每个数组元素的 [1, arr[i]] 范围内的非互质数对的计数

给定一个由N个整数组成的数组arr[] ，数组中每个第 i^个元素的任务是从[1, arr[i]]范围内找到非互质数对的数量。

例子：

Input: N = 2, arr[] = {3, 4}
Output: 2 4
Explanation:

1. All non-co-prime pairs from the range [1, 3] are (2, 2) and (3, 3).
2. All non-co-prime pairs from the range [1, 4] are (2, 2), (2, 4), (3, 3) and (4, 4).

Input: N = 3, arr[] = {5, 10, 20}
Output: 5 23 82

朴素的方法：解决问题的最简单方法是迭代每个第 i^个数组元素，然后从范围[1, arr[i]]中生成每个可能的对，并且对于每一对，检查它是否是非 co -prime，即该对的gcd是否大于1 。

请按照以下步骤解决此问题：

• 使用变量i遍历范围[0, N – 1] ，并执行以下步骤：  
  • 将变量lastEle初始化为arr[i]并计数为0以分别存储当前范围的最后一个值和互质数对的数量。
  • 使用变量x和y遍历范围[1, arr[i]]中的每一对，然后执行以下操作：
    • 如果GCD(x, y)大于1 ，则递增计数1 。
  • 最后，打印计数作为答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Recursive function to
// return gcd of two numbers
int gcd(int a, int b)
{
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to count the number of
// non co-prime pairs for each query
void countPairs(int* arr, int N)
{
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // Stores the count of
        // non co-prime pairs
        int count = 0;
 
        // Iterate over the range [1, x]
        for (int x = 1; x <= arr[i]; x++) {
 
            // Iterate over the range [x, y]
            for (int y = x; y <= arr[i]; y++) {
 
                // If gcd of current pair
                // is greater than 1
                if (gcd(x, y) > 1)
                    count++;
            }
        }
        cout << count << " ";
    }
}
 
// Driver Code
int main()
{
    // Given Input
    int arr[] = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Recursive function to
// return gcd of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to count the number of
// non co-prime pairs for each query
static void countPairs(int[] arr, int N)
{
     
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // Stores the count of
        // non co-prime pairs
        int count = 0;
 
        // Iterate over the range [1, x]
        for(int x = 1; x <= arr[i]; x++)
        {
             
            // Iterate over the range [x, y]
            for(int y = x; y <= arr[i]; y++)
            {
                 
                // If gcd of current pair
                // is greater than 1
                if (gcd(x, y) > 1)
                    count++;
            }
        }
        System.out.print(count + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Input
    int arr[] = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
}
}
 
// This code is contributed by subhammahato348

Python3

# Python3 program for the above approach
 
# Recursive program to
# return gcd of two numbers
def gcd(a, b):
     
    if b == 0:
        return a
         
    return gcd(b, a % b)
 
# Function to count the number of
# non co-prime pairs for each query
def countPairs(arr, N):
     
    # Traverse the array arr[]
    for i in range(0, N):
       
        # Stores the count of
        # non co-prime pairs
        count = 0
         
        # Iterate over the range [1,x]
        for x in range(1, arr[i] + 1):
             
            # Iterate over the range [x,y]
            for y in range(x, arr[i] + 1):
               
                # If gcd of current pair
                # is greater than 1
                if gcd(x, y) > 1:
                    count += 1
                     
        print(count, end = " ")
 
# Driver code
if __name__ == '__main__':
   
    # Given Input
    arr = [ 5, 10, 20 ]
    N = len(arr)
 
    # Function Call
    countPairs(arr, N)
 
# This code is contributed by MuskanKalra1

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Recursive function to
// return gcd of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to count the number of
// non co-prime pairs for each query
static void countPairs(int[] arr, int N)
{
     
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // Stores the count of
        // non co-prime pairs
        int count = 0;
 
        // Iterate over the range [1, x]
        for(int x = 1; x <= arr[i]; x++)
        {
             
            // Iterate over the range [x, y]
            for(int y = x; y <= arr[i]; y++)
            {
                 
                // If gcd of current pair
                // is greater than 1
                if (gcd(x, y) > 1)
                    count++;
            }
        }
        Console.Write(count + " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given Input
    int []arr = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

 
// JavaScript program for the above approach
 
// Recursive function to
// return gcd of two numbers
function gcd(a, b)
{
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to count the number of
// non co-prime pairs for each query
function countPairs(arr, N)
{
     
    // Traverse the array arr[]
    for(var i = 0; i < N; i++)
    {
         
        // Stores the count of
        // non co-prime pairs
        var count = 0;
 
        // Iterate over the range [1, x]
        for(var x = 1; x <= arr[i]; x++)
        {
             
            // Iterate over the range [x, y]
            for(var y = x; y <= arr[i]; y++)
            {
                 
                // If gcd of current pair
                // is greater than 1
                if (gcd(x, y) > 1)
                    count++;
            }
        }
        document.write(count + " ");
    }
}
 
// Driver Code
 
// Given Input
var arr = [ 5, 10, 20 ];
var N = 3;
 
// Function Call
countPairs(arr, N);
 
// This code is contributed by shivanisinghss2110
 

输出

5 23 82 

时间复杂度： O(N*M ² *log(M))，其中 M 是 大批。
辅助空间： O(1)

高效方法：上述方法可以通过使用欧拉的totient函数和前缀和数组进行优化。请按照以下步骤解决问题：

• 初始化两个数组，比如phi[]和ans[]为0 ，其中第i^个 数组的元素表示与i互质的整数的计数和从范围[1, arr[i]] 形成的非互质对的计数。
• 使用变量 i 遍历范围[1, MAX] ，并将i分配给phi[i]。
• 使用变量 i 遍历范围[2, MAX] ，并执行以下步骤：
  • 如果phi[i] = i，则使用变量j遍历范围[i, MAX]并执行以下步骤：
    • 将phi[j] / i从phi[j]递减，然后将j增加i 。
• 使用变量 i 遍历范围[1, MAX] ，并执行以下步骤：
  • 将ans[i]更新为ans[i – 1] + (i – phi[i])。
• 最后，完成上述步骤后，打印数组ans[] 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
const int MAX = 1005;
 
// Auxiliary function to pre-compute
// the answer for each array
void preCalculate(vector& phi,
                  vector& ans)
{
    phi[0] = 0;
    phi[1] = 1;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++)
        phi[i] = i;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++) {
 
        // If the number is prime
        if (phi[i] == i) {
 
            for (int j = i; j <= MAX; j += i)
 
                // Subtract the number of
                // pairs which has i as one
                // of their factors
                phi[j] -= (phi[j] / i);
        }
    }
 
    // Iterate over the range [1, MAX]
    for (int i = 1; i <= MAX; i++)
        ans[i] = ans[i - 1] + (i - phi[i]);
}
 
// Function to count the number of
// non co-prime pairs for each query
void countPairs(int* arr, int N)
{
    // The i-th element stores
    // the count of element that
    // are co-prime with i
    vector phi(1e5, 0);
 
    // Stores the resulting array
    vector ans(1e5, 0);
 
    // Function Call
    preCalculate(phi, ans);
 
    // Traverse the array arr[]
    for (int i = 0; i < N; ++i) {
        cout << ans[arr[i]] << " ";
    }
}
 
// Driver Code
int main()
{
    // Given Input
    int arr[] = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
}

Java

// Java program for the above approach
import java.util.*;
class GFG {
     
static int MAX = 1005;
 
// Auxiliary function to pre-compute
// the answer for each array
static void preCalculate(int[] phi,
                  int[] ans)
{
    phi[0] = 0;
    phi[1] = 1;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++)
        phi[i] = i;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++) {
 
        // If the number is prime
        if (phi[i] == i) {
 
            for (int j = i; j <= MAX; j += i)
 
                // Subtract the number of
                // pairs which has i as one
                // of their factors
                phi[j] -= (phi[j] / i);
        }
    }
 
    // Iterate over the range [1, MAX]
    for (int i = 1; i <= MAX; i++)
        ans[i] = ans[i - 1] + (i - phi[i]);
}
 
// Function to count the number of
// non co-prime pairs for each query
static void countPairs(int[] arr, int N)
{
    // The i-th element stores
    // the count of element that
    // are co-prime with i
    int[] phi = new int[100000];
    Arrays.fill(phi, 0);
 
    // Stores the resulting array
    int[] ans = new int[100000];
    Arrays.fill(ans, 0);
     
    // Function Call
    preCalculate(phi, ans);
 
    // Traverse the array arr[]
    for (int i = 0; i < N; ++i) {
        System.out.print(ans[arr[i]] + " ");
    }
}
   
 
// Driver Code
public static void main(String[] args)
{
     // Given Input
    int arr[] = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
}
}
 
// This code is contributed by code_hunt.

Python3

MAX = 1005;
 
def preCalculate(phi,ans):
    phi[0] = 0
    phi[1] = 1
     
    # Iterate over the range [1, MAX]
    for i in range(2, MAX+1):
        phi[i] = i
     
    # Iterate over the range [1, MAX]
    for i in range(2, MAX+1):
        if (phi[i] == i):
            for j in range(i, MAX+1, i):
                 
                # Subtract the number of
                # pairs which has i as one
                # of their factors
                phi[j] -= (phi[j] // i);
                 
    # Iterate over the range [1, MAX]
    for i in range(1, MAX+1):
        ans[i] = ans[i - 1] + (i - phi[i]);
 
# Function to count the number of
# non co-prime pairs for each query
def countPairs(arr, N):
 
    # The i-th element stores
    # the count of element that
    # are co-prime with i
    phi = [0 for i in range(100001)]
 
    # Stores the resulting array
    ans = [0 for i in range(100001)]
 
    # Function Call
    preCalculate(phi, ans);
 
    # Traverse the array arr[]
    for i in range(N):
        print(ans[arr[i]],end=' ');
 
# Given Input
arr= [5, 10, 20]
N = 3;
 
# Function Call
countPairs(arr, N);
 
# This code is contributed by rutvik_56.

C＃

// C# program for the above approach
using System;
class GFG{
 
static int MAX = 1005;
 
// Auxiliary function to pre-compute
// the answer for each array
static void preCalculate(int[] phi,
                  int[] ans)
{
    phi[0] = 0;
    phi[1] = 1;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++)
        phi[i] = i;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++) {
 
        // If the number is prime
        if (phi[i] == i) {
 
            for (int j = i; j <= MAX; j += i)
 
                // Subtract the number of
                // pairs which has i as one
                // of their factors
                phi[j] -= (phi[j] / i);
        }
    }
 
    // Iterate over the range [1, MAX]
    for (int i = 1; i <= MAX; i++)
        ans[i] = ans[i - 1] + (i - phi[i]);
}
 
// Function to count the number of
// non co-prime pairs for each query
static void countPairs(int[] arr, int N)
{
    // The i-th element stores
    // the count of element that
    // are co-prime with i
    int[] phi = new int[100000];
    Array.Clear(phi, 0, 100000);
 
    // Stores the resulting array
    int[] ans = new int[100000];
    Array.Clear(ans, 0, 100000);
 
    // Function Call
    preCalculate(phi, ans);
 
    // Traverse the array arr[]
    for (int i = 0; i < N; ++i) {
        Console.Write(ans[arr[i]] + " ");
    }
}
 
// Driver Code
public static void Main()
{
    // Given Input
    int[] arr = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
}
}
 
// This code is contributed by sanjoy_62.

Javascript

 
// JavaScript program for the above approach
 
const MAX = 1005;
 
// Auxiliary function to pre-compute
// the answer for each array
function preCalculate(phi, ans) {
  phi[0] = 0;
  phi[1] = 1;
 
  // Iterate over the range [1, MAX]
  for (let i = 2; i <= MAX; i++) phi[i] = i;
 
  // Iterate over the range [1, MAX]
  for (let i = 2; i <= MAX; i++) {
    // If the number is prime
    if (phi[i] == i) {
      for (let j = i; j <= MAX; j += i)
        // Subtract the number of
        // pairs which has i as one
        // of their factors
        phi[j] -= Math.floor(phi[j] / i);
    }
  }
 
  // Iterate over the range [1, MAX]
  for (let i = 1; i <= MAX; i++)
  ans[i] = ans[i - 1] + (i - phi[i]);
}
 
// Function to count the number of
// non co-prime pairs for each query
function countPairs(arr, N) {
  // The i-th element stores
  // the count of element that
  // are co-prime with i
  let phi = new Array(1e5).fill(0);
 
  // Stores the resulting array
  let ans = new Array(1e5).fill(0);
 
  // Function Call
  preCalculate(phi, ans);
 
  // Traverse the array arr[]
  for (let i = 0; i < N; ++i) {
    document.write(ans[arr[i]] + " ");
  }
}
 
// Driver Code
 
// Given Input
let arr = [5, 10, 20];
let N = 3;
 
// Function Call
countPairs(arr, N);
 

输出

5 23 82 

时间复杂度： O(N+ M*log(N))，其中 M 是数组的最大元素。
辅助空间： O(M+N)