来自两个数组的完全不同的丑数对
给定大小为N和M的两个数组arr1[]和arr2[] ，其中0 ≤ arr1[i], arr2[i] ≤ 1000对于所有有效i ，任务是从第一个数组中取出一个元素，从第二个数组中取出一个元素这样他们两个都是丑陋的数字。我们称它为一对（a，b）。您必须找到所有这些不同对的计数。注意(a, b) 和 (b, a) 没有区别。
丑数是其唯一质因数为2、3 或 5的数。
序列1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, .....显示了前几个丑陋的数字。按照惯例，包括 1。
例子：
Input: arr1[] = {7, 2, 3, 14}, arr2[] = {2, 11, 10}
Output: 4
All distinct pairs are (2, 2), (2, 10), (3, 2) and (3, 10)
Input: arr1[] = {1, 2, 3}, arr2[] = {1, 1}
Output: 3
All distinct pairs are (1, 1), (1, 2) and (1, 3)
方法：
- 首先生成所有难看的数字并将它们插入到 unordered_set s1 中。
- 取另一个空集 s2。
- 运行两个嵌套循环以从两个数组中生成所有可能的对，从第一个数组中获取一个元素（称为 a），从第二个数组中获取一个元素（称为 b）。
- 检查 a 是否存在于 s1 中。如果是，则检查 arr2[] 的每个元素是否也存在于 s1 中。
- 如果 a 和 b 都是丑数，则如果 a 小于 b，则在 s2 中插入对 (a, b)，否则插入 (b, a)。这样做是为了避免重复。
- 总对是集合 s2 的大小。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to get the nth ugly number
unsigned uglyNumber(int n)
{
// To store ugly numbers
int ugly[n];
int i2 = 0, i3 = 0, i5 = 0;
int next_multiple_of_2 = 2;
int next_multiple_of_3 = 3;
int next_multiple_of_5 = 5;
int next_ugly_no = 1;
ugly[0] = 1;
for (int i = 1; i < n; i++) {
next_ugly_no = min(next_multiple_of_2,
min(next_multiple_of_3,
next_multiple_of_5));
ugly[i] = next_ugly_no;
if (next_ugly_no == next_multiple_of_2) {
i2 = i2 + 1;
next_multiple_of_2 = ugly[i2] * 2;
}
if (next_ugly_no == next_multiple_of_3) {
i3 = i3 + 1;
next_multiple_of_3 = ugly[i3] * 3;
}
if (next_ugly_no == next_multiple_of_5) {
i5 = i5 + 1;
next_multiple_of_5 = ugly[i5] * 5;
}
}
return next_ugly_no;
}
// Function to return the required count of pairs
int totalPairs(int arr1[], int arr2[], int n, int m)
{
unordered_set s1;
int i = 1;
// Insert ugly numbers in set
// which are less than 1000
while (1) {
int next_ugly_number = uglyNumber(i);
if (next_ugly_number > 1000)
break;
s1.insert(next_ugly_number);
i++;
}
// Set is used to avoid duplicate pairs
set > s2;
for (int i = 0; i < n; i++) {
// Check if arr1[i] is an ugly number
if (s1.find(arr1[i]) != s1.end()) {
for (int j = 0; j < m; j++) {
// Check if arr2[i] is an ugly number
if (s1.find(arr2[j]) != s1.end()) {
if (arr1[i] < arr2[j])
s2.insert(make_pair(arr1[i], arr2[j]));
else
s2.insert(make_pair(arr2[j], arr1[i]));
}
}
}
}
// Return the size of the set s2
return s2.size();
}
// Driver code
int main()
{
int arr1[] = { 3, 7, 1 };
int arr2[] = { 5, 1, 10, 4 };
int n = sizeof(arr1) / sizeof(arr1[0]);
int m = sizeof(arr2) / sizeof(arr2[0]);
cout << totalPairs(arr1, arr2, n, m);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to get the nth ugly number
static int uglyNumber(int n)
{
// To store ugly numbers
int []ugly = new int[n];
int i2 = 0, i3 = 0, i5 = 0;
int next_multiple_of_2 = 2;
int next_multiple_of_3 = 3;
int next_multiple_of_5 = 5;
int next_ugly_no = 1;
ugly[0] = 1;
for (int i = 1; i < n; i++)
{
next_ugly_no = Math.min(next_multiple_of_2,
Math.min(next_multiple_of_3,
next_multiple_of_5));
ugly[i] = next_ugly_no;
if (next_ugly_no == next_multiple_of_2)
{
i2 = i2 + 1;
next_multiple_of_2 = ugly[i2] * 2;
}
if (next_ugly_no == next_multiple_of_3)
{
i3 = i3 + 1;
next_multiple_of_3 = ugly[i3] * 3;
}
if (next_ugly_no == next_multiple_of_5)
{
i5 = i5 + 1;
next_multiple_of_5 = ugly[i5] * 5;
}
}
return next_ugly_no;
}
// Function to return the required count of pairs
static int totalPairs(int arr1[], int arr2[],
int n, int m)
{
HashSet s1 = new HashSet();
int i = 1;
// Insert ugly numbers in set
// which are less than 1000
while (true)
{
int next_ugly_number = uglyNumber(i);
if (next_ugly_number > 1000)
break;
s1.add(next_ugly_number);
i++;
}
// Set is used to avoid duplicate pairs
HashSet s2 = new HashSet();
for (i = 0; i < n; i++)
{
// Check if arr1[i] is an ugly number
if (s1.contains(arr1[i]))
{
for (int j = 0; j < m; j++)
{
// Check if arr2[i] is an ugly number
if (s1.contains(arr2[j]))
{
if (arr1[i] < arr2[j])
s2.add(new pair(arr1[i], arr2[j]));
else
s2.add(new pair(arr2[j], arr1[i]));
}
}
}
}
// Return the size of the set s2
return s2.size();
}
// Driver code
public static void main(String[] args)
{
int arr1[] = { 3, 7, 1 };
int arr2[] = { 5, 1, 10, 4 };
int n = arr1.length;
int m = arr2.length;
System.out.println(totalPairs(arr1, arr2, n, m));
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 implementation of the approach
# Function to get the nth ugly number
def uglyNumber(n):
# To store ugly numbers
ugly = [None] * n
i2 = i3 = i5 = 0
next_multiple_of_2 = 2
next_multiple_of_3 = 3
next_multiple_of_5 = 5
next_ugly_no = 1
ugly[0] = 1
for i in range(1, n):
next_ugly_no = min(next_multiple_of_2,
min(next_multiple_of_3,
next_multiple_of_5))
ugly[i] = next_ugly_no
if (next_ugly_no == next_multiple_of_2):
i2 = i2 + 1
next_multiple_of_2 = ugly[i2] * 2
if (next_ugly_no == next_multiple_of_3):
i3 = i3 + 1
next_multiple_of_3 = ugly[i3] * 3
if (next_ugly_no == next_multiple_of_5):
i5 = i5 + 1
next_multiple_of_5 = ugly[i5] * 5
return next_ugly_no
# Function to return the required count of pairs
def totalPairs(arr1, arr2, n, m):
s1 = set()
i = 1
# Insert ugly numbers in set
# which are less than 1000
while True:
next_ugly_number = uglyNumber(i)
if (next_ugly_number > 1000):
break
s1.add(next_ugly_number)
i += 1
# Set is used to avoid duplicate pairs
s2 = set()
for i in range(0, n):
# Check if arr1[i] is an ugly number
if arr1[i] in s1:
for j in range(0, m):
# Check if arr2[i] is an ugly number
if arr2[j] in s1:
if (arr1[i] < arr2[j]):
s2.add((arr1[i], arr2[j]))
else:
s2.add((arr2[j], arr1[i]))
# Return the size of the set s2
return len(s2)
# Driver code
if __name__ == "__main__":
arr1 = [3, 7, 1]
arr2 = [5, 1, 10, 4]
n = len(arr1)
m = len(arr2)
print(totalPairs(arr1, arr2, n, m))
# This code is contributed by Rituraj JainC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
public class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to get the nth ugly number
static int uglyNumber(int n)
{
// To store ugly numbers
int []ugly = new int[n];
int i2 = 0, i3 = 0, i5 = 0;
int next_multiple_of_2 = 2;
int next_multiple_of_3 = 3;
int next_multiple_of_5 = 5;
int next_ugly_no = 1;
ugly[0] = 1;
for (int i = 1; i < n; i++)
{
next_ugly_no = Math.Min(next_multiple_of_2,
Math.Min(next_multiple_of_3,
next_multiple_of_5));
ugly[i] = next_ugly_no;
if (next_ugly_no == next_multiple_of_2)
{
i2 = i2 + 1;
next_multiple_of_2 = ugly[i2] * 2;
}
if (next_ugly_no == next_multiple_of_3)
{
i3 = i3 + 1;
next_multiple_of_3 = ugly[i3] * 3;
}
if (next_ugly_no == next_multiple_of_5)
{
i5 = i5 + 1;
next_multiple_of_5 = ugly[i5] * 5;
}
}
return next_ugly_no;
}
// Function to return the required count of pairs
static int totalPairs(int []arr1, int []arr2,
int n, int m)
{
HashSet s1 = new HashSet();
int i = 1;
// Insert ugly numbers in set
// which are less than 1000
while (true)
{
int next_ugly_number = uglyNumber(i);
if (next_ugly_number > 1000)
break;
s1.Add(next_ugly_number);
i++;
}
// Set is used to avoid duplicate pairs
HashSet s2 = new HashSet();
for (i = 0; i < n; i++)
{
// Check if arr1[i] is an ugly number
if (s1.Contains(arr1[i]))
{
for (int j = 0; j < m; j++)
{
// Check if arr2[i] is an ugly number
if (s1.Contains(arr2[j]))
{
if (arr1[i] < arr2[j])
s2.Add(new pair(arr1[i],
arr2[j]));
else
s2.Add(new pair(arr2[j],
arr1[i]));
}
}
}
}
// Return the size of the set s2
return s2.Count;
}
// Driver code
public static void Main(String[] args)
{
int []arr1 = { 3, 7, 1 };
int []arr2 = { 5, 1, 10, 4 };
int n = arr1.Length;
int m = arr2.Length;
Console.WriteLine(totalPairs(arr1, arr2, n, m));
}
}
// This code is contributed by Rajput-Ji Javascript
输出：
8