给定一个自然数数组，请为数组中的每个元素计算其适当除数的总和。
Input : int arr[] = {8, 13, 24, 36, 59, 75, 87}
Output : 7 1 36 55 1 49 21
Number 8 has 3 proper divisors 1, 2, 4
and their sum comes out to be 7.
在下面的文章中讨论了这个问题的幼稚解决方案。
自然数的所有适当除数的总和
我们可以通过使用Eratosthenes筛子来更有效地做到这一点。
这个想法是基于数字的素因数分解。通过使用筛子，我们可以存储一个数的所有素数及其幂。
To find all divisors, we need to consider
all powers of a prime factor and multiply
it with all all powers of other prime factors.
(For example, if the number is 36, its prime
factors are 2 and 3 and all divisors are 1,
2, 3, 4, 6, 9, 12 and 18.
Consider a number N can be written
as P1^Q1 * P2^Q2 * P3^Q3 (here only 3
prime factors are considered but there can
be more than that) then sum of its divisors
will be written as:
= P1^0 * P2^0 * P3^0 + P1^0 * P2^0 * P3^1 +
P1^0 * P2^0 * P3^2 + ................ +
P1^0 * P2^0 * P3^Q3 + P1^0 * P2^1 * P3^0 +
P1^0 * P2^1 * P3^1 + P1^0 * P2^1 * P3^2 +
................ + P1^0 * P2^1 * P3^Q3 +
.
.
.
P1^Q1 * P2^Q2 * P3^0 + P1^Q1 * P2^Q2 * P3^1 +
P1^Q1 * P2^Q2 * P3^2 + .......... +
P1^Q1 * P2^Q2 * P3^Q3
Above can be written as,
(((P1^(Q1+1)) - 1) /
(P1 - 1)) * (((P2^(Q2+1)) - 1) /
(P2 - 1)) * (((P3^(Q3 + 1)) - 1) /
(P3 - 1))
以下是基于以上公式的实现。
C++
// C++ program to find sum of proper divisors for
// every element in an array.
#include
using namespace std;
#define MAX 1000001
#define pii pair
#define F first
#define S second
// To store prime factors and their
// powers
vector factors[MAX];
// Fills factors such that factors[i] is
// a vector of pairs containing prime factors
// (of i) and their powers.
// Also sets values in isPrime[]
void sieveOfEratothenese()
{
// To check if a number is prime
bool isPrime[MAX];
memset(isPrime, true, sizeof(isPrime));
isPrime[0] = isPrime[1] = false;
for (int i = 2; i < MAX; i++)
{
// If i is prime, then update its
// powers in all multiples of it.
if (isPrime[i])
{
for (int j = i; j < MAX; j += i)
{
int k, l;
isPrime[j] = false;
for (k = j, l = 0; k % i == 0; l++, k /= i)
;
factors[j].push_back(make_pair(i, l));
}
}
}
}
// Returns sum of proper divisors of num
// using factors[]
int sumOfProperDivisors(int num)
{
// Applying above discussed formula for every
// array element
int mul = 1;
for (int i = 0; i < factors[num].size(); i++)
mul *= ((pow(factors[num][i].F,
factors[num][i].S + 1) - 1) /
(factors[num][i].F - 1));
return mul - num;
}
// Driver code
int main()
{
sieveOfEratothenese();
int arr[] = { 8, 13, 24, 36, 59, 75, 91 };
for (int i = 0; i < sizeof(arr) / sizeof(int); i++)
cout << sumOfProperDivisors(arr[i]) << " ";
cout << endl;
return 0;
} Java
// Java program to find sum of proper divisors for
// every element in an array.
import java.util.*;
class GFG
{
static final int MAX = 100001;
static class pair
{
int F, S;
public pair(int f, int s) {
F = f;
S = s;
}
}
// To store prime factors and their
// powers
static Vector []factors = new Vector[MAX];
// Fills factors such that factors[i] is
// a vector of pairs containing prime factors
// (of i) and their powers.
// Also sets values in isPrime[]
static void sieveOfEratothenese()
{
// To check if a number is prime
boolean []isPrime = new boolean[MAX];
Arrays.fill(isPrime, true);
isPrime[0] = isPrime[1] = false;
for (int i = 2; i < MAX; i++)
{
// If i is prime, then update its
// powers in all multiples of it.
if (isPrime[i])
{
for (int j = i; j < MAX; j += i)
{
int k, l;
isPrime[j] = false;
for (k = j, l = 0; k % i == 0; l++, k /= i)
;
factors[j].add(new pair(i, l));
}
}
}
}
// Returns sum of proper divisors of num
// using factors[]
static int sumOfProperDivisors(int num)
{
// Applying above discussed formula for every
// array element
int mul = 1;
for (int i = 0; i < factors[num].size(); i++)
mul *= ((Math.pow(factors[num].get(i).F,
factors[num].get(i).S + 1) - 1) /
(factors[num].get(i).F - 1));
return mul - num;
}
// Driver code
public static void main(String[] args)
{
for (int i = 0; i < MAX; i++)
factors[i] = new Vector();
sieveOfEratothenese();
int arr[] = { 8, 13, 24, 36, 59, 75, 91 };
for (int i = 0; i < arr.length; i++)
System.out.print(sumOfProperDivisors(arr[i])+ " ");
System.out.println();
}
}
// This code is contributed by 29AjayKumar C#
// C# program to find sum of proper divisors for
// every element in an array.
using System;
using System.Collections.Generic;
class GFG
{
static readonly int MAX = 100001;
class pair
{
public int F, S;
public pair(int f, int s)
{
F = f;
S = s;
}
}
// To store prime factors and their
// powers
static List []factors = new List[MAX];
// Fills factors such that factors[i] is
// a vector of pairs containing prime factors
// (of i) and their powers.
// Also sets values in isPrime[]
static void sieveOfEratothenese()
{
// To check if a number is prime
bool []isPrime = new bool[MAX];
for (int i = 0; i < MAX; i++)
isPrime[i] = true;
isPrime[0] = isPrime[1] = false;
for (int i = 2; i < MAX; i++)
{
// If i is prime, then update its
// powers in all multiples of it.
if (isPrime[i])
{
for (int j = i; j < MAX; j += i)
{
int k, l;
isPrime[j] = false;
for (k = j, l = 0; k % i == 0; l++, k /= i)
;
factors[j].Add(new pair(i, l));
}
}
}
}
// Returns sum of proper divisors of num
// using factors[]
static int sumOfProperDivisors(int num)
{
// Applying above discussed formula for every
// array element
int mul = 1;
for (int i = 0; i < factors[num].Count; i++)
mul *= (int)((Math.Pow(factors[num][i].F,
factors[num][i].S + 1) - 1) /
(factors[num][i].F - 1));
return mul - num;
}
// Driver code
public static void Main(String[] args)
{
for (int i = 0; i < MAX; i++)
factors[i] = new List();
sieveOfEratothenese();
int []arr = { 8, 13, 24, 36, 59, 75, 91 };
for (int i = 0; i < arr.Length; i++)
Console.Write(sumOfProperDivisors(arr[i])+ " ");
Console.WriteLine();
}
}
// This code is contributed by 29AjayKumar 输出：
7 1 36 55 1 49 21