We introduce an integer. Count the number of times the software in its decimal notation includes some figures, let us say 3. That's all the condition. For example: 123 -> 1; 54321345 -> 2; 3333 -> 4 etc.
The task is very simple … at the high school level. But in order, to do the task is not quite so trivial – complicate:
– offer several (as much as possible) different ways to implement;
– for each realization reduce code writing so, that it, more concise as possible.
Solutions (It is where to turn because of the simplicity of the task, solutions can be very, Fill your):
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | #include <iostream> using namespace std; int main() { setlocale(LC_ALL, "rus"); unsigned long e; unsigned n = 0; cout << "Ввод числа: "; cin >> e; do if (3 == e % 10) n++; while ((e /= 10) > 0); cout << "Цифр 3 в числе " << n << " штук" << endl; } |
It all fits into the solution one operator do … while. But, pay attention to, what “injected integer” – This input is always a string, representing a number (Here the trick in the formulation of the problem). Then:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | #include <iostream> using namespace std; int main() { setlocale(LC_ALL, "rus"); char e[80], *p = e; unsigned n = 0; cout << "Ввод числа: "; cin >> e; while ((p = strchr(p, '3')) != NULL) p++, n++; cout << "Цифр 3 в числе " << n << " штук" << endl; } |
Or so:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | #include <iostream> //#include <string> using namespace std; int main() { setlocale(LC_ALL, "rus"); char e[80]; unsigned n = 0, i; cout << "Ввод числа: "; cin >> e; for (i = 0; i < strlen(e); i++) if (e[i] == '3') n++; cout << "Цифр 3 в числе " << n << " штук" << endl; } |
Or even so:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | #include <iostream> #include <string> using namespace std; int main() { setlocale(LC_ALL, "rus"); string e; unsigned n = 0; cout << "Ввод числа: "; cin >> e; for (string::iterator i = e.begin();; n++, i++) if ((i = find(i, e.end(), '3')) == e.end()) break; cout << "Цифр 3 в числе " << n << " штук" << endl; } |
And that's how it looks:
1 2 3 4 5 6 7 8 9 10 11 12 | $ ./mul3_1 Ввод числа: 1234531 Цифр 3 в числе 2 штук $ ./mul3_2 Ввод числа: 12341231 Цифр 3 в числе 2 штук $ ./mul3_3 Ввод числа: 12341231 Цифр 3 в числе 2 штук $ ./mul3_4 Ввод числа: 333 Цифр 3 в числе 3 штук |
P.S. This problem is a good illustration of the fundamental programming principles, what any the problem can be solved many and very various ways.
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