Sum of NDefine ๐(๐)f(X) as the largest divisor of ๐X other than ๐X itself. For example, ๐(12)=6f(12)=6 and ๐(7)=1f(7)=1.Note that ๐(1)f(1) is defined to be 00.Given an integer ๐พK, find the sum of all ๐N such that ๐(๐)=๐พf(N)=K.It can be shown that the answer does not exceed 101810 18 under the given constraints.Input FormatThe first line of input will contain a single integer ๐T, denoting the number of test cases.Each test case consists of a single integer ๐พK.Output FormatFor each test case, output on a new line, the sum of all ๐N which satisfy ๐(๐)=๐พf(N)=K.Constraints1โค๐โค1041โคTโค10 4 2โค๐พโค1062โคKโค10 6 Sample 1:InputOutput3231841536Explanation:Test Case 1 : The only number which has ๐(๐)=2f(N)=2 is 44.Test Case 2 : The only numbers which have ๐(๐)=3f(N)=3 are 66 and 99. The sum of them is 1515.
Question
Sum of NDefine ๐(๐)f(X) as the largest divisor of ๐X other than ๐X itself. For example, ๐(12)=6f(12)=6 and ๐(7)=1f(7)=1.Note that ๐(1)f(1) is defined to be 00.Given an integer ๐พK, find the sum of all ๐N such that ๐(๐)=๐พf(N)=K.It can be shown that the answer does not exceed 101810 18 under the given constraints.Input FormatThe first line of input will contain a single integer ๐T, denoting the number of test cases.Each test case consists of a single integer ๐พK.Output FormatFor each test case, output on a new line, the sum of all ๐N which satisfy ๐(๐)=๐พf(N)=K.Constraints1โค๐โค1041โคTโค10 4 2โค๐พโค1062โคKโค10 6 Sample 1:InputOutput3231841536Explanation:Test Case 1 : The only number which has ๐(๐)=2f(N)=2 is 44.Test Case 2 : The only numbers which have ๐(๐)=3f(N)=3 are 66 and 99. The sum of them is 1515.
Solution
The problem is asking to find the sum of all numbers N such that the largest divisor of N, other than N itself, is equal to a given number K.
Here are the steps to solve this problem:
-
Initialize a variable, say
total, to store the sum of all numbers N. Settotalto 0. -
For each test case, do the following:
a. Read the input number K.
b. For each number i from 1 to K, do the following:
i. If K is divisible by i, then add 2i to
total. This is because if K is a divisor of N, then N could be 2K or any multiple of K. But since we are looking for the largest divisor other than N itself, we only consider 2*K. -
Print
total.
This algorithm works because for any number N, the largest divisor other than N itself is N/2. So, we only need to consider the numbers that are twice the divisors of K.
Please note that this algorithm assumes that the input numbers are within the given constraints. If the input numbers can be larger, then this algorithm may not work correctly due to integer overflow. In that case, you may need to use a data type that can handle larger numbers, or modify the algorithm to avoid overflow.
Similar Questions
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