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Streamlining of

rielyn

int col = 0;
for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
        if (matrix[i][j] != 0) { 
            col++; 
        }
        if (col == m) {
            for (i = 0; i < n; i++) {
                for (int j1 = 0; j1 < m; ++j1) {
                    int pos = j1;
                    int tmp = matrix[i][pos];
                    for (j = j1 + 1; j < m; ++j) {
                        if (matrix[i][j] < tmp) {
                            pos = j;
                            tmp = matrix[i][j];
                        }
                    }
                    matrix[i][pos] = matrix[i][j1];
                    matrix[i][j1] = tmp; // меняем местами наименьший с a[j1]
                }
            }
        } 
    }
}

This slice of the code shall be sorted by the height of the line elements in a matrix where there are no zero elements (if any one is left unchanged). But in fact, he's sorting all the lines, I can't figure out what I'm wrong about, say thank you in advance! Choice.

Laycee last edited by

Move the energie col in the first cycle
And why do you need a second cycle of i?
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This method is repeated in several activities. You can't streamline? Trying to solve the plague.
Software Programming • java android • • Mystic

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You create a class that runs an interface. View.OnClickListener, describe the logic of clicks in it, and you'll be happy:class FabClickListener implements View.OnClickListener { @Override public void onClick(View v) { Intent sendIntent = new Intent(); sendIntent.setAction(Intent.ACTION_SEND); sendIntent.putExtra(Intent.EXTRA_TEXT, "Вы можете скачать приложение по ссылке: https://play.google.com/store/apps/details?id=" + BuildConfig.APPLICATION_ID); sendIntent.setType("text/plain"); v.getContext().startActivity(sendIntent); Toast.makeText(v.getContext(), "Поделиться приложением", Toast.LENGTH_SHORT).show(); } } Next in the code, you just write.fab_home.setOnClickListener(new FabClickListener());
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Streamlining of the range of disposal facilities
Software Programming • java choice gradcation • • chanisef

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I think it's sorted.public static void main(String[] args) { Car[] cars = new Car[7]; cars[0] = new Car("Skoda"); cars[1] = new Car("FIAT"); cars[2] = new Car("FORD"); cars[3] = new Car("MAN"); cars[4] = new Car("KIA"); cars[5] = new Car("BMW"); cars[6] = new Car("LADA"); for (int i = 0; i < cars.length; i++) { for (int j = 0; j < cars.length - i - 1; j++) { int q = cars[j].compareTo(cars[j + 1]); if (q < 0) { Car temp = cars[j + 1]; cars[j + 1] = cars[j]; cars[j] = temp; } } } for (int i = 0; i < cars.length; i++) { System.out.println(cars[i]); } } exhaust{name='Skoda'} {name='MAN'} {name='LADA'} {name='KIA'} {name='FORD'} {name='FIAT'} {name='BMW'}
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Stability
Software Programming • c language • • juvenalb

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If a consistent incremental increation of number two is considered N Bits starting with zero combination 0..0the sequence with the two neighbouring single battles will arise in the form of increments xxx110..0a couple of neighbouring units followed by zeros.It is clear that, no matter how many of us have incremented this sequence, we will not be interested (i.e. maintaining this couple of neighbouring units) until we get a transfer to the younger unit of this couple. I mean, we don't know if we're interested in any of the results of the timeline until we get to the sequence. xxx111..1 And we're not gonna add a unit to it, having xxx000..0♪ This unnecessary results don't have to be generated at all - it's easy to jump in sequence generation.So, in order to generate the desired sequences, we need to implement the usual increments. N- a significant binary number with one chemical modification: i In the meantime, it became important 1We need to check the next senior level. i + 1♪ If it does, 1Then we must artificially continue to carry out the algorithm of increments. i#include <stdio.h> #include <stdbool.h> #define N 5u bool increment_no_11(unsigned char (*n)[N]) { unsigned i; for (i = 0; i < N; ++i) if ((*n)[i] == 1) (*n)[i] = 0; else if (i + 1 == N || (*n)[i + 1] == 0) { (*n)[i] = 1; break; } return i < N; } int main(void) { unsigned char n[N] = { 0 }; do { for (unsigned i = N; i-- > 0;) printf("%d", n[i]); printf("\n"); } while (increment_no_11(&n)); } For N = 100 Number of sequences will be equal 927372692193078999176♪
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Streamline the mass by binary box
Software Programming • c++ algorithm sorting • • Avante

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I guess that's what this is about. https://ru.wikipedia.org/wiki/%D0%A1%D0%BE%D1%80%D1%82%D0%B8%D1%80%D0%BE%D0%B2%D0%BA%D0%B0_%D0%B2%D1%81%D1%82%D0%B0%D0%B2%D0%BA%D0%B0%D0%BC%D0%B8 ♪In C+++, boxing is done with two standard algorithms - std::upper_bound and std::rotate♪template<typename ForwardIterator> void insertion_sort(ForwardIterator first, ForwardIterator last) { for (auto it = first; it != last; ++it) { auto insert_to = std::upper_bound(first, it, *it); std::rotate(insert_to, it, std::next(it)); } } it indicates the current element.upper_bound Makes binary search within range [first; it)and finds a position for the stand. it♪rotate Turn the range [insert_to; it]moving the element to the position it location insert_to♪
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Relocation of elements from a symbolic mass to a two-dimensional mass
Software Programming • c++ c language liabilities char • • authera

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If you count the whitespace divider, that's to say "words" what you think by the specification of the format. %s Total scanfYou can use it. sscanf and develop a similar structure char *argv[]#include <assert.h> #include <stdio.h> #include <stdlib.h> #include <string.h> char **split_words(const char *s) { assert(s != NULL); size_t n_words = 0, n_words_mem = 0; size_t n_lead, n_word; for (const char *p = s; sscanf(p, " %zn%*s%zn", &n_lead, &n_word) == 0 && *(p + n_lead) != '\0'; p += n_word) { assert(n_lead < n_word); ++n_words; n_words_mem += n_word - n_lead + 1; } if (n_words == 0) return NULL; char **words = malloc((n_words + 1) * sizeof *words); assert(n_words_mem > 0); char *words_mem = malloc(n_words_mem); if (words == NULL || words_mem == NULL) { free(words); free(words_mem); return NULL; } size_t i_word = 0; char *word = words_mem; for (const char *p = s; sscanf(p, " %zn%s%zn", &n_lead, word, &n_word) == 1 && *(p + n_lead) != '\0'; p += n_word, ++i_word, word += n_word - n_lead + 1) { assert(i_word < n_words); assert(word < words_mem + n_words_mem); assert(n_lead < n_word && strlen(word) == n_word - n_lead); words[i_word] = word; } assert(i_word == n_words); assert(word == words_mem + n_words_mem); words[i_word] = NULL; return words; } int main(void) { char **words = split_words(" Mama myla raammu "); for (char **pword = words; *pword != NULL; ++pword) printf("'%s'\n", *pword); } Do you have a structure like this as a result of a two-storey mass, you're more visible. Because the details of your questioning aren't very rich.
Classes. Streamlining, addition. C+++
Software Programming • c++ classes • • Demir

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#include <iostream> #include <string> #include <cstring> #include <windows.h> #include <vector> //подключаем векторы using namespace std; class Book { string name; int k; vector<string> author; //поменяли string на vector<string> int year; int pages; public: Book() : name(""), k(0), author(vector<string>(1,"")), year(0), pages(0) {} //authot("") => author(vector<string>(1,"")) ~Book() {} void data(); void out(); friend void sort1(Book* a, int n); const char* getname(void) { return name.c_str(); // костыль, что бы не менять код sort1, возвращаем строку в виде const char* } }; void sort1(Book* a, int n) { for (int i = 0; i < n - 1; i++) for (int j = i + 1; j < n; j++) if (strcmp(a[i].getname(), a[j].getname()) > 0) { Book t = a[i]; a[i] = a[j]; a[j] = t; } } void Book::data() { setlocale(LC_ALL, "Russian"); cin.ignore(100, '\n'); cout << "Название книги: " << endl; getline(cin, name); //изменено cin>>name; cout << "Количество авторов: " << endl; cin >> k; cin.ignore(100, '\n'); cout << "Автор(ы): " << endl; string temp; int i = k; // while (i--!=0) { // getline(cin, temp); // построчно считываем авторов author.push_back(temp); // } // cout << "Год издания: " << endl; cin >> year; cout << "Количество страниц: " << endl; cin >> pages; } void Book::out() { cout << "Название книги: " << name << endl; cout << "Количество авторов: " << k << endl; cout << "Автор(ы): "; // for (auto elem : author) // выводим всех авторов одной строкой cout << elem << " "; // cout << endl; // cout << "Год издания: " << year << endl; cout << "Количество страниц: " << pages << endl; } int main() { SetConsoleOutputCP(1251); SetConsoleCP(1251); int n, i; cout << "Введите количество книг: "; cin >> n; Book* a = new Book[n]; for (i = 0; i < n; i++) { cout << "Книга " << i + 1 << endl; a[i].data(); } for (i = 0; i < n; i++) { a[i].out(); } sort1(a, n); cout << endl; cout << "Названия книг по алфавиту: " << endl; for (i = 0; i < n; i++) { a[i].out(); } return 0; }
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What's the CSS inherit and initial for?
Software Programming • c language • • emran

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InheritInherit simply means that the style will be inherited from the parent element.According to the https://www.w3.org/TR/css3-cascade/#inheriting the Inherit:Specifies that a property must inherit its value from its parent element. The word inherit can be used for any property CSS, and in any element HTML.Can read more about Inherit https://developer.mozilla.org/pt-BR/docs/Web/CSS/inherit .InitialInitial denotes the initial value of the property, as defined in the https://www.w3.org/TR/css3-values/#common-keywords :Represents the specified value which is designated as the initial value of the property. The initial word can be used for any property CSS, and in any element HTML.In short, Initial is used to set the property CSS as your default value.You can see an example of Initial https://developer.mozilla.org/pt-BR/docs/Web/CSS/initial .I formulated the answer based on me https://stackoverflow.com/questions/15564650/what-does-inherit-mean-in-css and https://stackoverflow.com/questions/18534561/what-is-use-of-initial-value-in-css SO issuesen, and I used free translation in some quotes.
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How is the simple numbers program working?
Software Programming • c language • • jules

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for (i = 3; isPrime && i * i <= n; i += 2) isPrime = n % i; There's an insidious algorithm of the species, "take all the odd numbers until we find the one that shares. n Or we don't get to sqrt(n)i=sqrt(n) i2=n() " approved by a small number of Silence Magic.From magic here, it seems only that in C, all non-nulled meanings are logical truth and zero, respectively, lies. And that's the cycle. for Not only to circumvent the mass.Nor should we forget the banal fact that the balance of separation is zero at the time and only when the divided aim is divided into a businessman.And to understand the logic and to act consistently, all this line can be rewritten by while:i = 3; while (isPrime && i*i <= n) { int div = n % i; if (div!=0) { isPrime = 1; // true } else { isPrime = 0; // false } i += 2; }
Someone can explain to me about the order of the departure of the following program
Software Programming • c++ c language • • Mystic

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I think it can help clarify things that in the messages you're printing for debugging include printing the value of this. For example:MiClase::MiClase(){ dato = 0; cout << "Soy el constructor por defecto " << this; cout << " Mi Dato es " << dato << '\n'; } And so with the rest of the functions that are implemented. The value of this is unique for each of the objects that are being created, which allows us to see in the messages about which objects these operations are acting. Thus, for example, the first message we would see when running the program would be:Soy el constructor comun 0x7fff785fee68 Mi dato es 10 Number 0x7fff785fee68 is the direction of the object being created. Somehow we can think of that number as a kind of "unique identity" of the object. When later we see that same number appear again in another message, we will know that it is acting on the same object.On the other hand, comparing with the code, we know that the first object to be created is m1so we can establish the association and say that the number 0x7fff785fee68 That's it. (in a sense) object m1.Execute the complete program and see what output it produces (I will stop the output after the function returns funcion3(), since that is the doubt that concerns us:Soy el constructor comun 0x7fff785fee68 Mi dato es 10 Soy el constructor de copia 0x7fff785fee60 Mi Dato es 10 Soy el constructor comun 0x7fff785fee50 Mi dato es 30 antes de la asignacion Soy el constructor de copia 0x7fff785fee40 Mi Dato es 10 Soy el operador de asignacion0x7fff785fee60 Soy el constructor de copia 0x7fff785fee48 Mi Dato es 10 Soy el destructor 0x7fff785fee48 Mi dato era 10 Soy el destructor 0x7fff785fee40 Mi dato era 10 despues de la asignacion Antes de invocar a funcion3 Soy el constructor de copia 0x7fff785fee28 Mi Dato es 30 Soy el constructor por defecto 0x7fff785fee30 Mi Dato es 0 Esta es la funcion 3 El dato del objeto local es 300 Soy el operador de asignacion0x7fff785fee60 Soy el constructor de copia 0x7fff785fee38 Mi Dato es 300 Soy el destructor 0x7fff785fee38 Mi dato era 300 Soy el destructor 0x7fff785fee30 Mi dato era 300 Soy el destructor 0x7fff785fee28 Mi dato era 30 Despues de invocar a funcion3 Now I'm going to rewrite that output by changing the mysterious numbers as 0x7fff785fee68 for the object that I know is actually. For example, we saw that 0x7fff785fee68 is the object m1. When that number comes back, I'll change it (m1) to make messages easier to read and follow.From time to time new numbers appear, which are neither m1 and m2 and m3. They are temporary objects that C++ is creating to make the passage of parameters, copies, assignments, etc. When there's a number of these I'll call you (tmp1) or (tmp2)as they are being created.The result of "translation" in this way is as follows:Soy el constructor comun (m1) Mi dato es 10 Soy el constructor de copia (m2) Mi Dato es 10 Soy el constructor comun (m3) Mi dato es 30 antes de la asignacion Soy el constructor de copia (tmp1) Mi Dato es 10 Soy el operador de asignacion(m2) Soy el constructor de copia (tmp2) Mi Dato es 10 Soy el destructor (tmp2) Mi dato era 10 Soy el destructor (tmp1) Mi dato era 10 despues de la asignacion Antes de invocar a funcion3 Soy el constructor de copia (m) Mi Dato es 30 Soy el constructor por defecto (local) Mi Dato es 0 Esta es la funcion 3 El dato del objeto local es 300 Soy el operador de asignacion(m2) Soy el constructor de copia (tmp3) Mi Dato es 300 Soy el destructor (tmp3) Mi dato era 300 Soy el destructor (local) Mi dato era 300 Soy el destructor (m) Mi dato era 30 Despues de invocar a funcion3 Let's go study this way.Initial creation of objectsSoy el constructor comun (m1) Mi dato es 10 Soy el constructor de copia (m2) Mi Dato es 10 Soy el constructor comun (m3) Mi dato es 30 There are no mysteries here, every object is created in its own direction.Allocation m2=m1Here is one thing that may seem surprising:Soy el constructor de copia (tmp1) Mi Dato es 10 Soy el operador de asignacion(m2) Soy el constructor de copia (tmp2) Mi Dato es 10 Soy el destructor (tmp2) Mi dato era 10 Soy el destructor (tmp1) Mi dato era 10 We see that in the first place a copy of m1 a temporary object (tmp1), and it is that object that is passed to the assignment operator. The assignment operator seems to create an additional copy internally (tmp2) used to copy the data hypothetically in m2. Then the two temporary objects are destroyed.There. Two temporary copies and that's why the copy operator is called twice. These two copies occur due to the implementation you have made of the assignment operator, which is basically:MiClase MiClase::operator=(MiClase m){ dato = m.dato; return *this; } We see that the function receives m and upon return *thisThe return is also worthwhile. So it is necessary to invoke the copy operator to copy the argument m2 in the parameter m and again the copy operator for the returned value *this (with which nothing is done afterwards). Somehow (so I interpret it), the assignment has created a temporary object "to the right of the equal sign" to copy m1 in it and another "left of the equal sign" to pass it to the copy builder m2.So what I called (tmp1) It's actually m and what I called (tmp2) is the value returned by the operator =.Allocation m2=m1.funcion3(m3)Here the message sequence tells another story:Soy el constructor de copia (m) Mi Dato es 30 Soy el constructor por defecto (local) Mi Dato es 0 Esta es la funcion 3 El dato del objeto local es 300 Soy el operador de asignacion(m2) Soy el constructor de copia (tmp3) Mi Dato es 300 Soy el destructor (tmp3) Mi dato era 300 Soy el destructor (local) Mi dato era 300 Soy el destructor (m) Mi dato era 30 The thing starts as it can be expected, creating a new object I've called (m) because it is the formal parameter of the function funcion3(). Then you create another one that has to be the local variable. But when we go to the assignment we see that in this case only a temporary object is created. This object corresponds to return *thisbut missing the copy builder that would be necessary to pass the parameter to operator=(m).ConclusionSo it is not that the messages come out in reverse order as you expected, but one of those you expected is not. You expected this:[..omitido..] Esta es la funcion 3 El dato del objeto local es 300 Soy el constructor de copia Mi Dato es 300 <----- ESTE EN REALIDAD NO SALE Soy el operador de asignacion Soy el constructor de copia Mi Dato es 300 <----- ESTE SI The first would be to pass the value local returned by funcion3() as parameter to operator=()and that not produced. The rest is normal.Why doesn't that copy occur in this case? This is definitely a compiler optimization, but I can't give you more details about it because it depends on how it is implemented.
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scanf reading the variable twice!
Software Programming • c language sc languageanf • • juvenalb

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Luis,The problem is that space you left in scanf:scanf("%d ", &valor); Remove the same, leaving as follows:scanf("%d", &valor); See online: https://repl.it/repls/EasygoingBothDemand
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Streamline lock-free
Software Programming • c++ algorithm list winapi freelock • • obi

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What's the fundamental difference between AVL and red black? I think it doesn't matter.The result may be one, but it's easier for me to clearly record the difference in the height of the support. In my view, this approach is the way in which the balance sheet algorithm is maximized.It's interesting. When can lock-free algorithms be applied to sufficiently complex structures?Pretty complicated, I'm not sure.Only in cases where simultaneous access is unlikely.That's what I'm interested in right now.Available N flows and one parallel task. The challenge is solved by its irration. Each iteration shall be carried out at approximately fixed times (regulation of the length of the irration [discretion of the task] shall be made from outside, static). Every flow, it's a waste. And since, initially, one task should therefore be divided into the remaining flows. May each flow that has not been tasked with the maximum resource and divide it into two. If the flow has chosen its irration, it seeks a new free task or tries to divide the maximum if the number of tasks is less than the number of flows. The task is considered to be available for division if it is above the specified limit. Thus, as long as the task is divided, flows will always be dealt with.That's the algorithm I'm trying to do without locking. The list is a list of free tasks. Challenges that have not yet taken on any flow. The free task can either be divided or one of its determinants. Naturally, the number of tasks should not exceed the number of flows, but there may be little logical exceptions. Since the number of flows is equal to the number of kernels, the list is small enough to be very frustrated in the grading, but it may be necessary to think about the optimization of the grading if there is greater competition between flows.Theoretically, I can count the time spent on competition between interlock flow, because I can count the number of idle cycles and I can figure out how many tacts the cycles are running.That is, ultimately, it will be a percentage of the time spent on the combined lock-free synchronization.I can't just count the time spent on the synchronization of standard tools (critical sections, me). If it's obvious that tact measurements are made in front of these locks, it'll be very inaccurate. It seems that the most logical way is to measure the time of the task on both modes of synchronization and to determine whether lock-free wins in this case. But such experiments, if I do, are not short of time.Update: To be honest with you, SMP I never managed to stabilize this algorithm.But after a long and hard study. lock-free concluded that it was pure (i)inlinesuch methods are incapable. For further experiments, they need them. display barrier as at the competitor level _ReadBarrier/_WriteBarrier/_ReadWriteBarrierand at the equipment level _mm_lfence/_mm_sfence/_mm_mfence♪I would be very grateful if there were any working examples with that organization of the set of tasks, since I still have a very specific expectation. lock-free:I've started a test task, "having out the set of points with a rectangular window." Each of these alternatives considers the time spent on the solution and accumulates it in the battery of the nucleus (the fixed flow). After completion of the calculations, the key parameter is calculated, the number of points per second equals S / Twhere T - is the total time of all the irrations at the slowest kernel (the flow fixed on the nucleus) and S - total number of points processed. In the case of a lock-free This characteristic once deteriorated to 300 million in sections on 8,3GHz yards, and I'd like to keep that speed in return for 20-50 million blocks.I mean, "depressed" is exactly the word, because I've set this target 100 times, and every subsequent irration has received an increase in speed.I never managed to stabilize this algorithm.About a month ago, I've been doing this. In the 3rd option: http://msdn.microsoft.com/ru-ru/library/windows/desktop/ms683648(v=vs.85).aspx on dynamics (when the number of tasks is unpredictablely increasing) and on static glazing (when the maximum number of tasks is known in advance or a fixed set of tasks is available). Productivity API Lower, as we spend time challenging the function and we have no opportunity to sort the summit.Static glazing is the highest. Although it has been implemented lock-free-стека free SIMD- I didn't have to. I'm not gonna say the exact numbers right now, but... my preliminary assessment The use of this sorting is ~3-6% Gains against the basic algorithm lock-free♪ ~20-30% - I didn't even think about "pine," and "nuclear" synchronizations. lock-free)I will describe in greater detail the implementation in a single step, a little later when the time for a simple test will be taken.
How do you streamline the program?
Software Programming • c sharp • • Marcee

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6,000 images in the database are stored in full. But in dataGridView1, they're small, but they're all loaded. So load the image not entirely, but the reduced copies and show it in dataGridView1.
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How do we streamline the code? JS
Software Programming • javascript javascript events • • esther

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I would do that.function refreshAll(){ refresh('.fadeUp'); refresh('.slideDown'); refresh('.run-text'); } window.onload = function(){ setTimeout(function () { refreshAll(); window.onresize = refreshAll; window.onscroll = refreshAll; }, 1500); } But it's not the optimization, it's the code refacturing. Optimization implies a target function (time, memory). And refacturing means simplifying the unification of the code for further expansion and use
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Is there any plugin or CSS code generator for Shapes?
Software Programming • c language • • shizuka

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Paul you can use http://www.samuelrossille.com/css-shape/ , it is in beta yet but it is very useful and seems to be what you need.
Help me streamline the code. Python
Software Programming • python python 3.x optimization math • • Bogopo

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I moved the code. https://ru.stackoverflow.com/users/176262/igor With JS on Python on this issue, you might be able to: https://ru.stackoverflow.com/questions/1152154/%D0%9F%D0%BE%D0%BC%D0%BE%D0%B3%D0%B8%D1%82%D0%B5-%D1%80%D0%B5%D1%88%D0%B8%D1%82%D1%8C-%D0%B7%D0%B0%D0%B4%D0%B0%D1%87%D1%83-%D0%9F%D1%80%D1%8F%D0%BC%D0%BE%D1%83%D0%B3%D0%BE%D0%BB%D1%8C%D0%BD%D0%B8%D0%BA%D0%B8/1152589#1152589 This is the code.def maxArea(rs): a = 0 for i in range(len(rs)): r = rs[i] ai = r[0] * r[1] + back(rs, r[1], i) + forward(rs, r[1], i) a = max(a, ai) return a def back(rs, h, idx): a = 0 for j in range(idx-1, -1, -1): if rs[j][0] >= h: a += rs[j][0] * h else: break return a def forward(rs, h, idx): a = 0 for j in range(idx+1, len(rs)): if rs[j][1] >= h: a += rs[j][0] * h else: break return a rs1 = [[4, 3], [2, 1], [2, 5]] rs2 = [[4, 3], [2, 1], [3, 5]] print(maxArea(rs1)) print(maxArea(rs2))

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Software Programming • c sharp • • Marcee

How do we streamline the code? JS
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Is there any plugin or CSS code generator for Shapes?
Software Programming • c language • • shizuka

Help me streamline the code. Python
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