Editorial for NEWPRL

PROBLEM LINK:Contest Page | CodeChef

Practice

Author: targer_5star
Tester: targer_5star
Editorialist: targer_5star

DIFFICULTY:

SIMPLE

PREREQUISITES:

Math

PROBLEM:

Analyze the Pattern of the following series Which is Increasing in a particular way and Find the Nth term of the series.

6,24,60,120,210,……

Find and Print The Nth Term of the series

QUICK EXPLANATION:

The Answer for the Nth term can be obtained by using then following formula
Answer=(n) x (n+1) x (n+2).

EXPLANATION:

Find the Nth term of the series.

6,24,60,120,210,……

So if you observe the pattern carefully you will find that the Nth Term is been obtained by the following formula:- Answer=(n) x (n+1) x (n+2).

SOLUTIONS:

Setter's Solution

//Prem se bolo Radhe Radhe

#include <bits/stdc++.h>

const long long SZ = 4e3 + 7;

const long long inf = 1e18;

const long long MOD = 1e9 + 7;

const long long mod = 1e9 + 7;

long long opnmbr = 1;

#define ll long long

#define ld long double

#define pb push_back

#define mp make_pair

#define eb emplace_back

#define pll pair<ll, ll>

#define vi vector

#define vs vector

#define vpl vector

#define qi queue

#define si set

#define map map<ll, ll>

#define umap unordered_map<ll, ll>

#define fi first

#define se second

#define sz(x) (ll)x.size()

#define all© ©.begin(), ©.end()

#define allr© ©.rbegin(), ©.rend()

#define Max(a,b) ((a > b) ? a : b)

#define Min(a,b) ((a < b) ? a : b)

#define ci(X) ll X; cin>>X

#define cii(X, Y) ll X, Y; cin>>X>>Y

#define ciii(X, Y, Z) ll X, Y, Z; cin>>X>>Y>>Z

#define ciiii(W, X, Y, Z) ll W, X, Y, Z; cin>>W>>X>>Y>>Z

#define co cout<<

#define in cin>>

#define Jivan_ka_asli_aadhar_to_prem_hai_na ll ___T; cin>>___T; while (___T-- > 0)

#define Code_Wode_mai_kya_rakha_hai ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

#define inf 1e18

#define ed endl

#define abhi(i, n) for(ll i = 0; i < (n); i++)

#define abhi1(i, n) for(ll i = 1; i <= (n); i++)

#define abhi2(i, l, r) for(ll i = l; i < ®; i++)

#define rabhi(i, l, r) for(ll i = l; i >= ®; i–)

#define ms0(X) memset((X), 0, sizeof((X)))

#define ms1(X, V) memset((X), 1, sizeof((X)))

#define ms2(X, V) memset((X), 1, sizeof((X)))

#define flv(X, V) fill(all((X)), V)

#define gcd(a,b) __gcd(a,b)

using namespace std;

ll prmod(ll a, ll b)

{

ll ans = 1;

while (b)

{

    if (b & 1) ans = (ans * a) % MOD;

    b = b / 2;

    a = (a * a) % MOD;

}

return ans;

}

int pri(ll n)

{

int flag=0;

for(ll i = 2; i <= n / 2; ++i)

{

 if(n % i == 0)

 {

    flag = 1;

    break;

  }

}

if(flag==0)

{

  return 1;

}

else

{

   return 0;

}

}

ll lcm(ll a, ll b)

{

 return (a / gcd(a, b)) * b;

}

void Radhe_Radhe()

{

ll n;

cin>>n;

ll ans=(n)(n+1)(n+2);

cout<<ans<<"\n";

}

int main()

{

   Code_Wode_mai_kya_rakha_hai;

   Jivan_ka_asli_aadhar_to_prem_hai_na

   {

          Radhe_Radhe();

   }

}

Tester's Solution

void Radhe_Radhe()

{

ll n;

cin>>n;

ll ans=(n)(n+1)(n+2);

cout<<ans<<"\n";

}

Editorialist's Solution

void Radhe_Radhe()

{

ll n;

cin>>n;

ll ans=(n)(n+1)(n+2);

cout<<ans<<"\n";

}

1 Like