C++ crisis HELP!!!!

I program, but not in C++ specifically. If I was solving a problem like this, the quickest and dirtiest way of doing it would be to take all the values entered, and then loop through from 1-99. If the code can find that value of your current loop integer in that array of saved values, then carry on to the next. Otherwise exit the loop and fail. If it gets to 99 and none have failed, then add the score.

You could also find all the unique values first, do a quick test to make sure there are 99 values there before entering the loop just to avoid wasting time.
As I say, don't code in C++ specifically, but I'm sure the functions to do this are easy enough to find.

Make a boolean array with length 99 (all 'false').
Instead of saving all the inputs, and wasting space & looping time - set the corresponding cell in the array to 'true' for each input.
After all inputs are done, go over the array and see if everything's true or not.

(i dont do c++ btw) i would probably go for something like:
kinda similar to MjrPITA but like reverse. set array with 99 length and on input unset inputted values. if array is empty at the end, you are good. this way you kinda won't need to loop

You could sort it, making things simpler.

Sorry if I misunderstood your problem:

You put the numbers into a textbox right? Maybe add create an INT named <textbox_value> or something (without the "<>" obv>, and let that have the entered value of the textbox.

Then let the program check the <textbox_value> and if it's within the allowed value range say to add the value of <textbox_value> to the vector and reset the <textbox_value>.

Do you only get 1000 points, if you enter all the numbers? What if I enter 98 unique numbers? Do I get 0 points? Or do I get 10 points per unique number?

I think you are looking for a frequency vector. It's purpose it's storing the number of appearances of a specific number in a list of numbers.
The flow goes like this:
make a vector the size of 100.(0-99)
1.input numbers(ex.1 , 88, 23)
2.your vector will look like this
3.{0,0,0,0,0,0,0,....,0,0} . You go through it and add +1 on that position
{0,1(position 1),0,0,.....,1(position 23),0,...,1(position 88),0,...}
4.check vector if all numbers are bigger than 0.
5.Repeat if 4 returns 4 from step 1

Here is an example with 2 input numbers for number 0-9

1. check if vector has all elements >0
   Vector(0,0,0,0,0,0,0,0,0,0)
   input 2,5

2.check if vector has all elements >0
Vector(0,0,1,0,0,1,0,0,0,0)
input 5 ,9
3.check if vector has all elements >0
Vector(0,0,1,0,0,2,0,0,0,1)
input 0,1
4.check if vector has all elements >0
Vector(1,1,1,0,0,2,0,0,0,1)
input 3,9
5.check if vector has all elements >0
Vector(1,1,1,1,0,2,0,0,0,2)
input 9,9
6.check if vector has all elements >0
Vector(1,1,1,1,0,2,0,0,0,4)
input 4,6
7.check if vector has all elements >0
Vector(1,1,1,1,1,2,1,0,0,2)
input 7,8
8.check if vector has all elements >0
Vector(1,1,1,1,1,2,1,1,1,2)
Done

I didn't use vectors in a long time but this might help
https://www.geeksforgeeks.org/program-to-find-frequency-of-each-element-in-a-vector-using-map-in-c/

Feel free to ask for any more help.

So many helpful people here! I think I'm too tired to understand anything right now. Hopefully I can fix this thing in the 4-5 hours I have when I wake up tomorrow...

https://pastebin.com/z8MRfNbb
UPD: I'm using std::vector<bool> and std::all_of. It's simple and short.
UPD2: It's similar to https://www.steamgifts.com/discussion/Bz8So/c-crisis-help#4wmtWYk.
UPD3: std::vector<bool> v(99); is better than std::vector<bool> v(99); v.resize(99);, of course.

I haven't written a simple C++ program in a while, so here you go:
Solution using array.
Same solution but using Vector.

If you need anything explained in more detail, feel free to add me and ask, though do keep in mind that I haven't done proper programming in years. :/

Looks like others have beat me to it. I program in Java at the enterprise level, but I have not had the immense pain joy of dealing with C++. I wish you the best with your assignment and studies though :)

My program had a problem with another aspect so I figured out that I had to skip the big ass vector (I used push_back to store the inputted values at the beginning of the loop and then used the vector to calculate values... my god).

Is it good to have a big ass vector in this case though? Right now the thing kind of loops, so every sequence of 10 is not stored in a vector. I'd like to store all the sequences in one vector if that's necessary, but that means to use push_back in a new vector to store the values of the first vector every time... or something?!

The cin for the numbers looks like this from a tip I found, is it possible to store this value in a vector?
https://imgur.com/Y1yb7xo

Tbh i dont really understand the question. Do you want to check if the user inputted all numbers from 1-99 in the input? Or just want to check if the user inputted number no more and no less than that number?

First of all, sorry about my english.

To solve the problem using only loops, without arrays or vectors, i would code a new function called "check" who recives an array as parameter that function iterate over integers between 1 and 99, in each loop check if the number is in the array, if that funcion pass all iterations then meet the condition.
in pseudocode it can be

function check(array) {
for (int i = 1; i <= 99; i += 1) {
if (!array.contains(i)) {
return false
}
}
return true
}

This is a very rudimentary code. I didn't make anything with fancy library functions like find() and arrays. Purely loops and a single vector. I also haven't made anything that makes it possible to input 10 numbers at once, so you'll have to figure that one yourself. Also check for exceptions like 102 and -2 in the input and make an error if you want to.

So here's the thing, why not check the vector array one by one. It's not optimized but you only need 99 intergers to check, so it shouldn't take too much CPU. The idea is, if your input isn't found in the vector array, it's ignored. Only unique intergers got inserted into it. Once you stop saying yes to the repetition, it checks if your vector is 99 in size. If it is, that means you get 1000 points!

#include <iostream>
#include <string>
#include <vector>
using namespace std;

int main()
{
    vector<int> vect;
    int inputnum;
    int vectorsize;
    string newline;
    bool userinput = true;
    while(userinput)
    {
        bool numexist = false;
        cout << "Enter your numbers: ";
        cin >> inputnum;
        if(inputnum) //check if inputnum is valid
        {
            for(int i = 0; i<vect.size(); i++)
            {
                if(vect[i] == inputnum)
                {
                    numexist = true;
                }
            }
            if(!numexist)
            {
                vect.push_back(inputnum);
                vectorsize++;
            }
        }

        cout << "One more time? ";
        cin >> newline;
        if(newline == "N" || newline == "n")
        {
            userinput = false;
        }
    }
    int winamount = 0;
    if(vect.size() > 98)
    {
        winamount = 1000;
    }
    cout << "Winning amount: " << winamount<< endl;
}

Hope this code I lazily put together in 10 minutes help.

What about, making a boolean to check the final sentence, something like

Make a loop to check number by numer if X value its found, for example

I will try to write code, but i'm very rusty, i hope you can get the idea, i will add notes of what i'm trying to do

if (j=0; j>101; j++){ //a loop to advance in the postition of the vector, self adding a value to advance in position

       if ( i=0; i>101;i++){                                         //a loop to check every number from 0-100

                   if (i=vector[j]) {                                   //this if checks if the new number it's on the vector, with it should, at least one time

                                countBoolean = 1                 //just a flag like i was told, a int variable to check if something it's ok

                    }

          i++

        }

j++
}

if (countBoolean <=2){ //if the sentence found the same number more than 1 time, the score it's 0, you can add
another variable to make sure that are only 100 numbers in the vector
score=0

}
else {

score = 1000 //but if everything it's ok it should give 1000 points
}
:

again, sorry for my crappy english and really poor code, hope it helps