PROBLEM LINK:CodeChef: Practical coding for everyone
Author: targer_5star
Tester: targer_5star
Editorialist: targer_5star
DIFFICULTY:
CAKEWALK
PREREQUISITES:
Math
PROBLEM:
Analyze the following pattern and print the Nth Term.
10,37,82,145……
You have to print the Nth term after analyzing the following series.
QUICK EXPLANATION:
The answer of this series can be obtained by :- ((3^2)+1), ((6^2)+1), ((9^2)+1), ((12^2)+1)…and soo on.
EXPLANATION:
The answer of this series can be obtained by :- ((3^2)+1), ((6^2)+1), ((9^2)+1), ((12^2)+1)…and soo on.
We can use any kind for loop to reach till nth term and increasing the answer in the above way view the solution to understand the pattern and the implementation clearly.
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(c) (c).begin(), (c).end()
#define allr(c) (c).rbegin(), (c).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 < (r); i++)
#define rabhi(i, l, r) for(ll i = l; i >= (r); 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,ans=0;
cin>>n;
int j=3;
for(int i=1;i<=n;i++)
{
ans=(j*j)+1;
if(i==n)
{
cout<<ans<<"\n";
}
j=j+3;
}
}
int main()
{
Code_Wode_mai_kya_rakha_hai;
Jivan_ka_asli_aadhar_to_prem_hai_na
{
Radhe_Radhe();
}
}
Tester'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(c) (c).begin(), (c).end()
#define allr(c) (c).rbegin(), (c).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 < (r); i++)
#define rabhi(i, l, r) for(ll i = l; i >= (r); 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,ans=0;
cin>>n;
int j=3;
for(int i=1;i<=n;i++)
{
ans=(j*j)+1;
if(i==n)
{
cout<<ans<<"\n";
}
j=j+3;
}
}
int main()
{
Code_Wode_mai_kya_rakha_hai;
Jivan_ka_asli_aadhar_to_prem_hai_na
{
Radhe_Radhe();
}
}
Editorialist'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(c) (c).begin(), (c).end()
#define allr(c) (c).rbegin(), (c).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 < (r); i++)
#define rabhi(i, l, r) for(ll i = l; i >= (r); 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,ans=0;
cin>>n;
int j=3;
for(int i=1;i<=n;i++)
{
ans=(j*j)+1;
if(i==n)
{
cout<<ans<<"\n";
}
j=j+3;
}
}
int main()
{
Code_Wode_mai_kya_rakha_hai;
Jivan_ka_asli_aadhar_to_prem_hai_na
{
Radhe_Radhe();
}
}