给定数字N，任务是找出不同数字的最大和，以使所有这些数字的最小公倍数为N。
Input : N = 12
Output : 20
Maximum sum which we can achieve is,
1 + 2 + 3 + 4 + 6 + 12 = 28
Input : N = 15
Output : 24 我们可以通过观察某些情况来解决该问题，因为N必须是所有数字的LCM，所以它们都将是N的除数，但是由于一个数字只能相加一次，因此所有被取数字应该是不同的。这个想法是将N的每个除数求和一次，以使结果最大化。
我们怎么能说我们得到的和是最大和?原因是，我们将N的所有除数都纳入了总和，现在，如果我们再将一个非N的除数的和加入一个总和，则总和将增加，但LCM属性将不会被所有这些整数所保持。因此，除了N的所有除数之外，不可能再增加一个数，因此，给定N个所有除数的总和，我们的问题归结为这一点，这可以在O(sqrt(N))时间内解决。
因此，解决方案的总时间复杂度将为O(sqrt(N))，其中O(1)为额外空间。
下面是上述概念的代码。请参阅此帖子以查找数字的所有除数。
C++
// C/C++ program to get maximum sum of Numbers
// with condition that their LCM should be N
#include
using namespace std;
// Method returns maximum sum f distinct
// number whose LCM is N
int getMaximumSumWithLCMN(int N)
{
int sum = 0;
int LIM = sqrt(N);
// find all divisors which divides 'N'
for (int i = 1; i <= LIM; i++) {
// if 'i' is divisor of 'N'
if (N % i == 0) {
// if both divisors are same then add
// it only once else add both
if (i == (N / i))
sum += i;
else
sum += (i + N / i);
}
}
return sum;
}
// Driver code to test above methods
int main()
{
int N = 12;
cout << getMaximumSumWithLCMN(N) << endl;
return 0;
} Java
// Java program to get
// maximum sum of Numbers
// with condition that
// their LCM should be N
class GFG {
// Method returns maximum
// sum f distinct number
// whose LCM is N
static int getMaximumSumWithLCMN(int N)
{
int sum = 0;
int LIM = (int)Math.sqrt(N);
// find all divisors which divides 'N'
for (int i = 1; i <= LIM; i++) {
// if 'i' is divisor of 'N'
if (N % i == 0) {
// if both divisors are same then add
// it only once else add both
if (i == (N / i))
sum += i;
else
sum += (i + N / i);
}
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int N = 12;
System.out.println(getMaximumSumWithLCMN(N));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python program to get
# maximum sum of Numbers
# with condition that
# their LCM should be N
import math
# Method returns maximum sum f distinct
# number whose LCM is N
def getMaximumSumWithLCMN(N):
sum = 0
LIM = int(math.sqrt(N))
# find all divisors which divides 'N'
for i in range(1, LIM + 1):
# if 'i' is divisor of 'N'
if (N % i == 0):
# if both divisors are same then add
# it only once else add both
if (i == (N // i)):
sum = sum + i
else:
sum = sum + (i + N // i)
return sum
# driver code
N = 12
print(getMaximumSumWithLCMN(N))
# This code is contributed
# by Anant Agarwal.C#
// C# program to get maximum sum
// of Numbers with condition that
// their LCM should be N
using System;
class GFG {
// Method returns maximum sum f
// distinct number whose LCM is N
static int getMaximumSumWithLCMN(int N)
{
int sum = 0;
int LIM = (int)Math.Sqrt(N);
// Find all divisors which divides 'N'
for (int i = 1; i <= LIM; i++) {
// if 'i' is divisor of 'N'
if (N % i == 0) {
// if both divisors are same then
// add it only once else add both
if (i == (N / i))
sum += i;
else
sum += (i + N / i);
}
}
return sum;
}
// Driver code
public static void Main()
{
int N = 12;
Console.Write(getMaximumSumWithLCMN(N));
}
}
// This code is contributed by nitin mittalPHP
Javascript
输出：
28