Question

Вася написал код для решения некоторой задачи. Приводим его на трех языках программирования: Python: n = int(input()) ans = 0 i = 0 while n > 0: if i % 2 == 1 and n % 10 % 2 == 0: ans += 1 n //= 10 i += 1 print(ans) Pascal: var n, i, ans: longint; begin readln(n); ans := 0; i := 0; while n > 0 do begin if (i mod 2 = 1) and (n mod 10 mod 2 = 0) then ans := ans + 1; n := n div 10; i := i + 1; end; writeln(ans); end. C++: #include using namespace std; int main() { int n, ans = 0, i = 0; cin >> n; while (n > 0){ if (i % 2 == 1 and n % 10 % 2 == 0) ans++; n /= 10; i++; } cout << ans; return 0; } А теперь ему стало интересно: при каком наименьшем n его программа выведет в качестве ответа число 3?

Answer 1

200000, насколько я понял.Если вы дочитаете это - респект.%)

 При первом запуске условие не выполниться, поэтому переменная ans останется прежней( в таком случае нам безразлично значение единиц числа n, и мы ставим наименьшее - 0), в этом же цикле от числа n убираются единицы и к переменная i увеличивается на 1.
Цикл начинается по новой, теперь условие выполняется( i делится на 2 с остатком 1), но также на два должны без остатка делится и десятки числа n(опять берем наименьшее число 0) переменная ans увеличивается на 1.
В следующем цикле условие опять не выполняется, i увеличивается на 1, от n убираются сотни(сотни числа n равны 0)
В четвертом цикле asn и i увеличиваются на 1, тысячи числа n опять же равны 0.
В 5-ом цикле условие не выполняется, i увеличивается на 1, десятки тысяч числа n равны 0.
И наконец в шестом цикле переменная asn увеличится на 1, но при том условии,что сотни тысяч числа n делятся на 2 без остатка - ноль здесь уже не подходит, поскольку число не может начинаться с нуля. Берем 2.

Answer 2

#### Python Solution: The Python code provided by Vasya is as follows: ```python n = int(input()) ans = 0 i = 0 while n > 0: if i % 2 == 1 and n % 10 % 2 == 0: ans += 1 n //= 10 i += 1 print(ans) ``` This code takes an integer input `n` and initializes `ans` and `i` to 0. It then enters a while loop that continues until `n` becomes 0. Inside the loop, it checks if `i` is odd and the last digit of `n` is even. If both conditions are true, it increments `ans` by 1. Finally, it divides `n` by 10 and increments `i` by 1. The code prints the final value of `ans` as the output. To determine the smallest value of `n` that will output 3, we need to analyze the code. The code increments `ans` by 1 whenever `i` is odd and the last digit of `n` is even. Since `ans` needs to be 3, we can deduce that there must be at least 3 occurrences of this condition in the while loop. Let's break down the conditions: - `i` needs to be odd: This means that `i % 2` should be equal to 1. - The last digit of `n` needs to be even: This means that `n % 10 % 2` should be equal to 0. Based on these conditions, we can try different values of `n` to find the smallest one that satisfies the conditions. Starting from 1, we can increment `n` by 1 until we find a value that satisfies the conditions. #### Solution Analysis: To find the smallest value of `n` that will output 3, we can analyze the conditions in the code: - `i % 2 == 1`: This condition is satisfied when `i` is an odd number. - `n % 10 % 2 == 0`: This condition is satisfied when the last digit of `n` is an even number. We need to find a value of `n` that satisfies both conditions at least 3 times. Let's analyze the code further to understand how the conditions are checked. The code uses a while loop that continues until `n` becomes 0. Inside the loop, it checks the conditions and increments `ans` if they are satisfied. Then, it divides `n` by 10 and increments `i` by 1. To satisfy the conditions at least 3 times, we need to find a value of `n` that has at least 3 even digits in odd positions. For example, if `n` is 2468, the conditions will be satisfied for the digits 4, 6, and 8 (at positions 1, 3, and 5, respectively). #### Finding the Smallest Value of n: To find the smallest value of `n` that will output 3, we can start from 1 and increment `n` by 1 until we find a value that satisfies the conditions. Let's try different values of `n` and check if they satisfy the conditions: - For `n = 1`, the conditions are not satisfied because the last digit of 1 is odd. - For `n = 2`, the conditions are not satisfied because the last digit of 2 is even, but `i` is not odd. - For `n = 3`, the conditions are not satisfied because the last digit of 3 is odd. - For `n = 4`, the conditions are satisfied once because the last digit of 4 is even and `i` is odd. - For `n = 5`, the conditions are not satisfied because the last digit of 5 is odd. - For `n = 6`, the conditions are satisfied once because the last digit of 6 is even and `i` is odd. - For `n = 7`, the conditions are not satisfied because the last digit of 7 is odd. - For `n = 8`, the conditions are satisfied once because the last digit of 8 is even and `i` is odd. - For `n = 9`, the conditions are not satisfied because the last digit of 9 is odd. - For `n = 10`, the conditions are not satisfied because the last digit of 10 is even, but `i` is not odd. - For `n = 11`, the conditions are not satisfied because the last digit of 11 is odd. - For `n = 12`, the conditions are satisfied twice because the last digit of 12 is even and `i` is odd. - For `n = 13`, the conditions are not satisfied because the last digit of 13 is odd. - For `n = 14`, the conditions are satisfied twice because the last digit of 14 is even and `i` is odd. - For `n = 15`, the conditions are not satisfied because the last digit of 15 is odd. - For `n = 16`, the conditions are satisfied twice because the last digit of 16 is even and `i` is odd. - For `n = 17`, the conditions are not satisfied because the last digit of 17 is odd. - For `n = 18`, the conditions are satisfied twice because the last digit of 18 is even and `i` is odd. - For `n = 19`, the conditions are not satisfied because the last digit of 19 is odd. - For `n = 20`, the conditions are not satisfied because the last digit of 20 is even, but `i` is not odd. Based on the above analysis, the smallest value of `n` that will output 3 is **18**. #### Answer: The smallest value of `n` that will output 3 when executing the provided code is **18**.