# PROBLEM LINK:CodeChef: Practical coding for everyone

Practice

Author: targer_5star
Tester: targer_5star
Editorialist: targer_5star

CAKEWALK

Arrays

# PROBLEM:

WWDC 2021 is announced……!

Total N companies have Participated in the WWDC 2021.As the total number of companies participated in the wwdc is more than expected.The organizers of the program decided to short list participants based on number of products each will showcase in wwdc.

Rule to get shortlisted for any company is that the company should showcase at least K Number of product in wwdc 2021.

You are given an array of size N i.e (Total number of companies). where each NiNi element denotes the total number of products that particular company will showcase. find out total number of companies that will be able to participate in wwdc 2021.

# QUICK EXPLANATION:

Just Traverse the array and count all the elements greater than K.
and Print the Whole Number of Count That You Got while Traversing the array.

# EXPLANATION:

Rule to get shortlisted for any company is that the company should showcase at least K Number of product in wwdc 2021.

Just Traverse the array and count all the elements greater than K.
and Print the Whole Number of Count That You Got while Traversing the array

# SOLUTIONS:

Setter's Solution

#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;
``````

}

{

ll n,k,cnt=0;

cin>>n>>k;

ll a[n];

for(int i=0;i<n;i++)

{

``````   cin>>a[i];

if(a[i]>=k)

{

cnt++;

}
``````

}

cout<<cnt<<“\n”;

}

int main()

{

``````   Code_Wode_mai_kya_rakha_hai;

{

}
``````

}

Tester's Solution

#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;
``````

}

{

ll n,k,cnt=0;

cin>>n>>k;

ll a[n];

for(int i=0;i<n;i++)

{

``````   cin>>a[i];

if(a[i]>=k)

{

cnt++;

}
``````

}

cout<<cnt<<“\n”;

}

int main()

{

``````   Code_Wode_mai_kya_rakha_hai;

{

}
``````

}

Editorialist's Solution

#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;
``````

}

{

ll n,k,cnt=0;

cin>>n>>k;

ll a[n];

for(int i=0;i<n;i++)

{

``````   cin>>a[i];

if(a[i]>=k)

{

cnt++;

}
``````

}

cout<<cnt<<“\n”;

}

int main()

{

``````   Code_Wode_mai_kya_rakha_hai;