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# Special Triplets CodeChef Solution – Queslers

## Problem: Special Triplets CodeChef Solution

Gintoki has been very lazy recently and he hasn’t made enough money to pay the rent this month. So the old landlady has given him a problem to solve instead, if he can solve this problem the rent will be waived. The problem is as follows:

A triplet of integers (A,B,C)(A,B,C) is considered to be special if it satisfies the following properties for a given integer NN :

• AmodB=CAmodB=C
• BmodC=0BmodC=0
• 1≤A,B,C≤N1≤A,B,C≤N

Example: There are three special triplets for N=3N=3.

• (1,3,1)(1,3,1) is a special triplet, since (1mod3)=1(1mod3)=1 and (3mod1)=0(3mod1)=0.
• (1,2,1)(1,2,1) is a special triplet, since (1mod2)=1(1mod2)=1 and (2mod1)=0(2mod1)=0.
• (3,2,1)(3,2,1) is a special triplet, since (3mod2)=1(3mod2)=1 and (2mod1)=0(2mod1)=0.

The landlady gives Gintoki an integer NN. Now Gintoki needs to find the number of special triplets. Can you help him to find the answer?

### Input Format

• The first line of the input contains a single integer TT denoting the number of test cases. The description of TT test cases follows.
• The first and only line of each test case contains a single integer NN.

### Output Format

• For each testcase, output in a single line the number of special triplets.

### Constraints

• 1≤T≤101≤T≤10
• 2≤N≤1052≤N≤105

Subtask #3 (75 points): Original constraints

### Sample Input 1

``````3
3
4
5
``````

### Sample Output 1

``````3
6
9
``````

### Explanation

Test case 11: It is explained in the problem statement.

Test case 22: The special triplets are (1,3,1)(1,3,1), (1,2,1)(1,2,1), (3,2,1)(3,2,1), (1,4,1)(1,4,1), (4,3,1)(4,3,1), (2,4,2)(2,4,2). Hence the answer is 66.

Test case 33: The special triplets are (1,3,1)(1,3,1), (1,2,1)(1,2,1), (3,2,1)(3,2,1), (1,4,1)(1,4,1), (4,3,1)(4,3,1), (1,5,1)(1,5,1), (5,4,1)(5,4,1), (5,2,1)(5,2,1), (2,4,2)(2,4,2). Hence the answer is 99.

## Special Triplets CodeChef Solution Using Python

``````# cook your dish here
t = int(input())

for i in range(t):
n = int(input())
c = 0
for j in range(1, n // 2 + 1):
for k in range(2 * j, n + 1, j):
a = n // k
b = n % k
if b >= j:
c += a + 1
else:
c += a
print(c)``````

## Special Triplets CodeChef Solution Using C++

``````#include <bits/stdc++.h>
using namespace std;

#define ll long long

ll m = 1e5+5;
vector<vector<ll>>v(m);

void preprocess(){
for(ll i = 1;i<m;i++){
for(ll j = 2*i;j<m;j+=i){
v[j].push_back(i);
}
}
}

int main() {
ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
preprocess();
ll t;
cin >> t;
while(t--){
ll n;
cin >> n;
ll ans = 0;
for(ll i = 2;i<=n;i++){
ll fact = v[i].size();
ll last = n-(n%i);
ans += fact*(last/i);
for(ll j = 0;j<fact;j++){
if(last+v[i][j] <= n)
ans++;
}
}
cout << ans << "\n";
}
return 0;
}
``````

## Special Triplets CodeChef Solution Using Java

``````import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Main {
StringTokenizer st;

{
}

String next()
{
while (st == null || !st.hasMoreElements()) {
try {
}
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());
}

String nextLine()
{
String str = "";
try {
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) throws java.lang.Exception
{
int t = sc.nextInt();
while(t-- > 0)
{
int n = sc.nextInt();
int cnt = 0;
for(int i = 1; i <= n; i++)
{
for(int j = i; j <= n; j += i)
{
if(j % i == 0)
{
for(int k = i; k <= n; k += j)
{
if(k % j == i)
cnt++;
}
}
}
}
System.out.println(cnt);
}
}
}``````
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