Ticket Office Run

Question

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

Here is other's code:
#include <iostream>
using namespace std;

int main()
{
int ages[5];

for (int i = 0; i < 5; ++i)
{
cin >> ages[i];
}
double m;
double p;
m = ages[0];
for (int x = 1; x < 5; ++x) {
if ( m > ages[x]) {
m = ages[x];
}
}
p = 50 - (50*m/100);
cout << p << endl;

return 0;
}

I don't really get the part start from double m and p.
Why m = ages[0] ( is this means that it don't have any elements in this arrays ?)
why int x equal to 1 ?
what does the ( ( m > ages[x]) {
m = ages[x]; } means ?

Answer 1

Big Fish i thinks the logic is wrong but I will Still explain u what is happening here.

part 1 : Data Input from user to an array "ages[5]" using a for loop
eg. ages[5] = {23,17,65,47,15}

part 2 : Finding the Youngest aged person.
explained :-
a. They declared "m" and "p" variable where m means minimum age and p means price of tickets
b. They give the value of first element of ages array (which is the age of first person ) to m variable
c. then they run a loop to find the smallest element of array ( minimum aged person ) or discount %
d. as we know the array element starts from 0 and we already stored the value of 0th element in m variable so they started the loop from second element using int i = 1

part 3 : Caluclation logics to get price with discount and print the price with cout

Answer 2

Big Fish this is not ur fault
I means you are just a beginner that is not a big deal
U just need to practise, I know you had paid attention to lesson but this things ( having bugs and unable to catch up the code fastly ) happens with every beginner even with me also about 3 years ago when I am in School.
So Just keep practicing.😁

Answer 3

Thanks guys! yah I think I need to do recap again :(

Answer 4

Abhay Bro

Logic is absolutely right. There is no any issue.

Answer 5

🅰🅹 🅐🅝🅐🅝🅣 yeah u r right

Answer 6

:"D Thanks QQQQ !! I will practice as more as I can

Answer 7

Big Fish

ages[0] means we are getting 1st element to compare with next elements in an array to get youngest age.

We have started x with 1 because we have to compare 1st element with next elements so doesn't make sense to compare 1st element with 1st element.

Answer 8

For that you should have paid attention in lessons before this module question.

This specific part is covert here 👇
https://www.sololearn.com/learning/1625/

Answer 9

Ya I pay attention but somehow brain stuck btw thanks!

Answer 10

Ticket Office

Why test case 3? I need an answer for this?
include <iostream>
using namespace std;
int main(){
int ages[5];

for (int i + 0; <5; i++){
cin>>ages[i];
}

int a= ages[0];
int b= ages[1];
int c= ages[2];
int d= ages[3];
int e= ages[4];

if ((a<b)&&(a<c)&&(a<d)&&(a<e)){
float w=a;
float p=(w/100)*50;
float y=50-p;
cout<<y<<endl;
}
if ((b<a)&&(b<c)&&(b<d)&&(b<e)){
float w=b;
float p=(w/100)*50;
float y=50-p;
cout<<y<<endl;
}
if ((c<b)&&(c<a)&&(c<d)&&(c<e)){
float w=c;
float p=(w/100)*50;
float y=50-p;
cout<<y<<endl;
}
if ((d<b)&&(d<c)&&(d<a)&&(d<e)){
float w=d;
float p=(w/100)*50;
float y=50-p;
cout<<y<<endl;
}
if ((e<b)&&(e<c)&&(e<d)&&(e<a)){
float w=e;
float p=(w/100)*50;
float y=50-p;
cout<<y<<endl;
}

return 0;
}

Answer 11

Ticket Office

Why test case 3? I need an answer for this?
include <iostream>
using namespace std;
int main(){
int ages[5];

for (int i + 0; <5; i++){
cin>>ages[i];
}

int a= ages[0];
int b= ages[1];
int c= ages[2];
int d= ages[3];
int e= ages[4];

if ((a<b)&&(a<c)&&(a<d)&&(a<e)){
float w=a;
float p=(w/100)*50;
float y=50-p;
cout<<y<<endl;
}
if ((b<a)&&(b<c)&&(b<d)&&(b<e)){
float w=b;
float p=(w/100)*50;
float y=50-p;
cout<<y<<endl;
}
if ((c<b)&&(c<a)&&(c<d)&&(c<e)){
float w=c;
float p=(w/100)*50;
float y=50-p;
cout<<y<<endl;
}
if ((d<b)&&(d<c)&&(d<a)&&(d<e)){
float w=d;
float p=(w/100)*50;
float y=50-p;
cout<<y<<endl;
}
if ((e<b)&&(e<c)&&(e<d)&&(e<a)){
float w=e;
float p=(w/100)*50;
float y=50-p;
cout<<y<<endl;
}

return 0;
}

Answer 12

#include <iostream>
using namespace std;

int main() {
int ages[5];
int total = 50;
float final, discount;

for (int i = 0; i < 5; i++) {
cin >> ages[i];
}

int min = ages[0];

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

if (ages[i] < min)
min = ages[i];
}

discount =(min * 0.01) * total;

final = total - discount;

cout << final;

return 0;
}

Answer 13

Hi guys! My code variant.
#include <iostream>
using namespace std;

int main()
{
int ages[5];
int min_age;
min_age = ages[0];
float t=50.0;
float price;
for (int i = 0; i < 5; ++i)
{
cin >> ages[i];
if(ages[i] < min_age)
{
min_age = ages[i];
}
}
price=t-((t*min_age)/100);
cout<<"Price with discount "<< price<<endl;
return 0;
}