Cycle: for (j = 2; j < i; j++) , what is j<i . How to understand this? Where did we come up with j ?

 public class Za { public static void main(String args[]) { int i, j; // Объявление переменных i и j. boolean isprime; // Объявление переменной isprime for (i = 2; i < 100; i++) { // Объявление цикла isprime = true; //Вопрос: Зачем здесь это? for (j = 2; j < i; j++) /* Не ясно объявление цикла? Ведь i=2, а тут опять пишем j=2 и еще пишем j<i (как это возможно если Они оба равны 2). Вообще зачем, что написали внутри цикла */ if ((i % j) == 0) isprime = false; // Непонятно, почему мы делим по модулю и // если без остатка вернется false if (isprime) // Не понятно эта управляющая конструкция System.out.println(i + " is prime."); } } } // Почему создается это: isprime = true; // Проверить, делится ли число без остатка. for (j = 2; j < i; j++) // Если число делится без остатка, оно простое if ((i % j) == 0) isprime = false;
  • 2
    The title should write a short title, and the questions are already in the body of the question. - Unick
  • 2
    Have you taken the code somewhere and cannot understand what it is doing? - iksuy
  • code took in the internet - KR88OS

4 answers 4

I added explanations with comments in the code:

 for (i = 2; i < 100; i++) { // перебираем все числа от двух до ста isprime = true; // если не доказано обратное - число простое for (j = 2; j < i; j++) // перебираем все числа от двух до текущего числа (хотя достаточно до Sqrt(i)) if ((i % j) == 0) isprime = false; // если остаток равен нулю, то текущее число i делится нацело на делитель j и простым быть не может. После этого должен стоять break, т.к. дальше искать делители бессмісленно if (isprime) // если не было найдено ни одного делителя, то число простое System.out.println(i + " is prime."); }
  • isprime = true; // if the opposite is not proved - the number is simple, but why do we write this after for (i = 2; i <100; i ++) and not before - KR88OS
  • Because we need to check each of the numbers 2-100 for simplicity - Anton Shchyrov
  • I need a tutor upholstered free of charge urgently ( - KR88OS

This is an algorithm to search for primes up to 100, the second loop is all numbers less than the i-th to check (therefore, j <i), and isprime is a regular flag, which is finally checked to display the number in the console.

    // and it is not clear why we did that? because i = 2 and here we write again j = 2

    You have i=2 only in the first iteration of the cycle in i, in all subsequent ones there is no.

    // and it is not clear why we divide by module and say if without a trace the error (false)

    Again it is wrong, if it divides without remainder, then we assign a false value to the Boolean variable, and not an error. But in general:

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      The point is that these are cycles, and you say that perform an internal cycle while j <i, as soon as this inequality stops, then exit the cycle. This makes sense since the next time i will be 3 4 5, etc. to 10