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📜  [L, R] 范围内的数字计数,可以表示为两个完美幂的和

📅  最后修改于: 2022-05-13 01:56:09.803000             🧑  作者: Mango

[L, R] 范围内的数字计数,可以表示为两个完美幂的和

给定一个范围[L, R] ,任务是找到范围[L, R]中可以 表示为两个完美幂的和。

例子:

方法:给定的问题可以通过使用一些数学观察来解决。请按照以下步骤解决问题:

  • 2生成小于R的所有数字的所有可能幂,并将这些数字存储在数组pow[]中。
  • 初始化一个布尔数组,例如arr[]大小(R + 1)0
  • 生成数组pow[]的所有可能的不同对,如果对的总和最多为 R ,则在数组arr[]中将其标记为1
  • 现在,找到数组arr[]的前缀和。
  • 完成上述步骤后,打印arr[R] – arr[L – 1]的值作为结果。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the number of numbers
// that can be expressed in the form of
// the sum of two perfect powers
int TotalPerfectPowerSum(long long L,
                         long long R)
{
    // Stores all possible powers
    vector pows;
 
    // Push 1 and 0 in it
    pows.push_back(0);
    pows.push_back(1);
 
    // Iterate over all the exponents
    for (int p = 2; p < 25; p++) {
 
        // Iterate over all possible numbers
        long long int num = 2;
 
        // This loop will run for a
        // maximum of sqrt(R) times
        while ((long long int)(pow(num, p) + 0.5) <= R) {
 
            // Push this power in
            // the array pows[]
            pows.push_back(
                (long long int)(pow(num, p) + 0.5));
 
            // Increase the number
            num++;
        }
    }
 
    // Stores if i can be expressed as
    // the sum of perfect power or not
    int ok[R + 1];
    memset(ok, 0, sizeof(ok));
 
    // Iterate over all possible pairs
    // of the array pows[]
    for (int i = 0;
         i < pows.size(); i++) {
 
        for (int j = 0;
             j < pows.size(); j++) {
 
            if (pows[i] + pows[j] <= R
                and pows[i] + pows[j] >= L) {
 
                // The number is valid
                ok[pows[i] + pows[j]] = 1;
            }
        }
    }
 
    // Find the prefix sum of the
    // array ok[]
    for (int i = 0; i <= R; i++) {
        ok[i] += ok[i - 1];
    }
 
    // Return the count of required number
    return ok[R] - ok[L - 1];
}
 
// Driver Code
signed main()
{
    int L = 5, R = 8;
    cout << TotalPerfectPowerSum(L, R);
 
    return 0;
}


Java
// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to find the number of numbers
    // that can be expressed in the form of
    // the sum of two perfect powers
    static int TotalPerfectPowerSum(int L, int R)
    {
        // Stores all possible powers
        ArrayList pows = new ArrayList();
 
        // Push 1 and 0 in it
        pows.add(0);
        pows.add(1);
 
        // Iterate over all the exponents
        for (int p = 2; p < 25; p++) {
 
            // Iterate over all possible numbers
            int num = 2;
 
            // This loop will run for a
            // maximum of sqrt(R) times
            while ((int)(Math.pow(num, p) + 0.5) <= R) {
 
                // Push this power in
                // the array pows[]
                pows.add((int)(Math.pow(num, p) + 0.5));
 
                // Increase the number
                num++;
            }
        }
 
        // Stores if i can be expressed as
        // the sum of perfect power or not
        int[] ok = new int[R + 2];
        // memset(ok, 0, sizeof(ok));
 
        // Iterate over all possible pairs
        // of the array pows[]
        for (int i = 0; i < pows.size(); i++) {
 
            for (int j = 0; j < pows.size(); j++) {
 
                if (pows.get(i) + pows.get(j) <= R
                    && pows.get(i) + pows.get(j) >= L) {
 
                    // The number is valid
                    ok[pows.get(i) + pows.get(j)] = 1;
                }
            }
        }
 
        // Find the prefix sum of the
        // array ok[]
        for (int i = 1; i <= R; i++) {
            ok[i] += ok[i - 1];
        }
 
        // Return the count of required number
        return ok[R] - ok[L - 1];
    }
   
    // Driver Code
    public static void main(String args[])
    {
 
        int L = 5, R = 8;
        System.out.print(TotalPerfectPowerSum(L, R));
    }
}
 
// This code is contributed by avijitmondal1998.


Python3
# python program for the above approach
 
# Function to find the number of numbers
# that can be expressed in the form of
# the sum of two perfect powers
def TotalPerfectPowerSum(L, R):
   
    # Stores all possible powers
    pows = []
 
    # Push 1 and 0 in it
    pows.append(0)
    pows.append(1)
 
    # Iterate over all the exponents
    for p in range(2, 25):
 
                # Iterate over all possible numbers
        num = 2
 
        # This loop will run for a
        # maximum of sqrt(R) times
        while ((int)(pow(num, p) + 0.5) <= R):
 
                        # Push this power in
                        # the array pows[]
            pows.append((int)(pow(num, p) + 0.5))
 
            # Increase the number
            num = num + 1
 
        # Stores if i can be expressed as
        # the sum of perfect power or not
    ok = [0 for _ in range(R + 1)]
     
    # int ok[R + 1];
    # memset(ok, 0, sizeof(ok));
    # Iterate over all possible pairs
    # of the array pows[]
    for i in range(0, int(len(pows))):
        for j in range(0, len(pows)):
            if (pows[i] + pows[j] <= R and pows[i] + pows[j] >= L):
 
                                # The number is valid
                ok[pows[i] + pows[j]] = 1
 
        # Find the prefix sum of the
        # array ok[]
    for i in range(0, R+1):
        ok[i] += ok[i - 1]
 
        # Return the count of required number
    return ok[R] - ok[L - 1]
 
# Driver Code
if __name__ == "__main__":
    L = 5
    R = 8
    print(TotalPerfectPowerSum(L, R))
     
    # This code is contributed by rakeshsahni


C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
 
    // Function to find the number of numbers
    // that can be expressed in the form of
    // the sum of two perfect powers
    static int TotalPerfectPowerSum(long L, long R)
    {
        // Stores all possible powers
        List pows = new List();
 
        // Push 1 and 0 in it
        pows.Add(0);
        pows.Add(1);
 
        // Iterate over all the exponents
        for (int p = 2; p < 25; p++) {
 
            // Iterate over all possible numbers
            long num = 2;
 
            // This loop will run for a
            // maximum of sqrt(R) times
            while ((long)(Math.Pow(num, p) + 0.5) <= R) {
 
                // Push this power in
                // the array pows[]
                pows.Add((long)(Math.Pow(num, p) + 0.5));
 
                // Increase the number
                num++;
            }
        }
 
        // Stores if i can be expressed as
        // the sum of perfect power or not
        int[] ok = new int[R + 2];
        // memset(ok, 0, sizeof(ok));
 
        // Iterate over all possible pairs
        // of the array pows[]
        for (int i = 0; i < pows.Count; i++) {
 
            for (int j = 0; j < pows.Count; j++) {
 
                if (pows[i] + pows[j] <= R
                    && pows[i] + pows[j] >= L) {
 
                    // The number is valid
                    ok[pows[i] + pows[j]] = 1;
                }
            }
        }
 
        // Find the prefix sum of the
        // array ok[]
        for (int i = 1; i <= R; i++) {
            ok[i] += ok[i - 1];
        }
 
        // Return the count of required number
        return ok[R] - ok[L - 1];
    }
 
    // Driver Code
    public static void Main()
    {
        int L = 5, R = 8;
        Console.WriteLine(TotalPerfectPowerSum(L, R));
    }
}
 
// This code is contributed by ukasp.


Javascript


输出:
2

时间复杂度: O(R*log(R))
辅助空间: O(R)