PROBLEM LINK:
Contest Division 1
Contest Division 2
Contest Division 3
Practice
Setter: Ashish Gupta
Tester: Samarth Gupta
Editorialist: Taranpreet Singh
DIFFICULTY
Cakewalk
PREREQUISITES
None
PROBLEM
The current price of petrol is X rupees, and the current price of diesel is Y rupees. At the start of each month, the price of petrol increases by A rupees and the price of diesel increases by B rupees.
Chef is curious to know which fuel costs less at the end of the K^{th} month. If petrol is cheaper than diesel at the end of K^{th} month, then print \verb+PETROL+. If diesel is cheaper than petrol at the end of the K^{th} month, then print \verb+DIESEL+. If both the fuels have the same price at the end of the K^{th} month, then print \verb+SAME PRICE+.
QUICK EXPLANATION
- The price of Petrol after K days is X+A*K and the price of Diesel after K days is Y+B*K
- Compare which is higher and print the answer accordingly.
EXPLANATION
Every day, the price of Petrol rises by A, and the price of diesel rises by B. This repeated for K days.
So the price of Petrol after K days is X+(A+A+A +\ldots K \text{times}) = X + A*K.
Similarly, the price of Diesel after $K$$ days is Y+(B+B+B +\ldots K \text{times}) = Y + B*K.
Now we can compare the values X+A*K and Y+B*K and print the fuel name accordingly.
The following snippet describes the solution except for input.
Petrol = X+A*K
Diesel = Y+B*K
if petrol < diesel: print("PETROL");
else if petrol > diesel: print("DIESEL");
else: print("SAME PRICE");
Constraints were low enough to actually simulate the prices of K days one by one, with time complexity O(K) per test case.
TIME COMPLEXITY
The time complexity is O(1) per test case.
SOLUTIONS
Setter's Solution
#include <bits/stdc++.h>
using namespace std;
void solve(){
int X; cin >> X;
int Y; cin >> Y;
int A; cin >> A;
int B; cin >> B;
int K; cin >> K;
int petrol = X + A*K;
int diesel = Y + B*K;
if(petrol < diesel){
cout << "PETROL" << '\n';
}else if(diesel < petrol){
cout << "DIESEL" << '\n';
}else{
cout << "SAME PRICE" << '\n';
}
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin >> t;
for(int i = 0; i < t; i++){
solve();
}
return 0;
}
Tester's Solution
#include <bits/stdc++.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) {
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,' ');
}
void readEOF(){
assert(getchar()==EOF);
}
int main() {
// your code goes here
int t;
t = readIntLn(1, 1000);
while(t--){
int x, y, a, b, k;
x = readIntSp(0, 1000);
y = readIntSp(0, 1000);
a = readIntSp(0, 1000);
b = readIntSp(0, 1000);
k = readIntLn(1, 1000);
if(x + a*k < y + b*k)
cout << "PETROL\n";
else if(x + a*k > y + b*k)
cout << "DIESEL\n";
else
cout << "SAME PRICE\n";
}
readEOF();
return 0;
}
Editorialist's Solution
import java.util.*;
import java.io.*;
class CHEAPFUEL{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int X = ni(), Y = ni(), A = ni(), B = ni(), K = ni();
int petrol = X+K*A, diesel = Y+K*B;
if(petrol < diesel)pn("PETROL");
else if(petrol > diesel)pn("DIESEL");
else pn("SAME PRICE");
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
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 CHEAPFUEL().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. 