给定一个范围[l, r] ,任务是找到给定范围内数字的所有偶数因子的总和。
例子:
Input: l = 6, r = 8
Output: 22
factors(6) = 1, 2, 3, 6, evenfactors(6) = 2, 6 sumEvenFactors(6) = 2 + 6 = 8
factors(7) = 1, 7, No even factors
factors(8) = 1, 2, 4, 8, evenfactors(8) = 2, 4, 8 sumEvenFactors(8) = 2 + 4 + 8 = 14
Therefore sum of all even factors = 8 + 14 = 22
Input: l = 1, r = 10
Output: 42
方法:我们可以修改 Sieve Of Eratosthenes 来存储一个数字的所有偶数因子的总和在它的相应索引处。然后我们将创建一个前缀数组来存储该索引的总和。现在每个查询都可以使用 prefix[r] – prefix[l – 1] 在 O(1) 中回答。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
#define ll long long int
const int MAX = 100000;
ll prefix[MAX];
// Function to calculate the prefix sum
// of all the even factors
void sieve_modified()
{
for (int i = 2; i < MAX; i += 2) {
// Add i to all the multiples of i
for (int j = i; j < MAX; j += i)
prefix[j] += i;
}
// Update the prefix sum
for (int i = 1; i < MAX; i++)
prefix[i] += prefix[i - 1];
}
// Function to return the sum of
// all the even factors of the
// numbers in the given range
ll sumEvenFactors(int L, int R)
{
return (prefix[R] - prefix[L - 1]);
}
// Driver code
int main()
{
sieve_modified();
int l = 6, r = 10;
cout << sumEvenFactors(l, r);
return 0;
}
Java
// Java implementation of the approach
class GFG {
static final int MAX = 100000;
static long prefix[] = new long[MAX];
// Function to calculate the prefix sum
// of all the even factors
static void sieve_modified()
{
for (int i = 2; i < MAX; i += 2) {
// Add i to all the multiples of i
for (int j = i; j < MAX; j += i)
prefix[j] += i;
}
// Update the prefix sum
for (int i = 1; i < MAX; i++)
prefix[i] += prefix[i - 1];
}
// Function to return the sum of
// all the even factors of the
// numbers in the given range
static long sumEvenFactors(int L, int R)
{
return (prefix[R] - prefix[L - 1]);
}
// Driver code
public static void main(String args[])
{
sieve_modified();
int l = 6, r = 10;
System.out.print(sumEvenFactors(l, r));
}
}
Python3
# Python3 implementation of the approach.
# Function to calculate the prefix sum
# of all the even factors
def sieve_modified():
for i in range(2, MAX, 2):
# Add i to all the multiples of i
for j in range(i, MAX, i):
prefix[j] += i
# Update the prefix sum
for i in range(1, MAX):
prefix[i] += prefix[i - 1]
# Function to return the sum of
# all the even factors of the
# numbers in the given range
def sumEvenFactors(L, R):
return (prefix[R] - prefix[L - 1])
# Driver code
if __name__ == "__main__":
MAX = 100000
prefix = [0] * MAX
sieve_modified()
l, r = 6, 10
print(sumEvenFactors(l, r))
# This code is contributed by Rituraj Jain
C#
using System;
// C# implementation of the approach
using System;
class GFG
{
public const int MAX = 100000;
public static long[] prefix = new long[MAX];
// Function to calculate the prefix sum
// of all the even factors
public static void sieve_modified()
{
for (int i = 2; i < MAX; i += 2)
{
// Add i to all the multiples of i
for (int j = i; j < MAX; j += i)
{
prefix[j] += i;
}
}
// Update the prefix sum
for (int i = 1; i < MAX; i++)
{
prefix[i] += prefix[i - 1];
}
}
// Function to return the sum of
// all the even factors of the
// numbers in the given range
public static long sumEvenFactors(int L, int R)
{
return (prefix[R] - prefix[L - 1]);
}
// Driver code
public static void Main(string[] args)
{
sieve_modified();
int l = 6, r = 10;
Console.Write(sumEvenFactors(l, r));
}
}
// This code is contributed by Shrikant13
PHP
Javascript
输出:
34
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