给定大小为N的数组arr ，任务是对所有有效索引i的索引j的数量(j
例子：
Input: arr[] = {8, 1, 28, 4, 2, 6, 7}
Output: 0, 1, 0, 2, 3, 0, 1
No of multiples for each element before itself –
N(8) = 0 ()
N(1) = 1 (8)
N(28) = 0 ()
N(4) = 2 (28, 8)
N(2) = 3 (4, 28, 8)
N(6) = 0 ()
N(7) = 1 (28)
Input: arr[] = {1, 1, 1, 1}
Output: 0, 1, 2, 3
天真的方法：遍历每个索引i在[0，i-1]范围内的所有有效索引j ；并计算每个索引的除数。
时间复杂度： O(N 2 )
空间复杂度： O(1)
高效方法：这种方法是使用地图。在遍历数组时增加映射中因子的计数，并在映射中查找该计数以查找所有有效j(
下面是上述方法的实现。
C++
// C++ program to count of multiples
// in an Array before every element
#include
using namespace std;
// Function to find all factors of N
// and keep their count in map
void add_factors(int n,
unordered_map& mp)
{
// Traverse from 1 to sqrt(N)
// if i divides N,
// increment i and N/i in map
for (int i = 1; i <= int(sqrt(n)); i++) {
if (n % i == 0) {
if (n / i == i)
mp[i]++;
else {
mp[i]++;
mp[n / i]++;
}
}
}
}
// Function to count of multiples
// in an Array before every element
void count_divisors(int a[], int n)
{
// To store factors all of all numbers
unordered_map mp;
// Traverse for all possible i's
for (int i = 0; i < n; i++) {
// Printing value of a[i] in map
cout << mp[a[i]] << " ";
// Now updating the factors
// of a[i] in the map
add_factors(a[i], mp);
}
}
// Driver code
int main()
{
int arr[] = { 8, 1, 28, 4, 2, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
count_divisors(arr, n);
return 0;
} Java
// Java program to count of multiples
// in an Array before every element
import java.util.*;
class GFG{
// Function to find all factors of N
// and keep their count in map
static void add_factors(int n,
HashMap mp)
{
// Traverse from 1 to Math.sqrt(N)
// if i divides N,
// increment i and N/i in map
for (int i = 1; i <= (Math.sqrt(n)); i++) {
if (n % i == 0) {
if (n / i == i) {
if(mp.containsKey(i))
mp.put(i, mp.get(i) + 1);
else
mp.put(i, 1);
}
else {
if(mp.containsKey(i))
mp.put(i, mp.get(i) + 1);
else
mp.put(i, 1);
if(mp.containsKey(n / i))
mp.put(n / i, mp.get(n / i) + 1);
else
mp.put(n / i, 1);
}
}
}
}
// Function to count of multiples
// in an Array before every element
static void count_divisors(int a[], int n)
{
// To store factors all of all numbers
HashMap mp = new HashMap();
// Traverse for all possible i's
for (int i = 0; i < n; i++) {
// Printing value of a[i] in map
System.out.print(mp.get(a[i]) == null ? 0 + " " : mp.get(a[i]) + " ");
// Now updating the factors
// of a[i] in the map
add_factors(a[i], mp);
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 8, 1, 28, 4, 2, 6, 7 };
int n = arr.length;
// Function call
count_divisors(arr, n);
}
}
// This code is contributed by 29AjayKumar Python3
# Python 3 program to count of multiples
# in an Array before every element
from collections import defaultdict
import math
# Function to find all factors of N
# and keep their count in map
def add_factors(n, mp):
# Traverse from 1 to sqrt(N)
# if i divides N,
# increment i and N/i in map
for i in range(1, int(math.sqrt(n)) + 1,):
if (n % i == 0):
if (n // i == i):
mp[i] += 1
else :
mp[i] += 1
mp[n // i] += 1
# Function to count of multiples
# in an Array before every element
def count_divisors(a, n):
# To store factors all of all numbers
mp = defaultdict(int)
# Traverse for all possible i's
for i in range(n) :
# Printing value of a[i] in map
print(mp[a[i]], end=" ")
# Now updating the factors
# of a[i] in the map
add_factors(a[i], mp)
# Driver code
if __name__ == "__main__":
arr = [ 8, 1, 28, 4, 2, 6, 7 ]
n = len(arr)
# Function call
count_divisors(arr, n)
# This code is contributed by chitranayalC#
// C# program to count of multiples
// in an Array before every element
using System;
using System.Collections.Generic;
class GFG{
// Function to find all factors of N
// and keep their count in map
static void add_factors(int n,
Dictionary mp)
{
// Traverse from 1 to Math.Sqrt(N)
// if i divides N,
// increment i and N/i in map
for (int i = 1; i <= (Math.Sqrt(n)); i++) {
if (n % i == 0) {
if (n / i == i) {
if(mp.ContainsKey(i))
mp[i] = mp[i] + 1;
else
mp.Add(i, 1);
}
else {
if(mp.ContainsKey(i))
mp[i] = mp[i] + 1;
else
mp.Add(i, 1);
if(mp.ContainsKey(n / i))
mp[n / i] = mp[n / i] + 1;
else
mp.Add(n / i, 1);
}
}
}
}
// Function to count of multiples
// in an Array before every element
static void count_divisors(int []a, int n)
{
// To store factors all of all numbers
Dictionary mp = new Dictionary();
// Traverse for all possible i's
for (int i = 0; i < n; i++) {
// Printing value of a[i] in map
Console.Write(!mp.ContainsKey(a[i]) ? 0 + " " : mp[a[i]] + " ");
// Now updating the factors
// of a[i] in the map
add_factors(a[i], mp);
}
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 8, 1, 28, 4, 2, 6, 7 };
int n = arr.Length;
// Function call
count_divisors(arr, n);
}
}
// This code is contributed by sapnasingh4991 输出：
0 1 0 2 3 0 1
时间复杂度： O(N * sqrt(N))