给定正整数数组和许多除数查询。在每个查询中，我们都得到一个整数k(> 0)，我们需要计算数组中所有可以被’k’完全整除的元素。
例子：
Input:
2 4 9 15 21 20
k = 2
k = 3
k = 5
Output:
3
3
2
Explanation:
Multiples of '2' in array are:- {2, 4, 20}
Multiples of '3' in array are:- {9, 15, 21}
Multiples of '5' in array are:- {15, 20}
简单方法是遍历整个数组中的每个’k’值，并通过检查数组中每个元素的模数来计算总倍数，即对于i(0
高效的方法是使用Eratosthenes筛网的概念。让我们定义array []中的最大值为’Max’。由于array []中所有数字的倍数始终小于Max，因此我们仅迭代到“ Max”。
现在对于每个值(例如’q’)迭代q，2q，3q，… tk (tk <= MAX)，因为所有这些数字都是’ q ‘的倍数。 1，2，…MAX)在ans []数组中。之后，我们可以在O(1)时间内回答每个查询。
C++
// C++ program to calculate all multiples
// of integer 'k' in array[]
#include
using namespace std;
// ans is global pointer so that both countSieve()
// and countMultiples() can access it.
int* ans = NULL;
// Function to pre-calculate all multiples of
// array elements
void countSieve(int arr[], int n)
{
int MAX = *max_element(arr, arr + n);
int cnt[MAX + 1];
// ans is global pointer so that query function
// can access it.
ans = new int[MAX + 1];
// Initialize both arrays as 0.
memset(cnt, 0, sizeof(cnt));
memset(ans, 0, (MAX + 1) * sizeof(int));
// Store the arr[] elements as index
// in cnt[] array
for (int i = 0; i < n; ++i)
++cnt[arr[i]];
// Iterate over all multiples as 'i'
// and keep the count of array[] ( In
// cnt[] array) elements in ans[] array
for (int i = 1; i <= MAX; ++i)
for (int j = i; j <= MAX; j += i)
ans[i] += cnt[j];
return;
}
int countMultiples(int k)
{
// return pre-calculated result
return ans[k];
}
// Driver code
int main()
{
int arr[] = { 2, 4, 9, 15, 21, 20 };
int n = sizeof(arr) / sizeof(arr[0]);
// pre-calculate all multiples
countSieve(arr, n);
int k = 2;
cout << countMultiples(k) << "\n";
k = 3;
cout << countMultiples(k) << "\n";
k = 5;
cout << countMultiples(k) << "\n";
return 0;
} Java
// Java program to calculate all multiples
// of integer 'k' in array[]
class CountMultiples {
// ans is global array so that both
// countSieve() and countMultiples()
// can access it.
static int ans[];
// Function to pre-calculate all
// multiples of array elements
static void countSieve(int arr[], int n)
{
int MAX = arr[0];
for (int i = 1; i < n; i++)
MAX = Math.max(arr[i], MAX);
int cnt[] = new int[MAX + 1];
// ans is global array so that
// query function can access it.
ans = new int[MAX + 1];
// Store the arr[] elements as
// index in cnt[] array
for (int i = 0; i < n; ++i)
++cnt[arr[i]];
// Iterate over all multiples as 'i'
// and keep the count of array[]
// (In cnt[] array) elements in ans[]
// array
for (int i = 1; i <= MAX; ++i)
for (int j = i; j <= MAX; j += i)
ans[i] += cnt[j];
return;
}
static int countMultiples(int k)
{
// return pre-calculated result
return ans[k];
}
// Driver code
public static void main(String args[])
{
int arr[] = { 2, 4, 9, 15, 21, 20 };
int n = 6;
// pre-calculate all multiples
countSieve(arr, n);
int k = 2;
System.out.println(countMultiples(k));
k = 3;
System.out.println(countMultiples(k));
k = 5;
System.out.println(countMultiples(k));
}
}
/*This code is contributed by Danish Kaleem */Python3
# Python3 program to calculate all multiples
# of integer 'k' in array[]
# ans is global array so that both countSieve()
# and countMultiples() can access it.
ans = []
# Function to pre-calculate all multiples
# of array elements
# Here, the arguments are as follows
# a: given array
# n: length of given array
def countSieve(arr, n):
MAX=max(arr)
# Accessing the global array in the function
global ans
# Initializing "ans" array with zeros
ans = [0]*(MAX + 1)
# Initializing "cnt" array with zeros
cnt = [0]*(MAX + 1)
#Store the arr[] elements as index in cnt[] array
for i in range(n):
cnt[arr[i]] += 1
# Iterate over all multiples as 'i'
# and keep the count of array[] ( In
# cnt[] array) elements in ans[] array
for i in range(1, MAX+1):
for j in range(i, MAX+1, i):
ans[i] += cnt[j]
def countMultiples(k):
# Return pre-calculated result
return(ans[k])
# Driver code
if __name__ == "__main__":
arr = [2, 4, 9 ,15, 21, 20]
n=len(arr)
# Pre-calculate all multiples
countSieve(arr, n)
k=2
print(countMultiples(2))
k=3
print(countMultiples(3))
k=5
print(countMultiples(5))
# This code is contributed by Pratik SomwanshiC#
// C# program to calculate all multiples
// of integer 'k' in array[]
using System;
class GFG {
// ans is global array so that both
// countSieve() and countMultiples()
// can access it.
static int[] ans;
// Function to pre-calculate all
// multiples of array elements
static void countSieve(int[] arr, int n)
{
int MAX = arr[0];
for (int i = 1; i < n; i++)
MAX = Math.Max(arr[i], MAX);
int[] cnt = new int[MAX + 1];
// ans is global array so that
// query function can access it.
ans = new int[MAX + 1];
// Store the arr[] elements as
// index in cnt[] array
for (int i = 0; i < n; ++i)
++cnt[arr[i]];
// Iterate over all multiples as
// 'i' and keep the count of
// array[] (In cnt[] array)
// elements in ans[] array
for (int i = 1; i <= MAX; ++i)
for (int j = i; j <= MAX; j += i)
ans[i] += cnt[j];
return;
}
static int countMultiples(int k)
{
// return pre-calculated result
return ans[k];
}
// Driver code
public static void Main()
{
int[] arr = { 2, 4, 9, 15, 21, 20 };
int n = 6;
// pre-calculate all multiples
countSieve(arr, n);
int k = 2;
Console.WriteLine(countMultiples(k));
k = 3;
Console.WriteLine(countMultiples(k));
k = 5;
Console.WriteLine(countMultiples(k));
}
}
// This code is contributed by nitin mittalPHP
输出：
3
3
2
时间复杂度： O(M * log(M))，其中M是数组元素中的最大值。
辅助空间： O(MAX)