Adding a tutorial for Dimik-Even Odd 2

dimik-even-odd-2/en.md

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+# Dimik - Even Odd 2
+
+In this problem, you will be given `T` testcases.Each line of the testcase consist of an integer `n` which can be a number consisting of maximum 100 digits.We have to determine whether `n` is odd or even.
+
+## Hint: You just need to handle last bit 
+
+### Solution 
+* for each test case take the number as a string 
+* Check the last character of the string 
+* make it integer 
+* If its divisible by 2
+  * its a even number so print 'even'
+* if its not divisible by 2 
+  * its a odd number so print 'odd'
+
+### C++
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+int main()
+{
+    int t;
+    cin >> t;
+    for (int k = 1; k <= t; k++)
+    {
+        string s;
+        cin >> s;
+        string tmp = "";
+        tmp += s[s.size() - 1];
+        int val = stoi(tmp);
+        if (val % 2)
+            cout << "odd" << endl;
+        else
+            cout << "even" << endl;
+    }
+}
+```