使用数组计算质因数仅为 2 和 3 的范围内的数字 | 2套

给定两个正整数L 和 R ，任务是计算范围[L, R] 中质因数仅为 2 和 3的元素。
例子：

Input: L = 1, R = 10 
Output: 6 
Explanation: 
2 = 2 
3 = 3 
4 = 2 * 2 
6 = 2 * 3 
8 = 2 * 2 * 2 
9 = 3 * 3
Input: L = 100, R = 200 
Output: 5 
 

对于更简单的方法，请参阅从质因数仅为 2 和 3 的范围中计算数字。
方法：
要以优化的方式解决问题，请按照以下步骤操作：

• 将所有小于或等于 R的 2 的幂存储在数组power2[] 中。
• 类似地，将所有小于或等于 R的 3 的幂存储在另一个数组power3[] 中。
• 初始化第三个数组power23[]并存储power2[]的每个元素与power3[] 的每个元素小于或等于R的成对乘积。
• 现在对于任何范围[L, R] ，我们将简单地迭代数组power23[]并计算范围[L, R] 中的数字。

下面是上述方法的实现：

C++

// C++ program to count the elements
// in the range [L, R] whose prime
// factors are only 2 and 3.
 
#include 
using namespace std;
#define ll long long int
 
// Function which will calculate the
// elements in the given range
void calc_ans(ll l, ll r)
{
 
    vector power2, power3;
 
    // Store the current power of 2
    ll mul2 = 1;
    while (mul2 <= r) {
        power2.push_back(mul2);
        mul2 *= 2;
    }
 
    // Store the current power of 3
    ll mul3 = 1;
    while (mul3 <= r) {
        power3.push_back(mul3);
        mul3 *= 3;
    }
 
    // power23[] will store pairwise product of
    // elements of power2 and power3 that are <=r
    vector power23;
 
    for (int x = 0; x < power2.size(); x++) {
        for (int y = 0; y < power3.size(); y++) {
 
            ll mul = power2[x] * power3[y];
            if (mul == 1)
                continue;
 
            // Insert in power23][]
            // only if mul<=r
            if (mul <= r)
                power23.push_back(mul);
        }
    }
 
    // Store the required answer
    ll ans = 0;
    for (ll x : power23) {
        if (x >= l && x <= r)
            ans++;
    }
 
    // Print the result
    cout << ans << endl;
}
 
// Driver code
int main()
{
 
    ll l = 1, r = 10;
 
    calc_ans(l, r);
 
    return 0;
}

Java

// Java program to count the elements
// in the range [L, R] whose prime
// factors are only 2 and 3.
import java.util.*;
class GFG{
 
// Function which will calculate the
// elements in the given range
static void calc_ans(int l, int r)
{
 
    Vector power2 = new Vector(),
                    power3 = new Vector();
 
    // Store the current power of 2
    int mul2 = 1;
    while (mul2 <= r)
    {
        power2.add(mul2);
        mul2 *= 2;
    }
 
    // Store the current power of 3
    int mul3 = 1;
    while (mul3 <= r)
    {
        power3.add(mul3);
        mul3 *= 3;
    }
 
    // power23[] will store pairwise product of
    // elements of power2 and power3 that are <=r
    Vector power23 = new Vector();
 
    for (int x = 0; x < power2.size(); x++)
    {
        for (int y = 0; y < power3.size(); y++)
        {
            int mul = power2.get(x) *
                      power3.get(y);
            if (mul == 1)
                continue;
 
            // Insert in power23][]
            // only if mul<=r
            if (mul <= r)
                power23.add(mul);
        }
    }
 
    // Store the required answer
    int ans = 0;
    for (int x : power23)
    {
        if (x >= l && x <= r)
            ans++;
    }
 
    // Print the result
    System.out.print(ans + "\n");
}
 
// Driver code
public static void main(String[] args)
{
    int l = 1, r = 10;
 
    calc_ans(l, r);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to count the elements
# in the range [L, R] whose prime
# factors are only 2 and 3.
 
# Function which will calculate the
# elements in the given range
def calc_ans(l, r):
 
    power2 = []; power3 = [];
 
    # Store the current power of 2
    mul2 = 1;
    while (mul2 <= r):
        power2.append(mul2);
        mul2 *= 2;
 
    # Store the current power of 3
    mul3 = 1;
    while (mul3 <= r):
        power3.append(mul3);
        mul3 *= 3;
 
    # power23[] will store pairwise
    # product of elements of power2
    # and power3 that are <=r
    power23 = [];
 
    for x in range(len(power2)):
        for y in range(len(power3)):
 
            mul = power2[x] * power3[y];
            if (mul == 1):
                continue;
 
            # Insert in power23][]
            # only if mul<=r
            if (mul <= r):
                power23.append(mul);
 
    # Store the required answer
    ans = 0;
    for x in power23:
        if (x >= l and x <= r):
            ans += 1;
 
    # Print the result
    print(ans);
 
# Driver code
if __name__ == "__main__":
 
    l = 1; r = 10;
     
    calc_ans(l, r);
 
# This code is contributed by AnkitRai01

C#

// C# program to count the elements
// in the range [L, R] whose prime
// factors are only 2 and 3.
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function which will calculate the
// elements in the given range
static void calc_ans(int l, int r)
{
 
    List power2 = new List(),
              power3 = new List();
 
    // Store the current power of 2
    int mul2 = 1;
    while (mul2 <= r)
    {
        power2.Add(mul2);
        mul2 *= 2;
    }
 
    // Store the current power of 3
    int mul3 = 1;
    while (mul3 <= r)
    {
        power3.Add(mul3);
        mul3 *= 3;
    }
 
    // power23[] will store pairwise product of
    // elements of power2 and power3 that are <=r
    List power23 = new List();
 
    for (int x = 0; x < power2.Count; x++)
    {
        for (int y = 0; y < power3.Count; y++)
        {
            int mul = power2[x] *
                      power3[y];
            if (mul == 1)
                continue;
 
            // Insert in power23,]
            // only if mul<=r
            if (mul <= r)
                power23.Add(mul);
        }
    }
 
    // Store the required answer
    int ans = 0;
    foreach (int x in power23)
    {
        if (x >= l && x <= r)
            ans++;
    }
 
    // Print the result
    Console.Write(ans + "\n");
}
 
// Driver code
public static void Main(String[] args)
{
    int l = 1, r = 10;
 
    calc_ans(l, r);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript program to count the elements
// in the range [L, R] whose prime
// factors are only 2 and 3.
 
// Function which will calculate the
// elements in the given range
function calc_ans(l, r)
{
 
    var power2 = [], power3 = [];
 
    // Store the current power of 2
    var mul2 = 1;
    while (mul2 <= r) {
        power2.push(mul2);
        mul2 *= 2;
    }
 
    // Store the current power of 3
    var mul3 = 1;
    while (mul3 <= r) {
        power3.push(mul3);
        mul3 *= 3;
    }
 
    // power23[] will store pairwise product of
    // elements of power2 and power3 that are <=r
    var power23 = [];
 
    for (var x = 0; x < power2.length; x++) {
        for (var y = 0; y < power3.length; y++) {
 
            var mul = power2[x] * power3[y];
            if (mul == 1)
                continue;
 
            // Insert in power23][]
            // only if mul<=r
            if (mul <= r)
                power23.push(mul);
        }
    }
 
    // Store the required answer
    var ans = 0;
    power23.forEach(x => {
         
        if (x >= l && x <= r)
            ans++;
    });
 
    // Print the result
    document.write( ans );
}
 
// Driver code
var l = 1, r = 10;
calc_ans(l, r);
 
 

输出：

6

时间复杂度： O(log ₂ (R) * log ₃ (R))
注意：该方法可以进一步优化。存储 2 和 3 的幂后，可以使用两个指针计算答案，而不是生成所有数字

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