Therefore,
k*k = k*n1 × k*n2
or, n1*n2 = 1
Implies, n1 = n2 = 1, since n1, n2 are natural numbers.
Therefore,
a[i] = k*n1 = k, a[j] = k*n2 = k
i.e. the numbers must be equal.
因此,我们必须计算对中具有相同元素的对。
可以看出,只有arr [i] = arr [j]形式的对(arr [i],arr [j])会满足给定条件。因此,现在问题被简化为寻找对数(arr [i],arr [j]),从而使arr [i] = arr [j]。
下面是上述方法的实现:
C++
// C++ program to count number of pairs
// such that their hcf and lcm are equal
#include
using namespace std;
// Function to count number of pairs
// such that their hcf and lcm are equal
int countPairs(int a[], int n)
{
// Store frequencies of array elements
unordered_map frequency;
for (int i = 0; i < n; i++) {
frequency[a[i]]++;
}
int count = 0;
// Count of pairs (arr[i], arr[j])
// where arr[i] = arr[j]
for (auto x : frequency) {
int f = x.second;
count += f * (f - 1) / 2;
}
// Count of pairs (arr[i], arr[j]) where
// arr[i] = arr[j],
return count;
}
// Driver function
int main()
{
int arr[] = { 1, 1, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << countPairs(arr, n);
return 0;
}
Java
// Java program to count number of pairs
// such that their hcf and lcm are equal
import java.util.*;
class GFG{
// Function to count number of pairs
// such that their hcf and lcm are equal
static int countPairs(int arr[], int n)
{
// Store frequencies of array elements
HashMap frequency =
new HashMap();
for (int i = 0; i < n; i++) {
if(frequency.containsKey(arr[i])){
frequency.put(arr[i], frequency.get(arr[i])+1);
}else{
frequency.put(arr[i], 1);
}
}
int count = 0;
// Count of pairs (arr[i], arr[j])
// where arr[i] = arr[j]
for (Map.Entry x : frequency.entrySet()) {
int f = x.getValue();
count += f * (f - 1) / 2;
}
// Count of pairs (arr[i], arr[j]) where
// arr[i] = arr[j],
return count;
}
// Driver function
public static void main(String[] args)
{
int arr[] = { 1, 1, 1 };
int n = arr.length;
System.out.print(countPairs(arr, n));
}
}
// This code contributed by sapnasingh4991
C#
// C# program to count number of pairs
// such that their hcf and lcm are equal
using System;
using System.Collections.Generic;
class GFG{
// Function to count number of pairs
// such that their hcf and lcm are equal
static int countPairs(int []a, int n)
{
// Store frequencies of array elements
Dictionary frequency = new Dictionary();
for (int i = 0; i < n; i++)
{
if (frequency.ContainsKey(a[i]))
{
var val = frequency[a[i]];
frequency.Remove(a[i]);
frequency.Add(a[i], val + 1);
}
else
{
frequency.Add(a[i], 1);
}
}
int count = 0;
// Count of pairs (arr[i], arr[j])
// where arr[i] = arr[j]
foreach(KeyValuePair entry in frequency) {
int f = entry.Value;
count += f * (f - 1) / 2;
}
// Count of pairs (arr[i], arr[j]) where
// arr[i] = arr[j],
return count;
}
// Driver function
public static void Main(String[] args)
{
int []arr = { 1, 1, 1 };
int n = arr.Length;
Console.Write(countPairs(arr, n));
}
}
// This code is contributed by shivanisinghss2110
Python3
# Python 3 program to count number of pairs
# such that their hcf and lcm are equal
from collections import defaultdict
# Function to count number of pairs
# such that their hcf and lcm are equal
def countPairs(a, n):
# Store frequencies of array elements
frequency = defaultdict(int)
for i in range(n) :
frequency[a[i]] += 1
count = 0
# Count of pairs (arr[i], arr[j])
# where arr[i] = arr[j]
for x in frequency.keys():
f = frequency[x]
count += f * (f - 1) // 2
# Count of pairs (arr[i], arr[j]) where
# arr[i] = arr[j],
return count
# Driver function
if __name__ == "__main__":
arr = [ 1, 1, 1 ]
n = len(arr)
print(countPairs(arr, n))
# This code is contributed by chitranayal
输出:
3