The simplest way is to add a variable with an incumbency on every teration and check its conformity. Count container. However, in this case the cycle foreach actually turns into a normal one. for♪However, foreach is trapped on all elements and on the same footing, regardless of the element ' s position. If you have a question like this, then something's wrong with design. Better use the usual. for♪

As an option, we sort the collections and compare them:Collections.sort(list); Collections.sort(list001); System.out.println(list.equals(list001)); Help.

SaveBidAndOfferHandler a = new SaveBidAndOfferHandler(...); double Bigprice = a.BigBid;

If I understand correctly, this code has to do what you need://компаратор для сравнения объектов по полю description Comparator<Item> comparatorByDesc = (e1, e2) -> Comparator.<String>nullsFirst(Comparator.naturalOrder()).compare(e1.description, e2.description); //основной компаратор Comparator<Item> generalComparator = (e1, e2) -> { if (e1 == null || e2 == null) return Comparator.nullsLast(comparatorByDesc).compare(e1, e2); else return comparatorByDesc.compare(e1, e2); }; itemList.sort(generalComparator);

You have a problem with every flow reversing the whole set of lines. We need the overtake to work independently in every flow. This is the best use of the mass enumerator ReadFile♪ I mean, before the flux starts. var iterator = ReadFile.GetEnumerator() and transfer it as a parameter to function Scanner (or to preserve the style, declare globally). Internal functions Scanner instead foreach sequentially select the elements from the heterator. The terator is a shared resource, so the use of its methods should be protected by a operator lock:string nextString; while (true) { lock (iterator) { if (!iterator.MoveNext()) break; //если строки кончились - выходим из цикла nextString = iterator.Current; } doSomethingWith(nextString); }

Because the cycle is carried out:for ( i = 0; i < 10001; ) { if (is_prime(answer)) i++; answer += 2; } Or so:while( i < 10001 ) { if( is_prime( answer ) ) i++; answer += 2; } I mean, the way out of it will happen after it's done. answer Another one.As an option (not only):while( 1 ) { if( is_prime( answer ) ) { if( ++i == 10001 ) { break; } } answer += 2; }

B C+++ for floating cloths, no residual operator %♪Therefore, the condition in this proposal ifif (float(sqrt(b)) % 1 == 0) { wrong.I'll write the next cycle.unsigned int a; std::cin >> a; for ( unsigned int i = 1, b; ( b = i * i ) <= a; i++ ) std::cout << b << ' '; std::cout << std::endl; There is no need to include a heading cmath and use its declared function sqrt♪The programme can be made more meaningful. For example,#include <iostream> int main() { while ( true ) { std::cout << "Enter a non-negative number (0 - exit): "; unsigned int a; if ( not ( std::cin &gt;&gt; a ) || a == 0 ) break; for ( unsigned int i = 1, b; ( b = i * i ) &lt;= a; i++ ) std::cout &lt;&lt; b &lt;&lt; ' '; std::cout &lt;&lt; std::endl; } return 0; } Dialogue with the programme might look like that.Enter a non-negative number (0 - exit): 100 1 4 9 16 25 36 49 64 81 100 Enter a non-negative number (0 - exit): 50 1 4 9 16 25 36 49 Enter a non-negative number (0 - exit): 25 1 4 9 16 25 Enter a non-negative number (0 - exit): 0

I understand that your mission to work with masses and cycles and to standard containers and algorithms has not yet been accomplished.The programme may look as follows. It's only in the demonstration program that I don't introduce the elements of the keyboard, but generate their accidental values. You're going to have to change the program so that the user can give the size of the mass itself so that the mass can be dynamically released and then the user introduces the elements of the mass. Don't forget, at the end of the program, to remove this mass from the operator. delete []#include <iostream> #include <cstdlib> #include <ctime> int main() { const size_t N = 10; int a[N]; std::srand( ( unsigned int )std::time( nullptr ) ); for ( auto &amp;x : a ) x = std::rand() % N; for ( auto x : a ) std::cout &lt;&lt; x &lt;&lt; ' '; std::cout &lt;&lt; std::endl; for ( size_t i = 0; i &lt; N; i++ ) { size_t j = 0; while ( j != i &amp;&amp; a[j] != a[i] ) j++; if ( j++ == i ) { while ( j &lt; N &amp;&amp; a[j] != a[i] ) j++; if ( j != N ) std::cout &lt;&lt; a[i] &lt;&lt; ' '; } } std::cout &lt;&lt; std::endl; return 0; } The withdrawal of the programme this might look like.1 2 9 5 0 5 2 4 1 8 1 2 5 Working principle followed. If there's another element of the mass with the index iwe're looking at his meaning earlier, considering the elements in the range. [0, i)♪ If we've already had a case with him, we'll just miss him. Otherwise, let's see if it's in the range. [i+1, N) an additional element with the same value. If so, we'll get him to the console.If you're allowed to change the sequence of cells in the mass after they're injected, you can sort the mass using standard algorithms. std::sort as follows:std::sort( a, a + N ); insert a title#include <algorithm> and then sequentialize elements that coincide with a number of standing elements. For example#include <iostream> #include <cstdlib> #include <ctime> #include <algorithm> int main() { const size_t N = 10; int a[N]; std::srand( ( unsigned int )std::time( nullptr ) ); for ( auto &amp;x : a ) x = std::rand() % N; for ( auto x : a ) std::cout &lt;&lt; x &lt;&lt; ' '; std::cout &lt;&lt; std::endl; std::sort( a, a + N ); for ( size_t i = 0; i &lt; N; ) { size_t j = i + 1; while ( j &lt; N &amp;&amp; a[i] == a[j] ) j++; if ( j - i &gt; 1 ) { std::cout &lt;&lt; a[i] &lt;&lt; ' '; } i = j; } std::cout &lt;&lt; std::endl; return 0; } As for your code, at least this cycle.for (int i = 0; i < N; i++) { if (p[i] == p[i + 1]||p[i]==p[i+i]) { mas[i] = p[i]; } } leads to the uncertain conduct of the programme, as provided if (p[i] == p[i + 1]||p[i]==p[i+i]) which, in addition, makes no sense to cross the border of the area addressed by the indicator pWhen i equal N-1and not all elements of the body addressed by the indicator masvalues to be assigned. And this mass is not originally initialized, but therefore contains uncertain values. ♪

(1) You're moving all the elements of the mass in the cycle, but you're looking for "2" Only in the fourth element: strpbrk(a[3].mark,c) :-(2) Why? strpbrk()?(3) Why? c[] creates and initializes on every terrain, and not a constant at all?const char c[] = "2"; /* ... */ for (size_t i = 0; i < sizeof(a)/sizeof(a[0]); i++) { /* а если ищем только "2", то можно ещё проще: const char c = '2'; ... !strchr(a[i].mark, c) */ if ( !strstr(a[i].mark, c) ) { cout << a[i].fio << "\n\n"; } }

var arr = []; for (i = 0; i < 2; i++) { arr[i] = prompt('enter something', ''); } alert(arr);

int[] a = new int[] {1, 1, 1, 2, 1, 3, 3}; int[] b = new int[a.length]; int curIndex = 0; for (int i: a) { if (!IntStream.range(0, curIndex).anyMatch(num - > b[num] == i)) { b[curIndex++] = i; } } System.out.println(Arrays.toString(Arrays.copyOf(b, curIndex))); //[1, 2, 3] UPDATE!IntStream.range(0, b.length).anyMatch(num - > b[num] == i) to be replaced bypublic static boolean contains(int[] a, int num) { for (int i in a) { if (i == num) { return true; } } return false; } Then the main for the nextfor (int i: a) { if (!contains(b, i))) { b[curIndex++] = i; } }

This is a working option that reads Excel the file only once:import pandas as pd fn = r'C:\Temp.data\primer.xlsx' cols = ['label','x','y','z','value'] df = pd.read_excel(fn, skiprows=4, header=None, parse_cols='C:XFD') dfs = [] for i in range(df.columns.size//5): lbl_col = 5*i x = df.ix[(df[lbl_col] != 0) & (df[lbl_col] != 9999), lbl_col:lbl_col+4] #x.columns = pd.MultiIndex.from_tuples(list(product([i+1], cols))) x.columns = cols dfs.append(x.reset_index(drop=True)) result = pd.concat(dfs, axis=1) #result.to_excel('c:/temp/result.xlsx', startrow=3, index=True) result.to_excel('c:/temp/result.xlsx', startrow=3, index=False) Result: PS I wanted to add the numbers of the five-string blocks (as in your original file: 1-19), i.e. to create multilevel columns: x.columns = pd.MultiIndex.from_tuples(list(product([i+1], cols))) but this has not yet been implemented:NotImplementedError: Writing to Excel with MultiIndex columns and no index ('index'=False) is not yet implemented. That is, it is necessary to select, either multi-level (red) names of columns and unnecessary indexes or no index and single-level names of columns. ♪ ♪PPS if you need a hat, you can do it. https://stackoverflow.com/a/38075046/5741205

Let's see the next program.#include <iostream> int main() { int done = 5; int howmany = 3; for ( done = 1; done &lt; howmany; done++ ) { std::cout &lt;&lt; done &lt;&lt; std::endl; } std::cout &lt;&lt; done &lt;&lt; std::endl; return 0; } Her withdrawal to the console 1 2 3 In this program, in the functional code main declared variable done♪ This variable shall first be initiated by value 5int done = 5; and then in the proposal for value 1for ( done = 1; done < howmany; done++ ) ^^^^^^^^ After the cycle, this variable will be value 3.Now we'll consider the next program.#include <iostream> int main() { int done = 5; int howmany = 3; for ( int done = 1; done &lt; howmany; done++ ) { std::cout &lt;&lt; done &lt;&lt; std::endl; } std::cout &lt;&lt; done &lt;&lt; std::endl; return 0; } Her withdrawal to the console1 2 5 In this programme, the proposal for Name variable donematching the name of the variable declared in the code block main♪ for ( int done = 1; done < howmany; done++ ) ^^^^^^^^^^^^ This variable in the proposal for conceals a uniform variable with the same name declared in the outer code unit. Her area of definition is the body of the cycle. for♪ Once the cycle is out, the variable ceases to exist.From C+++ (6.5.3 The for statement)3 If the for-init-statement is a declaration, the scope of the name(s) declared extends to the end of the for statement.Therefore, the last line of programme withdrawal5 it's already turning the variable into a console. donewhich was declared before the cycle in the functional code unit mainIn general, the name declared in the internal field of the announcement conceals the same name as that declared in the foreign field of the declaration.From standard C++ (3.3.10 Name hiding)1 A name can be hidden by an explicit declaration of that same name in a nested declarative region or derived class (10.2).Note that there is an important difference in the definition of the proposal for between C++ and C.B C+++ in the second part of the proposal forwhere the condition is recorded, an announcement may also be made.Look at the next demonstration program.#include <iostream> int main() { int done = 5; int howmany = 3; for ( ; int howmany = done; done-- ) { std::cout &lt;&lt; done &lt;&lt; std::endl; } std::cout &lt;&lt; '\n' &lt;&lt; howmany &lt;&lt; std::endl; return 0; } Her withdrawal to the console 5 4 3 2 1 3 For each cycle reference forfor ( ; int howmany = done; done-- ) ^^^^^^^^^^^^^^^^^^ A variable declared howmanywhich conceals the variable with the same name declared in the function main♪ The value assigned to this variable shall be converted to a bulb type. If it is 0, the condition will be false and the cycle will come out. After the cycle, the variable value shall be added howmanydeclared in main♪Well, the last example is when in the proposal. for In its two parts, the variables shall be declared to hide the same variables with the same names declared in the same main♪ After leaving the cycle, these variables cease to exist and become visible variables declared in main♪#include <iostream> int main() { int done = 10; int howmany = 20; for ( int done = 5; int howmany = done; done-- ) { std::cout &lt;&lt; done &lt;&lt; std::endl; } std::cout &lt;&lt; '\n' &lt;&lt; done &lt;&lt; std::endl; std::cout &lt;&lt; howmany &lt;&lt; std::endl; return 0; } Conclusion of the programme5 4 3 2 1 10 20 Proposal for the programme can logically be presented as{ int done = 5; Label_repeat: { int howmany = done; if ( howmany != 0 ) { std::cout &lt;&lt; done &lt;&lt; std::endl; done--; goto Label_repeat; } else { goto Label_exit; } } } Label_exit: //...

Besides the mass of other nonsense, look, you didn't find it. int♪ What's going on? Condition if I'm not doing this. You're moving. else and remove the whole line. And what do you want to find in it after that? ♪ ♪And it well hurts to look at an extra search-- you'll find it before. if: int t = b.find(v[j]); int h = b.find(";"); if ((t != string::npos) && (h != string::npos)) But in any case, without a normal syntax analysis, you'll get a lot of crap. Not everything that starts with a type name and ends. ; - Announcement of variables. ♪ ♪

Let's say we need to read a whole number.int value; std::cin >> value; If the user does something wrong, the flow will change its condition. http://www.cplusplus.com/reference/ios/ios/good/ ♪ http://www.cplusplus.com/reference/ios/ios/fail/ ♪ You can check that:int value; std::cin >> value; if(std::cin.fail()){ //Как-то боремся с проблемой } Or a short recording:int value; std::cin >> value; if(!std::cin){ //Как-то боремся с проблемой } Now how to fight the problem:int value; std::cin >> value; if(!std::cin){ std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n'); //Выкидываем все что ввел пользователь до конца строки std::cin.clear(); //Убираем флаг ошибки. Теперь состояние потока снова good } That's all. Add that test and everything will work. There's still a misstatement. http://cpp.sh/9by3m

Example for 2x values - red and white: // красное - черное - зеро вероятность выпадения var arr = []; var rounds = 100; var n = 0; var i = 0; var max_length = 0; var length = 0; for (i = 0; i <= rounds; i++) { n = Math.round(Math.random() * 36); if ( n != 0) { n = (n % 2) * 2 - 1; } arr.push(n); } console.log(arr); for (i = 0; i < arr.length; i++) { if (arr[i] == 1) { length = 0; while (i < arr.length && arr[i] == 1) { i++; length++; } if (length > max_length) { max_length = length; } } } console.log('Максимально длинная красная последовательность: ' + max_length);

Can I help you? range(n) use it to repeat something. n once in Pitone:for _ in range(60): do_something() time.sleep(1) Right. and "effective" Is that a way? This depends on a specific task.The code is simple and understandable: do_something() 60 times, each next challenge occurs no earlier than a second after the completion of the previous challenge. At the end, we wait a second before we finish the cycle.Imagine. do_something() requires half a second on average, then the cycle will take a minute and a half or more. If that depends on the task.If you want the challenges to happen at the border of every second:for _ in range(60): do_something() time.sleep(1 - time.time() % 1) Precision before other processes, flows (specificity of process/flow planners in the operating system, GIL features of the Pitton Interpreter version), each challenge do_something() Except the first one comes close to time when time.time() the whole thing is returning.For example, if time.time() Come back. X.3 seconds. time.sleep() at least to sleep. 0.7 seconds (if the signal processor doesn't throw off) and means the next challenge. do_something() It's going to happen. X + 1 For a second in system time (if it didn't jump much while we were asleep). In this case, challenges do_something() may be more uniformly distributed and the cycle ends almost immediately in a minute, according to time.time() (if the time of implementation do_something() less than a second). If that depends on the task.If you don't want to start do_something()If more than a minute has passed in accordance with the chosen time limit, it is possible to use a clear condition:deadline = time.monotonic() + 60 while time.monotonic() < deadline: do_something() time.sleep(1) Items to be taken into account depending on the task:What a desirable attitude, if do_something() May be carried out more than a second: skip the irration, start a separate flow, flow pool/process?What's the desired behavior if the system time jumps (because someone shuts the laptop cover during the cycle or https://ru.wikipedia.org/wiki/%D0%93%D0%B8%D0%B1%D0%B5%D1%80%D0%BD%D0%B0%D1%86%D0%B8%D1%8F_%28%D0%BE%D0%BF%D0%B5%D1%80%D0%B0%D1%86%D0%B8%D0%BE%D0%BD%D0%BD%D1%8B%D0%B5_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D1%8B%29 For any other reason, or the mobile OS has downloaded your process to save energy in a while, or the code inside the VM, the big jumps are possible. Do you want to continue the cycle as if nothing happened when the system wakes up, or to cancel the subsequent challenges (outstandings) or to meet the remaining challenges as soon as possible, without pause or at all (deadline passed)?For example, if time changes during performance https://ru.stackoverflow.com/a/577286/23044 What do you think will happen? Whether this conduct is documented sched Module (is it not going to change between the different implementations/versions of Piton)? Is this important in your particular case? sleep() may become particularly attractive at this point.As a variation on the topic, https://stackoverflow.com/a/22498708/4279 or https://stackoverflow.com/a/14040516/4279 ♪ In more complex cases, if you don't have special preferences, you can. https://apscheduler.readthedocs.io ♪ In some cases, it makes sense to use the system planner (for large intervals) such as cron and Windows Task Scheduler.It is useful to know why call_repeatedly() removed https://www.python.org/dev/peps/pep-3156/#basic-callbacks :Note: A previous version of this PEP defined a method named call_repeatedly() , which promised to call a callback at regular intervals. This has been withdrawn because the design of such a function is overspecified. On the one hand, a simple timer loop can easily be emulated using a callback that reschedules itself using call_later() ; it is also easy to write coroutine containing a loop and a sleep() call (a toplevel function in the module, see below). On the other hand, due to the complexities of accurate timekeeping there are many traps and pitfalls here for the unaware ( https://www.python.org/dev/peps/pep-0418/ ), and different use cases require different behavior in edge cases. It is impossible to offer an API for this purpose that is bullet-proof in all cases, so it is deemed better to let application designers decide for themselves what kind of timer loop to implement.

// Функция возведения 10 в степень n function Extend(n: integer): integer; var i: integer; r: integer; begin r := 1; for i := 1 to n do r := r * 10; Result := r; end; var x: string; y: integer; z: string; m: integer; i: integer; begin // Читаем введенное число ReadLn(x); // Переводим его в целочисленный тип y := StrToInt(x); WriteLn('input: ', y); z := ''; for i := 0 to Length(x) - 1 do begin // Берем число по модулю 10 от остатка от деления на степень 10 // В виде строки z := z + IntToStr((y Div Extend(i)) Mod 10); // В виде числа m := m + ((y Div Extend(i)) Mod 10) * Extend(Length(x) - (i + 1)); end; WriteLn('output: ', z); WriteLn('output: ', m); end.

Flaw in the lineConsole.WriteLine("Дробь {0}/{1} = ", div(n, n1, sz)); Two parameters are expected in the format: {0}♪ {1}Transmitted, only one div(n, n1, sz)

Introduce the variable and break it.