PROBLEM LINK:
Author: Pradeep Kumar Sahu
Tester: Sumeet Pachauri
Editorialist: Pradeep Kumar Sahu
DIFFICULTY:
Cakewalk
PREREQUISITES:
Basic observations , Maths
PROBLEM:
You are given a positive integer N.
Your task is to calculate total number of pairs (i,j) such that 1\leq i, j \leq N and HCF(i,j) = LCM(i,j) .
Here HCF denotes the highest common factor and LCM denotes the lowest common multiple.
EXPLANATION:
We know that LCM of two numbers = HCF of same two numbers only when both numbers are equal.
So solution pairs will be : (1,1) , (2,2) , (3,3) , (4,4) , … , (N,N).
Hence the answer will be N itself.
TIME COMPLEXITY:
Time complexity is O(1) per test cases.
SOLUTIONS:
Setter's Solution
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin >> t;
while(t--)
{
int n;
cin >> n;
cout << n << "\n" ;
}
return 0;
}
Tester's Solution
/*
Author : Sumeet Pachauri
Codeforces : sumeet_4275
CodeChef : kaalbhairav01
________________________________________
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¦____-¦¦-¦¦-¦#¦-¦_¦-¦#¦-_¦¯¦¦-_¦¯¦¦¦-¦¦¦
¯_____¯¯____¯¯___¯___¯_____¯_____¯¯___¯¯
*/
#include "bits/stdc++.h"
using namespace std;
#define int long long
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define all(x) (x).begin(),(x).end()
#define uniq(v) (v).erase(unique(all(v)),(v).end())
#define sz(x) (int)((x).size())
#define fr first
#define sc second
#define pii pair<int,int>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define mem1(a) memset(a,-1,sizeof(a))
#define mem0(a) memset(a,0,sizeof(a))
#define r0 return 0
#define test int T; cin>>T; while(T--)
#define ppc __builtin_popcount
#define ppcll __builtin_popcountll
#define mod 1000000007
void solve()
{
int n;
cin>>n;
assert(n>=1);
assert(n<=1000000000);
cout<<n<<endl;
}
signed main(){
fastio
test
{ assert(T<=100000);
solve();
}
r0;
}
For doubts, please leave them in the comment section, I’ll address them.