-
#include <algorithm>
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#include <iomanip>
-
#include <ios>
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#include <iostream>
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#include <string>
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#include <vector>
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using namespace std;
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int main ()
-
{
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cout <<" plz enter your first name: ";
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string name;
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cin>>name;
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cout<<"Hello, "<<name<<"!"<<endl;
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cout<<plz enter ur midterm and final exam grades: ";
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double midterm, final;
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cin>>midterm>>final;
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cout<<"enter all ur homeworks grades, :
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"followed by end-of-file: ";
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vector<double> homework;
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double x;
-
while (cin>>x)
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homework.push_back(x);
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typedef vector<double>::size_type vec_sz;
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vec_sz size= homework.size();
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if(size==0) {
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cout<<endl<<"u must enter ur grades. "
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"plz try again."<<endl;
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return1; // does that mean if this statement fail it restart the if statement or the whole app
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sort(homework.begin(), homework.end() );
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vec_sz mid=size/2
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double median;
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median=size %2 ==0 ? (homework[mid]+homework[mid-1] /2
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:homework[mid];
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//isn't homework[mid]= to homework homework[size/2] which is how many numbers are contain in homework
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//just how does it works don't very get it
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streamsize prec=cout.precison(3)
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<<0.2+midterm+0.4 *final+0.4*median//isn't median =size%2
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<<setprecison(prec)< <end;
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return0;
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}
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I have read the book seven times, but still don't get it.
I know it sounds fishy, but I very need ur help.
Ur voice are super important to me.
Plz lend me a hand
20 1910
What problems are you having? are you getting an error message.? A little more information please.
no i just don't understand the script. Plz help
#
return1; // does that mean if this statement fail it restart the if statement
line 28
line 32-34
#
median=size %2 ==0 ? (homework[mid]+homework[mid-1] /2
#
:homework[mid];
#
//isn't homework[mid]= to homework homework[size/2] which is how many numbers
and line 37
<<0.2+midterm+0.4 *final+0.4*median//isn't median =size%2
no i just don't understand the script. Plz help
return1; // does that mean if this statement fail it restart the if statement
no no. I means that if no homeworks were inputed, stop the program. The return will break out of main. Basically shutting down your program.
median=size %2 ==0 ? (homework[mid]+homework[mid-1] /2
:homework[mid];
To take the median, you need to find the middle test. However, if there is an even number of tests, which one is the middle. Here's a visual example:
1 2 3 4 5 < the median is 3
1 2 3 4 < the median is (2 + 3)/2
This checks the size, if the size is even, do first command (between question mark and colon), if not, do what is after the colon. Basically it is a boolean if statement.
Hope that helped.
median=size %2 ==0 ? (homework[mid]+homework[mid-1] /2
:homework[mid];
sorry i still don't get it.
I get the non-pc version how to compute median.
Like if it is even divide the size /2
for example
10/2
the median will be 5 number +6 number /2
And the odd number is size/2
then the number to that is the median.
But the problem is isn't
median=size %2 ==0 ? (homework[mid]+homework[mid-1] /2
:homework[mid];
only calculate where is the median located ,but not the median itself???????
wow i must be slow 2 replies before i finished.
median will equal the value located in the homework vector
or it will equal the average of the two middle values in the homework vector.
Err don't quite get what do u mean by that.
Sorry about that.
okay
homework is a vector that contains doubles
if (size % 2 == 0) //there are an even number of grades
median = (homework[mid] + homework[mid-1] / 2) // (double + double) / 2
else // there are an odd number of grades
median = homework[mid] //equals the double value located in the middle of the vector
Maybe it's because it's Sunday overhere but I find this thread very chaotic.
Suppose the homework[] array is sorted in ascendng or descending order. Let the
number of elements in that homework[] array be 'n'.
There are just two cases:
1) n is odd; the median is homework[n/2];
2) n is even; the median is (homework[n/2]+homework[n/2+1])/2;
if n&1 == 1 then n is an odd number; there's not much more to it.
kind regards,
Jos
o so because size is vector double
Which means it actually contain the number itself and not how many numbers there is right???
It is Sunday, I know I am sorry for bugging u in Sunday.
any answer. Plz i need ur hand.
o so because size is vector double
Which means it actually contain the number itself and not how many numbers there is right???
It is Sunday, I know I am sorry for bugging u in Sunday.
You're not 'bugging' me; I'm just being lazy but you are over complicating matters.
What does 'actually contains the number itself' mean? Here's an example:
12, 14, 20, 100, 1000
There are five numbers, so n == 5 and n/2 == 2 (integer division which means
skip the remainder). C and C++ indexes start counting from 0 so, element #2
is the median, i.e. 20 is the median.
Another example:
12, 14, 20, 100
There are four numbers, so n== 4 and n/2 == 2 so element #2 and 2-1 == #1
make up the median (20+14)/2 == 17 is the median.
kind regards,
Jos
ps. I made a mistake in my previous reply (me being lazy); change the n/2+1
to n/2-1 and everything is fine again.
so size is actually
for example if i input these number.
1,2,3,4
size will contain 1,2,3,4
instead of just 4 like counting how many numbers a there
so size is actually
for example if i input these number.
1,2,3,4
size will contain 1,2,3,4
instead of just 4 like counting how many numbers a there
I'm sorry, I really don't understand what you're talking about; e.g. these numbers
10, 20, 30, 40
The size of the array/vector/whatever is still 4 but the number 4 isn't one of the
elements off the array/vector/whatever. The median will be the average of the
numbers stored at position 4/2 and 4/2-1 so (30+20)/2 == 25.
kind regards,
Jos
4/2 + 4/2-1 /2
because of order of operation so is the ((4/2+4/2)-1)/2
which is 2+2-1 which is three then divided by 2.
but that is the first number plus second /2.
not second number +the third one??????
4/2 + 4/2-1 /2
because of order of operation so is the ((4/2+4/2)-1)/2
which is 2+2-1 which is three then divided by 2.
but that is the first number plus second /2.
not second number +the third one??????
If there are four (4) numbers to be considered, take the 4/2 and 4/2-1, which are
the 2nd and 1st numbers and take the average of them. Remember C and C++
start counting array elements from zero: the 0th, 1st, 2nd, 3rd, 4th, 5th element etc.
kind regards,
Jos
so is the fifth number divided by 2 but rounded up the decimal.
so is 5/2 +5/2-1 then divided by 2
so is 2.5 + 1.5/2
but since they rounded up it is 3+2 /2 right?????
plz come back and help.
I need ur help plz.
or is it 4/2+1 and 4/2-1+1 then divided by two.
or is it 4/2+1 and 4/2-1+1 then divided by two.
What are you juggling? Here's another example:
0:10
1:20
2:30
3:40
4:60
The number left of the colon is the index of the number, right of the colon is
the number itself. This array has 5 elements, so 5/2 == 2 so that median is
found at index number 2, which is 30.
For an even number of elements:
0:10
1:20
2:30
3:40
There are four elements so take the average of the numbers at indexes 4/2
and 4/2-1, which are index values 2 and 1 so take the average of 30 and 20
which makes (30+20)/2 == 25.
In general: if n (the number of elements in the array) is odd, take the
element at index value n/2; if n is even take the average of the two numbers
at index positions n/2 and n/2-1. Of course the numbers in the array need
to be sorted, either ascending or descending.
kind regards,
Jos
I get it now thx
for ur great help
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