C++ Ticket Office problem solved but not marked as done

Question

So I wrote a very basic and ugly noob program to solve this problem.
The issue is it doesn't seem like giving me credit for. The program completed the first 2, and the 3rd and 4th hidden Test Cases. I wonder why isn't marked as done?! as you can see here:

https://i.postimg.cc/Dn46T8X7/Desktop-15-01-2021-00-48-30-380.png

The Problem if someone interested in in the Data Types, Arrays, Pointer's section:
Ticket Office

You are working on a ticketing system. A ticket costs $10.
The office is running a discount campaign: each group of 5 people is getting a discount, which is determined by the age of the youngest person in the group.
You need to create a program that takes the ages of all 5 people as input and outputs the total price of the tickets.

Sample Input:
55
28
15
38
63

Sample Output:
42.5

The youngest age is 15, so the group gets a 15% discount from the total price, which is $50 - 15% = $42.5
The given code is declaring an array of 5 elements and taking them from input using a loop.
-------------------------------------------------------------------------------------------------------
#include <iostream>
using namespace std;

int main() {
int ages[5];

for (int i = 0; i < 5; ++i) {
cin >> ages[i];
}
//your code goes here

return 0;
}
--------------------------------------------------------------------------------------------------------
Input is given.
Test case 1 : Test case 2: Rest is hidden as you can see on the link.
11 69
18 48
19 33
22 25
49 29

Answer 1

I've re-written in a similar manner Mateusz has mentioned and some others on the forums.

#include <iostream>
using namespace std;

int main() {
int ages[5];

for (int i = 0; i < 5; ++i) {
cin >> ages[i];
}
//your code goes here
float perc;
float min = ages[0];
for (int x = 1; x < 5; ++x){
if(min > ages[x])
min = ages[x];
}
perc = 50-(50*min/100);
cout << perc << endl;

return 0;
}

It works perfectly and looks great. I've also got the xp just as you mentioned, Arsenic.
I'd still like to find or find out new ways to do this.
(Under no circumstances I'd come up with this code by myself I feel like)

Answer 2

In the screenshot you posted it fails on the test case 3. There can be few problems.
First is the amount of code and how you search for min value. This is very inefficient and prone to errors. Better way would be to use a for loop. Something like that:

int min_age = ages[0];

for(int iage=1; iage < 5; ++iage) {
if (ages[iage] < min_age) {
min_age = ages[iage];
}
}

No unnecessary code repetition and less space for mistakes.
My guess is that the error comes from how you check for minimum value. Your if statement checks only if the value is smaller. What happens if the value is equal? Say you have 18, 18, 25. 18 is lesst than 25, but not less that 18, so it will evaluate to false and you will not find a minimum value.

Answer 3

I want to see what you submitted as solution of this problem fergoka

P.S. the problem is not considered solved till the time it doesn't pass all the tests.

check this out to know why SL hides them 👇
https://www.sololearn.com/Discuss/2662203/?ref=app

Answer 4

So basically what happeds there if I understand is that you check for the lowest numbers in a for loop and overwrite the lowest on top the rest in the min_age variable. It's really an elegant way to solve that. I couldn't find out a better way than that terrible one I jut did :)
It almost seems like your is a template type of way for checking for minimums and maximums in an array within a loop. Am I right?

Thanks for the reply.
Also, I don't think hidden Test Cases should be solved...until they become un-hidden at least.

Answer 5

Arsenic thanks for your answer and link. It's most likely the case Mateusz has mentioned and so there are at least 2 persons of the same age, my crappy and inefficient program is unable to check. This is what I've submitted://( I re-write it today soon and share it here.)

#include <iostream>
using namespace std;

int main() {
int ages[5];
float cost = 50; // total cost for 5 persons aka 50$
int x = 0; // youngest age
float per; // x age in percentage
float disc; // total discount

for (int i = 0; i < 5; ++i) {

cin >> ages[i];
}

if(ages[0] < ages[1] && ages[0] < ages[2] && ages[0] < ages[3] && ages[0] < ages[4])
{//checkin which number is the smallest in the array == youngest person in the group
x = ages[0];
per = (x*cost)/100; //Rule of third for percentage calculation
disc = cost - per;
cout << disc << endl;
}

if(ages[1] < ages[0] && ages[1] < ages[2] && ages[1] < ages[3] && ages[1] < ages[4])
{
x = ages[1];
per = (x*cost)/100; //Rule of third for percentage calculation
disc = cost - per;
cout << disc << endl;
}

if(ages[2] < ages[0] && ages[2] < ages[1] && ages[2] < ages[3] && ages[2] < ages[4])
{
x = ages[2];
per = (x*cost)/100; //Rule of third for percentage calculation
disc = cost - per;
cout << disc << endl;
}

if(ages[3] < ages[0] && ages[3] < ages[1] && ages[3] < ages[2] && ages[3] < ages[4])
{
x = ages[3];
per = (x*cost)/100; //Rule of third for percentage calculation
disc = cost - per;
cout << disc << endl;
}

if(ages[4] < ages[0] && ages[4] < ages[1] && ages[4] < ages[2] && ages[4] < ages[3])
{
x = ages[4];
per = (x*cost)/100; //Rule of third for percentage calculation
disc = cost - per;
cout << disc << endl;
}

return 0;
}

Answer 6

this is the right code :

#include <iostream>
using namespace std;

int main() {
int ages[5];

int mn=10000;

for (int i = 0; i < 5; ++i) {

int age;
cin >> age;
ages[i] = age;

mn = min(mn,age);
}
//your code goes here

float disc = float(mn)/float(100);

float price = 50 * (1-disc);

cout<<price<<endl;

return 0;
}

Answer 7

#include <iostream>
using namespace std;

int main() {
float ages[5],dis,min;

for (int i = 0; i < 5; ++i) {//getting values or input
cin >> ages[i];
}
//your code goes here
min=ages[0];//initialized minimum value to array's 0th location or first element
for(int i=0;i<5;i++)//getting the smallest element in array using for loop and if statement
{
if(ages[i]<min)
min=ages[i];
}
dis=(100-min)/100*50;
cout<<dis;

return 0;
}

Answer 8

For loop lagadiya hota ye sab kya kar diya bro

Answer 9

//HERE IS THE SOLUTION
#include <iostream>
using namespace std;

int main() {
int ages[5];

for (int i = 0; i < 5; ++i) {
cin >> ages[i];
}
//your code goes here
float temp = ages[0];
for(int i=0;i<5;i++)

{

if(temp>ages[i])
{
temp=ages[i];
}
}
float d = (50)*(temp/100);
float total = 50-d;
cout<<total;
return 0;
}