How can i complete ticket office module project in c++ ? Sololearn to achieve certificate....

Answer 1

I'm glad that it worked for you.
But you should try making a code which will find the smallest of the array instead of hard coding with all those if-else statements..

For your reference..
Hope you'll find some inspiration 😉

#include <iostream>
using namespace std;

double findYoungest(int arr[]){
double min = arr[0];
for(int i=0; i<5; i++) {
if(min>arr[i]) {
min=arr[i];
}
}
return min;
}

int main() {
int ages[5];

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

double discount = 50 - (findYoungest(ages)*0.5);
cout<< discount;

return 0;
}

Answer 2

Why the last line "everything is same" ?

You can't write anything which is not expected in the code coach.

Answer 3

You're welcome 😊

Answer 4

By making the appropriate code which passes all the testcases?

Answer 5

Can please you link your code here so others can see it and debug accordingly?

Answer 6

My code (I think it's as simple as it could get):

#include <iostream>
using namespace std;

int main() {
int ages [5], arrN;
for (arrN = 0; arrN < 5; arrN++) {
cin >> ages [arrN];
}
float min = ages [0];
for(int arrN = 0; arrN < 5 ;++arrN){
if(ages[arrN] <= min){
min = ages[arrN];
}}
cout << 50 -50 * min / 100 << endl;
return 0;}

Answer 7

This is what I used. Passed with it.

#include <iostream>
using namespace std;

int main() {
int ages[5];

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

double smallest = ages[0] ;
for ( int i=1; i < sizeof(ages)/sizeof(ages[0]); ++i )
if ( ages[i] < smallest )
smallest = ages[i] ;

cout << 50 - smallest / 100 * 50;

return 0;
}

Answer 8

I found the post with the 2nd array and using unnecessary doubles more confusing so here's how I did it:
#include <iostream>
using namespace std;

int main() {
int ages[5];
int i;
int min;

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

}
}
cout<<50-(min*0.5)<<endl;

return 0;
}

Answer 9

This is perfect code:

#include <iostream>
using namespace std;

int main() {
int ages[5];

for (int i = 0; i < 5; ++i) {
cin >> ages[i];
}
//your code goes here
int little = 9999999999999; //No one can be this age.
double k;
for (int i = 0; i < 5; i++) {
if (ages[i] < little) {
little = ages[i];
k = ages[i];
}
}

cout << 50 - 50 * k / 100;

return 0;
}

Answer 10

I have written a code which passes in 1st, 2nd, 4th and 5th test cases, but in 3rd case it is showing that i have written wrong code, and 3rd case is hidden, i can't even see what the case is...

Answer 11

#include <iostream>
using namespace std;

int main() {
float ages[5];

for (int i = 0; i < 5; ++i) {
cin >> ages[i];
}
//your code goes here
float doll=50;

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

float tem;
float p;

if(a<b&&a<c&&a<d&&a<e)
{
tem=doll*a/100;
p=doll-tem;
cout << p;
}
else if(b<a&&b<c&&b<d&&b<e)
{
tem=doll*b/100;
p=doll-tem;
cout << p;
}
else if(c<a&&c<b&&c<d&&c<e)
{
tem=doll*c/100;
p=doll-tem;
cout << p;
}
else if(d<a&&d<b&&d<c&&d<e)
{
tem=doll*d/100;
p=doll-tem;
cout << p;
}
else if(e<a&&e<b&&e<c&&e<d)
{
tem=doll*e/100;
p=doll-tem;
cout << p;
}
else
{
cout << "everthing is same";
}

return 0;
}

Answer 12

This is the code

Answer 13

It is like when all the values are provided in inputs are same.....

Answer 14

Hey thanks ❤😊 it's done
I just removed last else part and renamed else if as else and now all the test cases are satisfied😊....

Answer 15

Yes ✅it's good your given code is short and easy to understand.... Thank you ☺

Answer 16

Well arr is usual reserved for arrays not the name for a regular integer. Float is not needed for regular integers. Not like the cashier will type in they are 15.6 years old. Also <= is unnecessary as if the age is the same there is no reason to change the value of the variable. You only need < or > depending on which variable you put first. Otherwise it's the same as mine. I just spaced mine out for easy reading.

Answer 17

take a look in my codes saved u will find solution

Answer 18

#include <iostream>
using namespace std;

int main() {
int ages[5], arr;

for (int i = 0; i < 5; ++i) {
cin >> ages[i];
}
float min = ages[0];
for(int arr = 0; arr < 5 ;++arr) {
if(ages[arr] <= min) {
min = ages[arr];
}
}
cout << 50 - (min * 50) / 100;
return 0;
}

Answer 19

#include <iostream>
using namespace std;

int main() {
int ages[5];
double younger;

for (int i = 0; i < 5; ++i) {
cin >> ages[i];
}
//your code goes here
younger = ages[0];
for (int i = 0; i < 5; ++i) {
if (younger > ages[i]) {
younger = ages[i];
}
}
double discount = 50-((younger*50)/100);
cout << discount << endl;
return 0;
}

Answer 20

#include <iostream>
using namespace std;
int main() {
int ages[5];
for (int i = 0; i < 5; ++i) {
cin >> ages[i];
if (i>=1){
if (ages[i]<ages[0]){
ages[0]=ages[i];
}}
}
cout<<float(50-5*ages[0]/10.0);
}