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Dun radius of circumference R. Write a long-term program... I don't know what a mistake is.

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meriah last edited by
I'm going to stepic, there's a mission: Dun radius of circumference R. Write a program for the length and area of the area. Consider that the radius is a substance number and P = 3.14. The answer shall be accurate to 2 figures after the comma.
He wrote the following code:
```
#include <iostream>
#include <cmath>
using namespace std;
int main()

{

float R, pi=3.14, length, s;

cin >> R;

length = (pi*2)R;

R = pow(R, 2);

s = (pi)(R);
cout &lt;&lt; length &lt;&lt; " " &lt;&lt; s;

return 0;

}
```
But the test doesn't go through. What's the problem with the code?
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N
Nykeriab last edited by
The step is clearly written:
The accuracy is as follows:
cout << fixed << setprecision(i) << x
where i is the number of figures after the comma for expression x.
And the code.
```
cout << fixed << setprecision(2) << length << " " << s;
```
It's quite a test. ♪ ♪
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The logs don't show up if the test has been successful, moreover, INFO not shown if there was a mistake in another scope. https://github.com/catchorg/Catch2/blob/master/docs/logging.md#logging-macros Use it. UNSCOPED_INFO♪
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The case was still in a large amount of the Listener log file.The reason in Listener indicated that the connection had been set for a very long time and had worked tnsping for a very long time (sometimes 20 seconds). Once the connection was installed, the requests were carried out quickly.The method of abundance of symptoms revealed that this behaviour was most commonly caused by a large amount of log files.The file itself was found to be just aset of all the files with an extension of .log, he had a name listener.log and there were more than 4GBs.The file has been redesigned. Disconnected to listener for time: added LOGGING_LISTENER = OFF in the file listener.ora and reset Listener.
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That's it.if (hwnd = CreateWindow(wc.lpszClassName, L"Заголовок!", WS_OVERLAPPEDWINDOW, 0, 0, 600, 600, nullptr, nullptr, wc.hInstance, nullptr); hwnd == INVALID_HANDLE_VALUE) Replace:if ((hwnd = CreateWindow(wc.lpszClassName, L"Заголовок!", WS_OVERLAPPEDWINDOW, 0, 0, 600, 600, nullptr, nullptr, wc.hInstance, nullptr)) == INVALID_HANDLE_VALUE) You can still keep it as it is, but in the designs of the project, you can set a standard C+17.
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There's no list out, I think it's an addendum problem, in the cycle while
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You need to move one line to push_back.It was.while (current->next!=NULL) { current = current->next; current->next = uzel; } Stalowhile (current->next!=NULL) { current = current->next; } current->next = uzel;
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Mistake: NameError: name 'R' is not defined
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The problem this code is:def getNumRul(self): '''получение числа рулонов оболев для оклейки комнаты новой площади''' return print('You need ',R.getNewSqr()/Ob.getSqrOboi(),' rulonov of oboi') It's a class method. Roomand he doesn't know anything about a copy of this class announced in another module. He also doesn't know about the variable. Ob♪ If R can be replaced by selfthe creation of the facility Oboi It will be necessary to make this method (or even get rid of the object, turn it into a function).Yeah. return There's absolutely no need.
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https://stackoverflow.com/questions/31987331/silence-the-warning-non-conforming-tab-character-in-gfortran The tabular character is not part of the fortran standard.The compilation with -Wall supposes the following:-Wall.Enables commonly used warning options pertaining to usage that we recommend avoiding and that we believe are easy to avoid. This currently includes -Waliasing, -Wampersand, -Wconversion, -Wsurprising, -Wc-binding-type, -Wintrinsics-std, -Wno-tabs-Wintrinsic-shadow, -Wline-truncation, -Wtarget-lifetime, -Wreal-q-constant and -Wunused.The flag highlighted in black implies that the compiler does not allow to use tabulators in codification.To silence this error try to compile with -WtabsLet's go with the mistakes: integer :: i,j,suma integer, parameter :: N = 4, M = 20 real:: A(i,j), p(j),u(j) What dimensions will the arrays have A, p and u? The arrays should have known dimensions and certainly cannot come from variables except if they are declared as parameter: real:: A(4,20), p(20),u(20) real:: A(N,M), p(M),u(M) Another problem with arrays: u(3)= (/3.0,1.0,0.0/) If right on the previous line you have declared u Like a vector... Why are you trying to assign 3 values to one of the vector positions? That's not gonna work. You have to initialize. u with the game of values (without using indices): real:: u(3) u= (/3.0,1.0,0.0/) Another detail:suma=suma+(A(i,j)*u(j))= p(j) This instruction is wrong... You intend to assign to p(j) value suma? Otherwise? Is that the last assignment an operator of sum, record,...? It's complicated to put code at this point because I don't understand what you're trying to do with it. p(j) at this point.New mistakes:You're redeclaring ureal:: A(N,M), p(M),u(M) real:: u(3) The sum is still wrong.suma=suma+(A(i,j)*u(j))= p(j) If what you want is to assign the value of suma a p(j) Then you must do something like that:suma=suma+(A(i,j)*u(j)) p(j) = suma
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Recurring function - C+
Software Programming • c++ competition • • Burma

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Go to the recurring functions like, "How to accomplish this task, solve the task of smaller size"? And for all functions, with the requirement that the function be performed One logical action. Your function is looking for, and the answer is completely different.So there's a mass, its length, the meaning of the key. How are we looking for an element? Let's go to the end. If the last element of the mass is what we need, we'll get it back. If we don't, we'll find one less element in the mass:int find(int* array, int size, int key) { if (array[size-1] == key) return size-1; // Искомый элемент найден // Иначе ищем его среди `size-1` элементов return find(array,size-1,key); } Everyone? No, because we still have to decide what to do if the element is not found (d) if our function is to cause itself all the time, it'll work. endless) Since the indices are less than 0, let's go back -1 as a sign of absence. When do we know there's no element? In search of an empty space. So we need to add a test first.if (size == 0) return -1; // не найден So, finally,int find(int* array, int size, int key) { if (size == 0) return -1; // Не найден if (array[size-1] == key) return size-1; // Искомый элемент найден // Иначе ищем его среди `size-1` элементов return find(array,size-1,key); } That's it. The withdrawal is now organized in another function, for example,void find_message(int* array, int index) { if (index < 0) { cout << "Key not found\n"; return; } cout << "Key " << array[index] << : was found in position " << index << endl; } Well, main() Or where. ♪ ♪int array[SIZE] = ... ... int idx = find(array,SIZE,key); find_message(array,idx);
How do you put a function in the C++ code so it can be called from the C code?
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If the same headline file for C and C+++ is to be used, the definitions of functions can be removed in such a way.#ifdef __cplusplus extern "C" { #endif void doFirst(); void doSecond(); #ifdef __cplusplus } #endif It works because macroes. __cplusplus only in C++, but not in C.
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Virtually all the terms in programming are a professional fever plus an irony. In order to understand the terms, it is sufficient to have a vocabulary base in English, which is achieved by the practice of reading English-speaking texts in specialty.The work of the modern software player is closely related to the study of documentation, so you're just doing some sort of task. скилл (sighs)You get English.and related skills.In some cases, there has been a clash between Russian-speaking terms and English-speakers, such as:thread - flow - threadapplication - annex - appendixThe first column is an English-language term, the second is is a Russian term, the third is English, the English translation.The third column came from activities копирайтеров (sighs)Here's another Englishman.(d) Free from vocational education IT (sighs)You got it.that translated the special terms "to the forehead." In fact, the use of the terms in the second column and the lack of substantive education in Russia. This may be the expression of English speakers, for example, who have learned a little Russian, and this is Not quite correct.♪The very task of studying the terms makes sense if you're a linguist and a professional fever. For the programmer, the examination of terms is automatic in the process of professional communication with colleagues. Yeah, reading articles and documentation is also a communication, just a distance.
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Write a program to read 3 whole values (consider that equal values will not be read) and write them in increasing order. in JS
Software Programming • javascript • • jonetta

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The question is:Do you really know if your algorithm works for every possible case?By making a purely comparative sorting algorithm the first thing that should consideration is the amount of combinations that can be generated by the amount of elements in the input.To discover this number of combinations just calculate the factorial of the number of element that will be ordered, as are three element to be ordered then will be 3! = 6 different combinations that the input may be willing:smaller, medium, largersmaller, larger, mediummeans, smaller, largermeans, larger, smallerbigger, smaller, halfbigger, medium, smallerThe second thing you should consider is if your algorithm works in all cases.For testing purposes I built an array with numbers 1, 2 and 3 that will be used to test all possibilities of the algorithm:teste = [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]; Then I encapsulated your algorithm in a function to facilitate testing:function ordenar(entrada){ let maior, menor, meio; let n1 = entrada[0]; let n2 = entrada[1]; let n3 = entrada[2]; if ((n1 > n2) && (n1 > n3)) { maior = n1; if (n2 < n3) { meio = n3 menor = n2 } else { meio = n2 menor = n3 } } else if ((n2 > n1) && (n2 > n3)) { maior = n2; if (n1 < n3) { meio = n3 menor = n1 } else { meio = n1 menor = n3 } } return [menor,meio,maior] } It is the same algorithm of the question only that instead of receiving the entrance by the console it will receive at once one array of three elements with input, simulating the input by the three-number console.Then I created a test function that iterates about the different possible non-repeated combinations formed by numbers 1, 2 and 3://Teste de sucesso do algorítimo de ordenação. teste.forEach((caso) =>{ //Obtem o resultado do algorítimo de ordenação para o caso a ser analisado. let resultado = ordenar(caso); console.log(------------------------------); console.log(entrada: ${caso}); console.log(resultado: ${resultado}); //Faz uma comparação superficial entre o resultado esperado e o resultado console.log(teste ${JSON.stringify([1,2,3])==JSON.stringify(resultado)? 'APROVADO': 'REPROVADO'}); }); So I did the test here. https://repl.it/repls/LoneSoulfulMonad Resulting:------------------------------ entrada: 1,2,3 resultado: ,, teste REPROVADO entrada: 1,3,2 resultado: 1,2,3 teste APROVADO entrada: 2,1,3 resultado: ,, teste REPROVADO entrada: 2,3,1 resultado: 1,2,3 teste APROVADO entrada: 3,1,2 resultado: 1,2,3 teste APROVADO entrada: 3,2,1 resultado: 1,2,3 teste APROVADO How can your algorithm be seen failed in two cases:------------------------------ entrada: 1,2,3 resultado: ,, teste REPROVADO entrada: 2,1,3 resultado: ,, teste REPROVADO The first case is the simplest to solve, is when the entry is already ordered 1,2,3 just assign the variables menor, meio and maior the input values as set out://Solução para o caso da entrada já vier ordenada. let menor = n1 = entrada[0]; let meio = n2 = entrada[1]; let maior = n3 = entrada[2]; The second case in which the algorithmic failure is 2,1,3. Failure because the test condition for this entry is not present in your algorithm that would be (n1 > n2) && (n1 < n3). Making the changes your algorithm:function ordenar(entrada){ //Caso da entrada já vier ordenada. let menor = n1 = entrada[0]; let meio = n2 = entrada[1]; let maior = n3 = entrada[2]; if (n1 > n2){ if (n1 > n3) { maior = n1; if (n2 < n3) { meio = n3 menor = n2 } else { //meio = n2; <-- Se tornou redundante menor = n3 } } else { //caso (n1 < n3) menor = n2 meio = n1 } } else if ((n1 < n2) && (n2 > n3)) { maior = n2; if (n1 < n3) { meio = n3 //menor = n1 <-- Se tornou redundante } else { meio = n1 menor = n3 } } return [menor,meio,maior] } So I did this test: https://repl.it/repls/VisibleLemonchiffonForms Which resulted:------------------------------ entrada: 1,2,3 resultado: 1,2,3 teste APROVADO entrada: 1,3,2 resultado: 1,2,3 teste APROVADO entrada: 2,1,3 resultado: 1,2,3 teste APROVADO entrada: 2,3,1 resultado: 1,2,3 teste APROVADO entrada: 3,1,2 resultado: 1,2,3 teste APROVADO entrada: 3,2,1 resultado: 1,2,3 teste APROVADO Honestly I think that according to the given parameters you searched the most difficult path.The simplest way would be to declare an array to receive the entries and apply the method https://developer.mozilla.org/pt-BR/docs/Web/JavaScript/Reference/Global_Objects/Array/sort .var n= new Array(3); rs = require("readline-sync") n[0] = rs.questionInt("Digite o primeiro valor ") n[1] = rs.questionInt("Digite o segundo valor ") n[2] = rs.questionInt("Digite o terceiro valor ") n.sort(); console.log(n); If you do not want to use method sort() can use https://developer.mozilla.org/pt-BR/docs/Web/JavaScript/Reference/Operators/Atribuicao_via_desestruturacao to permute the array elements while testing them:var n= new Array(3); rs = require("readline-sync") n[0] = rs.questionInt("Digite o primeiro valor ") n[1] = rs.questionInt("Digite o segundo valor ") n[2] = rs.questionInt("Digite o terceiro valor ") if (n[0] > n[1]) [n[0],n[1]] = [n[1],n[0]]; if (n[1] > n[2]) [n[1],n[2]] = [n[2],n[1]]; if (n[0] > n[1]) [n[0],n[1]] = [n[1],n[0]]; console.log(n); The tests of these last two algorithmic: https://repl.it/repls/CompetentAdmiredEditors If you are curious about more elaborate ordination methods https://pt.wikipedia.org/wiki/Algoritmo_de_ordena%C3%A7%C3%A3o and search here on the site the names of the algorithms contained in this list.
An algorithm of the radius of the circumference into which the text can be written
Software Programming • algorithm • • jeanid

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I think it's easy. Knowing the height of the line and its serial number, we can calculate the required width (short of circle): ширина = корень из (высота*номер)*(2*радиус -высота*номер)♪ (I may be wrong, but I think you're capable of computing the choir length on your own)The challenge ahead is to write the text of this wide. If you don't know the radius in advance, you can count it on the basis that you need to write the area of the text into the circle: радиус = корень из (общая ширина * высота строки / пи) Add the tolerances, taking into account that part of the words will have to be moved to the next line so that everything fits. The text should be equated on both sides.
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How do you know the way to my files that lie next to the .pro?
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Use macro. __FILE__ You'll get the way to the file where you wrote this macroes in the form of a stable blue line, and then you can just get the substructure and get the way to the directory.
GetFocus on wxStringProperty
Software Programming • c++ wxwidgets • • briley

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Used SelectProperty()
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Don’t know the benefits of using DI containers
Software Programming • c sharp test • • derk

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I think it is not in history, but I thought as follows.There is a huge programCut it into components. But I can't change.インターフェイス the interface. You need to change the inside.Inject dependencies. Instead, object generation becomes complicated, and the generated logic is distributed.オブジェクトrate objects to DI containers using pre-registered generation rulesIt is easy to understand even small-scale code until dependency injection, and it is convenient for actual testing. On the other hand, it is difficult to feel that the same generation logic is dispersed in the code that is shown in the question, with few dependencies on each object, and the benefits of using the container may be thin.But IPerson Home PersonManager What if the number of places you need to generate? If you want to change it as new? In addition to simply different elephant classes, there may be different ways to generate. or you may want to use an existing instance.If you use the DI container, you just need to register the first pattern of "this can be generated" and the container will be combined properly. The B dependent on A is also dependent on C.In addition, kernel.Get<T>() That’s why you can create an object with a consistent procedure that can also lead to code readability.In addition, DI injects objects from outside, but there are also methods such as factory patterns and service locales only to avoid directly depending on the specific class.Reference https://github.com/ninject/Ninject/wiki/Dependency-Injection-By-Hand https://ja.wikipedia.org/wiki/%E4%BE%9D%E5%AD%98%E6%80%A7%E3%81%AE%E6%B3%A8%E5%85%A5 http://phpmentors.jp/post/100741112503/di-service-locator
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I can't understand why the cycle for doesn't take the last word of the line.
Software Programming • c++ lines • • gionna

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Because the last symbol of the line may not be a gap, if (s[i] == ' ') It won't work on the last word.If you're so eager to pick words,int main() { string s, n; if (getline(cin, s)) { for (int i = 0; i < s.size(); i++) { if (s[i] == ' ') { if (n.size()) { cout << n << endl; n.clear(); } } else n += s[i]; } if (n.size()) cout << n << endl; } } But you can be shorter...string s; if (getline(cin, s)) { istringstream ss(s); for(string n; ss >> n;) cout << n << endl; }

Dun radius of circumference R. Write a long-term program... I don't know what a mistake is.

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