dimik even-odd-2 added

dimik-even-odd-2/en.md

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+## Dimik Even or Odd 2
+
+### What the problem Wants
+In this problem, you will be given `T` testcases.Each line of the testcase consist of an integer `n`.We have to determine whether `n` is even or odd. The difference from dimik even odd 1 is that the number can have upto 100 digits which won't fit in long long data type of C++. 
+
+### Approach 
+- Take Input as a string so that it can take upto 100 digits 
+- Check the last digit of that string 
+- If its divisible by 2 then print 'even' otherwise print 'odd' 
+
+### Solution in C++
+
+> Step 1: Include necessary libraries
+
+```cpp
+#include<bits/stdc++.h>
+using namespace std;
+```
+
+The bits/stdc++.h library is included, which is a header file that includes all the standard libraries in C++.
+
+### Define the function to check if a number is even
+
+```cpp
+bool isEven(string number) {
+    int lastDigit = number[number.size() - 1] - '0';
+    return (lastDigit % 2 == 0);
+}
+```
+
+> Step 2: Function for identify Odd Even
+
+- Inside the isEven function, we extract the last digit of the number by accessing the last character of the string (number[number.size() - 1]) and subtracting the ASCII value of '0' from it. This converts the character representation of the digit to an integer.
+
+- We then check if the last digit is divisible by 2 using the modulo operator (%). If the remainder is 0, the number is even and we return true; otherwise, we return false
+
+> Step 3: Use the function from main 
+
+### Implement the main function
+```cpp
+int main() {
+    int numTestCases;
+    cin >> numTestCases;
+    cin.ignore(); // Ignore the newline character after numTestCases
+
+    for (int i = 0; i < numTestCases; i++) {
+        string number;
+        getline(cin, number);
+        if (isEven(number)) {
+            cout << "even" << endl;
+        } else {
+            cout << "odd" << endl;
+        }
+    }
+
+    return 0;
+}
+```
+
+- In the main function, we first read the number of test cases from the input using cin >> numTestCases.
+
+- Since we read an integer before reading strings, we need to ignore the newline character left in the input buffer. We do this using cin.ignore().
+
+- Next, we enter a loop to process each test case. For each test case, we read the number as a string using getline(cin, number). This allows us to handle numbers with multiple digits.
+
+- We then call the isEven function with the number as the argument. Depending on the return value, we output whether the number is even or odd using cout.
+
+### Full code 
+```cpp
+#include<bits/stdc++.h>
+using namespace std;
+
+bool isEven(string number) {
+    int lastDigit = number[number.size() - 1] - '0';
+    return (lastDigit % 2 == 0);
+}
+
+int main() {
+    int numTestCases;
+    cin >> numTestCases;
+    cin.ignore(); // Ignore the newline character after numTestCases
+
+    for (int i = 0; i < numTestCases; i++) {
+        string number;
+        getline(cin, number);
+        if (isEven(number)) {
+            cout << "even" << endl;
+        } else {
+            cout << "odd" << endl;
+        }
+    }
+
+    return 0;
+}
+```
+
+### Solution in python 
+
+As python can take upto 100 digit as input we will just check if its even or odd by just dividing it by 2
+
+```py
+testCase=eval(input())   # taking test case and eval converting it to int
+for i in range(testCase):  # for each test case
+    num=eval(input())  # take the numnber and convert it to int
+    if(num%2==0):   # check even or odd
+        print('even')
+    else:
+        print('odd')
+```