PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Setter: Hasan
Tester: Teja Vardhan Reddy
Editorialist: Taranpreet Singh
DIFFICULTY:
Cakewalk
PREREQUISITES:
None
PROBLEM:
There are N types of street food, and for each type of street food, we know the number of stores offering this type of food, the number of people interested in buying this type of food, and the profit per unit for this type of food.
We want to find the maximum profit we can get, by opening a store to offer the food of exactly one of above N types.
We can assume that If there are p people and s stores (including chef’s) for some food type, then Chef will get \lfloor \frac{p}{s} \rfloor people at his store.
QUICK EXPLANATION
- Each type of store is independent, we can calculate the maximum profit for each store and print maximum.
- For a food type with P_i people interested, S_i existing stores V_i profit per unit, the number of people at Chef’s store would be \lfloor \frac{P_i}{S_i+1} \rfloor and Chef’s profit would be V_i*\lfloor \frac{P_i}{S_i+1} \rfloor
EXPLANATION
The first thing to notice is that the profit from one type of food is independent of profit from other types of food.
Above allows us to compute profit by opening each type of store and calculating the maximum profit among them. So now we focus on calculating profit from opening a store of one type.
For a food type, P people are interested to buy, S stores already exist for the same type of food and Chef earns V per unit of food. So, after Chef opens a store of this food type, the number of stores become S+1 and the number of people buying from Chef’s store would be \lfloor \frac{P}{S+1} \rfloor Hence, the Chef’s profit becomes V*\lfloor \frac{P}{S+1} \rfloor
We can take the maximum over all the food types and print maximum.
Bonus Problem
In the same problem, Chef opens not one, but K stores, find the maximum profit Chef can get by opening K stores optimally, assuming he can only open at most one store of each type.
Harder version, he can now open multiple stores of the same type, calculate the maximum profit.
TIME COMPLEXITY
The time complexity is O(N) per test case.
SOLUTIONS:
Setter's Solution
#include <iostream>
#include <algorithm>
#include <string>
#include <assert.h>
using namespace std;
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
assert(cnt>0);
if(is_neg){
x= -x;
}
assert(l<=x && x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
int T;
int n;
int s,p,v;
int main(){
//freopen("00.txt","r",stdin);
//freopen("00o.txt","w",stdout);
T=readIntLn(1,100);
while(T--){
n=readIntLn(1,100);
int bst = 0;
for(int i=0;i<n;i++){
s=readIntSp(1,10000);
p=readIntSp(1,10000);
v=readIntLn(1,10000);
bst = max(bst, (p / (s+1)) * v);
}
cout<<bst<<endl;
}
assert(getchar()==-1);
}
Tester's Solution
//teja349
#include <bits/stdc++.h>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <utility>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <iomanip>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
//setbase - cout << setbase (16); cout << 100 << endl; Prints 64
//setfill - cout << setfill ('x') << setw (5); cout << 77 << endl; prints xxx77
//setprecision - cout << setprecision (14) << f << endl; Prints x.xxxx
//cout.precision(x) cout<<fixed<<val; // prints x digits after decimal in val
using namespace std;
using namespace __gnu_pbds;
#define f(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) f(i,0,n)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define pb push_back
#define mp make_pair
#define vi vector< int >
#define vl vector< ll >
#define ss second
#define ff first
#define ll long long
#define pii pair< int,int >
#define pll pair< ll,ll >
#define sz(a) a.size()
#define inf (1000*1000*1000+5)
#define all(a) a.begin(),a.end()
#define tri pair<int,pii>
#define vii vector<pii>
#define vll vector<pll>
#define viii vector<tri>
#define mod (1000*1000*1000+7)
#define pqueue priority_queue< int >
#define pdqueue priority_queue< int,vi ,greater< int > >
#define flush fflush(stdout)
#define primeDEN 727999983
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// find_by_order() // order_of_key
typedef tree<
int,
null_type,
less<int>,
rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;
int s[1234],p[1234],v[1234];
int main(){
std::ios::sync_with_stdio(false); cin.tie(NULL);
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int i;
int maxi=0;
rep(i,n){
cin>>s[i]>>p[i]>>v[i];
maxi=max(p[i]/(s[i]+1)*v[i],maxi);
}
cout<<maxi<<endl;
}
return 0;
}
Editorialist's Solution
import java.util.*;
import java.io.*;
import java.text.*;
class STFOOD{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int n = ni();
long ans = 0;
for(int i = 0; i< n; i++){
long s = nl(), p = nl(), v = nl();
ans = Math.max(ans, v*(p/(s+1)));
}
pn(ans);
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
DecimalFormat df = new DecimalFormat("0.00000000000");
static boolean multipleTC = true;
FastReader in;PrintWriter out;
void run() throws Exception{
in = new FastReader();
out = new PrintWriter(System.out);
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
new STFOOD().run();
}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str = "";
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}
Feel free to share your approach. Suggestions are welcomed as always. 
