Chef and Interview Codechef Solution: Chef wants to hire a new assistant. He published an advertisement regarding that in a newspaper. After seeing the advertisement, many candidates have applied for the job. Now chef wants to shortlist people for the interviews, so he gave all of them one problem which they must solve in order to get shortlisted.
The problem was : For a given positive integer N, what is the maximum sum of distinct numbers such that the Least Common Multiple of all these numbers is N.
Your friend Rupsa also applied for the job, but was unable to solve this problem and hence you’ve decided to help her out by writing a code for solving this problem.
Input
The first line of the input contains an integer T denoting the number of test cases.
Each test case contains a single integer N.
Output
For each test case, output a single line containing an integer corresponding to the answer for that test case.
Constraints
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
Subtask 1 (30 points):
- 1 ≤ T ≤ 100
- 1 ≤ N ≤ 105
Subtask 2 (70 points):
- original constraints
Sample Input 1
2 1 2
Sample Output 1
1 3
Explanation
Example 1 : Only possible number is 1, so the maximum sum of distinct numbers is exactly 1.
Example 2 : The distinct numbers you can have are just 1 and 2, so the sum is 3. If we consider any other number greater than 2, then the least common multiple will be more than 2.
Chef and Interview CodeChef Solution in JAVA
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
float nextFloat() {
return Float.parseFloat(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
char nextChar() {
char c = ' ';
try {
c = (char) br.read();
} catch (Exception e) {
e.printStackTrace();
}
return c;
}
}
public static void main(String[] args) {
FastReader sc = new FastReader();
int ntc = sc.nextInt();
while(ntc -- > 0)
{
int N = sc.nextInt();
long sum = 0;
for (int i = 1; i * i <=N ;i++) {
if(N % i == 0)
{
sum +=i;
if(i != N/i)
{
sum +=N/i;
}
}
}
System.out.println(sum);
}
}
}
Chef and Interview CodeChef Solution in CPP
#include <iostream>
using namespace std;
int main() {
int t;
cin>>t;
for(int i = 1; i <= t; i++){
int n, p;
long long sum = 0;
cin >> n;
for(p = 1; p * p <= n; p++){
if(n % p == 0){
sum += p;
if(p != n / p){
sum += n / p;
}
}
}
cout << sum << '\n';
}
return 0;
}
Chef and Interview CodeChef Solution in Python
import math
for _ in range(int(input())):
n=int(input())
p=int(math.sqrt(n))
ans=0
for i in range(1,p+1):
if n%i==0:
q=int(n/i)
if q==i:
ans+=i
else:
ans+=i+q
print(ans)Disclaimer: The above Problem (Chef and Interview) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.
