# MARARUN - Editorial

Setter: Daanish Mahajan
Tester: Istvan Nagy
Editorialist: Taranpreet Singh

Cakewalk

None

# PROBLEM

Given the distance covered per day in km and the number of days, determine the prize won by person, if a person can get prize A for covering at least 10 km, B for covering at least 21 km, and C for covering at least 42 km. Given A < B < C

# QUICK EXPLANATION

Compute T = d*D as the total distance covered, and compare T with 10, 21, and 42 to determine the category person can apply to, and find the maximum prize.

# EXPLANATION

Since person covers d km per day for D days, total distance covered is T = d*D. Now, the only thing we need to check is the largest prize category person can apply to.

Since we have 0 < A < B < C, we can see that

• If the distance covered is at least 42 km, a person can win prize C
• If the distance covered is between 21 km and 41 km, a person can win prize B
• If the distance covered is between 10 km and 20 km, a person can win prize A
• If the distance covered is less than 10 km, a person cannot win any prize.

We can write those as if conditions as represented by the following pseudocode.

``````T = d*D
if T >= 42: print(C)
if T >= 21 and T < 42: print(B)
if T >= 10 and T < 21: print(A)
if T < 10:print(0)
``````

We can also reduce the number of conditions by using an if-else ladder, as follows.

``````T = d*D
if T >= 42: print(C)
else if T >= 21: print(B)
else if T >= 10: print(A)
else: print(0)
``````

### Bonus problems

• What if we are not guaranteed A < B< C in above problem?

# TIME COMPLEXITY

The time complexity is O(1) per test case.

# SOLUTIONS

Setter's Solution
``````#include<bits/stdc++.h>
# define pb push_back
#define pii pair<int, int>
#define mp make_pair
# define ll long long int

using namespace std;

int main()
{
int t; cin >> t;
int days, d, a, b, c;
while(t--){
cin >> days >> d >> a >> b >> c;
int ans = 0;
if(days * d >= 42)ans = c;
else if(days * d >= 21)ans = b;
else if(days * d >= 10)ans = a;
cout << ans << endl;
}
}
``````
Tester's Solution
``````#include <iostream>
#include <cassert>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <random>

#ifdef HOME
#include <windows.h>
#endif

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

template<class T> bool umin(T& a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool umax(T& a, T b) { return a < b ? (a = b, true) : false; }

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) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

int main(int argc, char** argv)
{
#ifdef HOME
if(IsDebuggerPresent())
{
freopen("../in.txt", "rb", stdin);
freopen("../out.txt", "wb", stdout);
}
#endif
forn(tc, T)
{

int dist = d * D;
if (dist >= 42)
printf("%d\n", C);
else if (dist >= 21)
printf("%d\n", B);
else if (dist >= 10)
printf("%d\n", A);
else
printf("0\n");
}
assert(getchar() == -1);
return 0;
}
``````
Editorialist's Solution
``````import java.util.*;
import java.io.*;
class MARARUN{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
int D = ni(), d = ni(), A = ni(), B = ni(), C = ni();
int distance = D*d;
if(distance >= 42)pn(C);
else if(distance >= 21)pn(B);
else if(distance >= 10)pn(A);
else pn(0);
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
void run() throws Exception{
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 MARARUN().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());}

StringTokenizer st;
}

}

String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
}catch (IOException  e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}

String nextLine() throws Exception{
String str = "";
try{
Feel free to share your approach. Suggestions are welcomed as always. 