solution of small to large number

small_to_large/en.md

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+# Dimik - Small to Large
+
+There will be three different numbers. They have to be printed from small to large sizes.
+
+### Solution
+* Case number will be according to the number of test case.
+* Finding the middle number which is not min or max is the main task.
+
+### C++
+```cpp
+#include <iostream>
+#include <algorithm>
+using namespace std;
+
+int main(){
+    int test_case;
+    cin >> test_case;
+
+    for (int i = 1; i <= test_case; i++){
+        int n1, n2, n3;
+        cin >> n1 >> n2 >> n3;
+        int a = min(min(n1, n2), n3);
+        int b = max(max(n1, n2), n3);
+        if (n1 != a && n1 != b)
+        {
+            cout << "Case " << i <<": " << a << " " << n1 << " " << b << endl;
+        } else if (n2 != a && n2 != b){
+            cout << "Case " << i <<": " << a << " " << n2 << " " << b << endl;
+        } else if (n3 != a && n3 != b){
+            cout << "Case " << i <<": " << a << " " << n3 << " " << b << endl;
+        }
+    }
+    return 0;
+}
+```

square_number/en.md

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+# Dimik - Square Number
+
+You have to write a program to find out whether a number is a perfect square or not.
+
+### Solution
+* Store the square root of the number in a variable
+* check if the multiplication of that variable is equal to input the number
+
+# C++
+```cpp
+#include <iostream>
+#include <math.h>
+using namespace std;
+
+int main()
+{
+    int test_case;
+    cin >> test_case;
+
+    for (int i = 1; i <= test_case; i++)
+    {
+        int num;
+        cin >> num;
+        int a = sqrt(num);
+        if (a * a == num)
+        {
+            cout << "YES" << endl;
+        }
+        else
+        {
+            cout << "NO" << endl;
+        }
+    }
+
+    return 0;
+}
+```