Hi there. I'm Kurt!!
I have two algorithm problems.
I was wondering if you can check my answers for the first one and help me get a start with the second one:

1)Write an algorithm to compute the average of 5 numbers

Name: Ave5
Given: X1, X2, X3, X4, X5
Change: none
Intermediate: Total
Result: Average
Definition: Average:= Ave5 (X1, X2,X3,X4,X5)

METHOD

Get X1
Get X2
Get X3
Get X4
Get X5

Total = X1 + X2 + X3 + X4 + X5
Average = Total/5

Give Average

2) Write an algorithm to get the names and ages of two people. Return the name of the person who is older (in format "x is older than y", where x and y are the names of the two people), unless the two people are the same age, in which case, return the message "x is the same age as y.

I'm having trouble writing an algorithm for this particular case. It's more complicated than what I am used to. Can someone give me an idea on how to start a problem like this one ? I have made a couple of assumptions:
Name: PerAge
Givens: X, Y
Change: X,Y
Intermediate: Age
Result: none
Definition: Age =: PerAge (X,Y)

METHOD

Give X
Give Y

Well... I'm pretty lost at this point

>I was wondering if you can check my answers for the first one and help me get a start with the second one
The first one looks fine. Your problems with the second one makes me wonder.

>It's more complicated than what I am used to.
So basically anything that has more than one branch of execution is too much for you. You're looking at three branches: the ages are equal, x is less than y, and x is greater than y. Do you know how to use an if..else?

can i ask if what is the meaning of algorithmic problem cause thats my assignment im 1st year college in asian institute of computer studies.. plz help me

commented: Don't bump old threads -1

Check equal case, if not equal then, if a is older than b declare a oldest otherwise declare b older.

well all u gotto do , is to check the age of two people and print in the format asked
for example:

``````if (  x.age > y.age)
cout<<x.name<<"is older than <<y.name;
else if ( x.age == y.age)
cout<<x.name<<"is same age as "<<y.name;``````

well i am assuming ur doing using classes , it would be easier since the person will
share the same attributes so better to put in one class.

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