Editorial for THRFRND

editorial
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PROBLEM LINK:CodeChef: Practical coding for everyone

Practice

Author: targer_5star
Tester: targer_5star
Editorialist: targer_5star

DIFFICULTY:

CAKEWALK

PREREQUISITES:

Math.

PROBLEM:

There are three friend living on the straight line in chefland. The first friend lives at the point p1, the second friend lives at the point p2, and the third friend lives at the point p3.

They plan to go for a movie, so they need to meet at one point.

What is the minimum total distance (in Km) they have to travel in order to meet at some point and go for a movie.

QUICK EXPLANATION:

EXPLANATION:

You have given three integers if you observe and think about the testcase then you will find that we just have to find the maximum and minimum between the integers and print their positive difference.

SOLUTIONS:

There are three friend living on the straight line in chefland. The first friend lives at the point p1, the second friend lives at the point p2, and the third friend lives at the point p3.

You have given three integers let A,B,C. If you observe and think about the testcase then you will find that we just have to find the maximum and minimum between the integers and print their positive difference.

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); iā€“)

#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 a,b,c;

cin>>a>>b>>c;

ll x=max(a,b);

ll y=max(b,c);

ll z=max(x,y);

ll p=min(a,b);

ll q=min(b,c);

ll r=min(p,q);

cout<<z-r<<ā€œ\nā€;

}

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); iā€“)

#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 a,b,c;

cin>>a>>b>>c;

ll x=max(a,b);

ll y=max(b,c);

ll z=max(x,y);

ll p=min(a,b);

ll q=min(b,c);

ll r=min(p,q);

cout<<z-r<<ā€œ\nā€;

}

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); iā€“)

#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 a,b,c;

cin>>a>>b>>c;

ll x=max(a,b);

ll y=max(b,c);

ll z=max(x,y);

ll p=min(a,b);

ll q=min(b,c);

ll r=min(p,q);

cout<<z-r<<ā€œ\nā€;

}

int main()

{

   Code_Wode_mai_kya_rakha_hai;

   Jivan_ka_asli_aadhar_to_prem_hai_na

   {

          Radhe_Radhe();

   }

}

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