如何使用C ++计算1-100数组中的覆盖率百分比？

Question

This is for an assignment so I would appreciate no direct answers; rather, any logic help with my algorithms (or pointing out any logic flaws) would be incredibly helpful and appreciated!

I have a program that receives "n" number of elements from the user to put into a single-dimensional array. The array uses random generated numbers. IE: If the user inputs 88, a list of 88 random numbers (each between 1 to 100) is generated). "n" has a max of 100.

I must write 2 functions.

Function #1:

Determine the percentage of numbers that appear in the array of "n" elements.
So any duplicates would decrease the percentage.
And any missing numbers would decrease the percentage.
Thus if n = 75, then you have a maximum possible %age of 0.75 
(this max %age decreases if there are duplicates)

This function basically calls upon function #2. 

FUNCTION HEADER(GIVEN) = "double coverage (int array[], int n)"

Function #2:

Using a linear search, search for the key (key being the current # in the list of 1 to 100, which should be from the loop in function #1), in the array.

Return the position if that key is found in the array 
(IE: if this is the loops 40th run, it will be at the variable "39", 
and will go through every instance of an element in the array 
and if any element is equal to 39, all of those positions will be returned? 
I believe that is what our prof is asking)

Return -1 if the key is not found. 

Given notes = "Only function #1 calls function #2, 
and does so to find out if a certain value (key) is found among the first n elements of the array."

FUNCTION HEADER(GIVEN) = "int search (int array[], int n, int key)"

What I really need help with is the logic for the algorithm.

I would appreciate any help with this as I would approach this problem completely differently than our professor wants us.

My first thoughts would be to loop through function #1 for all variable keys of 1 through 100. And in that loop, go to the search function (function #2), in which a loop would go through every number in the array and add to a counter if a number was (1)a duplicate or (2) non-existent in the array. Then I would subtract that counter from 100. Thus if all numbers were included in the array except for the #40 and #41, and then #77 was a duplicate , the total percentage of coverage would be 100 - 3 = 97%.

Although as I type this I think that may in of itself be flawed? ^ Because with a max of 100 elements in the array, if the only number missing was 99, then you would subtract 1 for having that number missing, and then if there was a duplicate you would subtract another 1, and thus your percentage of coverage would be (100-2) = 98, when clearly it ought to be 99.

And this ^ is exactly why I would REALLY appreciate any logic help. :)

I know I am having some problems approaching this logically.

I think I can figure out the coding with a relative amount of ease; what I am struggling witht he most is the steps to take. So any pseudocode ideas would be amazing!

(I can post my entire program code so far if necessary for anyone, just ask, but it is rather long as of now as I have many other functions performing other tasks in the program) edit: I am seeing multiple people edit my post/tags. The tags are not ones I put xD So please don't downvote my question because you had to edit this?D: Someone just edited this back to the original way it was before someone else edited it. Ah

Answer 1

I may be mistaken, but as I read it all you need to do is:

• write a function that loops through the array of n elements to find a given number in it. It would return the index of first occurence, or a negative value in case the number cannot be found in the array.
• write a loop to call the function for all numbers 1 to 100 and count the finds. Then divide the result by 100.

Answer 2

I'm not sure if I understand this whole thing right, but 1 function you can do it, if you don't care about speed, it's better to put array into a vector , loop through 1..100 and use boost find function http://www.boost.org/doc/libs/1_41_0/doc/html/boost/algorithm/find_nth.html . There you can compare current value with the second entry value in the vector, if it contains you decrease, not not decrease, if you want to find if the unique number is in array, use http://www.cplusplus.com/reference/algorithm/find/ . I don't understand, how the percentage decreases, so it's on your own and I don't rly understand second function, but if its linear search use again find. PS Vector description http://www.cplusplus.com/reference/vector/vector/begin/ .

Answer 3

You want to know how many numbers in the range [1, 100] appear in your given array. You can search for each number in turn:

size_t count_unique(int array[], size_t n)
{
    size_t result = 0;

    for (int i = 1; i <= 100; ++i)
    {
        if (contains(array, n, i))
        {
            ++result;
        }
    }

    return result;
}

All you still need is an implementation of the containment check contains(array, n, i) , and to transform the unique count into a percentage (by using division).