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📜  使用数组| | | | | | | | | | | | | | | | | | | | | | | | | | | |套装2

📅  最后修改于: 2021-06-26 18:32:12             🧑  作者: Mango

给定两个正整数L和R ，任务是计算素数仅为2和3的范围[L，R]中的元素。

例子：

有关更简单的方法，请参阅对主要因子仅为2和3的范围中的数字进行计数。

方法：
要以优化的方式解决问题，请按照以下步骤操作：

  • 将小于或等于R的所有2的幂存储在数组power2 []中。
  • 类似地，将小于或等于R的所有3的幂存储在另一个数组power3 []中。
  • 初始化第三数组power23 []并存储power2 []的每个元素与power3 []的每个元素的小于或等于R的成对乘积。
  • 现在，对于任何范围[L，R] ，我们将简单地遍历数组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

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的乘方后，可以使用两个指针来计算答案，而不用生成所有数字

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