CHFPROFIT - Editorial

editorial
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PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3

Setter: Utkarsh Gupta
Tester: Manan Grover
Editorialist: Lavish Gupta

DIFFICULTY:

Cakewalk

PREREQUISITES:

None

PROBLEM:

Some time ago, Chef bought X stocks at the cost of Rs. Y each. Today, Chef is going to sell all these X stocks at Rs. Z each. What is Chef’s total profit after he sells them?

Chef’s profit equals the total amount he received by selling the stocks, minus the total amount he spent buying them.

EXPLANATION:

The Chef has bought X stocks at the cost of Rs. Y each, so the total amount of money spent is Rs X \cdot Y. Now, the chef is going to sell all these X stocks at Rs. Z each, and therefore the total amount of money that chef will get will be Rs X \cdot Z

So, Chef’s profit = total amount that chef received - total amount that chef spent = X\cdot Z - X \cdot Y = X \cdot (Z-Y).

TIME COMPLEXITY:

O(1) for each test case.

SOLUTION:

Setter's Solution 
//Utkarsh.25dec
#include <bits/stdc++.h>
#include <chrono>
#include <random>
#define ll long long int
#define ull unsigned long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(i,n) for(ll i=0;i<n;i++)
#define loop(i,a,b) for(ll i=a;i<=b;i++)
#define vi vector <int>
#define vs vector <string>
#define vc vector <char>
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
#define max3(a,b,c) max(max(a,b),c)
#define min3(a,b,c) min(min(a,b),c)
#define deb(x) cerr<<#x<<' '<<'='<<' '<<x<<'\n'
using namespace std;
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// ordered_set s ; s.order_of_key(val)  no. of elements strictly less than val
// s.find_by_order(i)  itertor to ith element (0 indexed)
typedef vector<vector<ll>> matrix;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
void solve()
{
    int x,y,z;
    cin>>x>>y>>z;
    cout<<(x*(z-y))<<'\n';
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int T=1;
    cin>>T;
    int t=0;
    while(t++<T)
    {
        //cout<<"Case #"<<t<<":"<<' ';
        solve();
        //cout<<'\n';
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Tester's Solution 
#include <bits/stdc++.h>
using namespace std;
int main(){
  ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
  int t;
  cin>>t;
  while(t--){
    int a, b, c;
    cin>>a>>b>>c;
    cout<<(c - b) * a<<"\n";
  }
  return 0;
}
Editorialist's Solution 
#include<bits/stdc++.h>
#define ll long long
using namespace std ;

const ll z = 1000000007 ;

void solve()
{
    int x , y , z ;
    cin >> x >> y >> z ;
    cout << max(0 , (z-y)*x) << endl ;
    return ;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    #ifndef ONLINE_JUDGE
    freopen("inputf.txt" , "r" , stdin) ;
    freopen("outputf.txt" , "w" , stdout) ;
    freopen("error.txt" , "w" , stderr) ;
    #endif

    int t;
    cin >> t ;
    while(t--)
        solve() ;

    return 0;
}