How to check if there is a remainder in the division of numbers in C ++?

very simple

if (x % t != 0) { cout << "при делении x на t есть остаток" << endl; } else { cout << "остатка нет" << endl; } 

From a mathematical point of view , the operation of calculating the remainder in C / C ++ for negative numbers, for example -11 % 8 == -3 , gives the wrong result to which we are accustomed from school. Of course, this is a matter of agreement , but ... This problem sometimes causes confusion and heated debate .

If the task is to determine only the absence of a residue, then % sufficient. However, there is a particular solution when the divisor is a number of degree 2. The same method is the easiest to determine the remainder of dividing by numbers greater than 1 equal degrees 2.

 int a; int b; //2 в степени n if (a & (b - 1)) //остаток от деления присутствует. Например, -11 & (8 - 1) == 5 else //остаток равен нулю. Например, -8 & (8 - 1) == 0 

Why is it important? Consider the sequence mod(n, 4)

For n & (4-1) the sequence is periodic

  n -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 n&3 3 * * * * 2 * * * * 1 * * * * 0 * * * * * 

For n% 4, the sequence is, generally speaking, non-periodic

  n -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 n%4 3 * * 2 * * 1 * * 0 * * * * * -1 * * -2 * * -3 * * 

Working with a periodic sequence is much more pleasant where it is necessary to observe equal intervals between events. You can build a type design

 switch (i & 3) { case 0: /*событие 1*/; break; case 1: /*событие 2*/; break; case 2: /*событие 3*/; break; case 3: /*событие 4*/; break; } 

and to be sure that each event will occur with an enviable regularity, regardless of the sign i and additional case not required.