From 217f51dd50b305b909757a82826e2b402bbd84d1 Mon Sep 17 00:00:00 2001
From: Nafis Fuad Pranta <nfpranta@gmail.com>
Date: Mon, 31 Jul 2023 16:45:29 +0600
Subject: [PATCH 1/3] Create en.md

---
 dimik-square-number/en.md | 26 ++++++++++++++++++++++++++
 1 file changed, 26 insertions(+)
 create mode 100644 dimik-square-number/en.md

diff --git a/dimik-square-number/en.md b/dimik-square-number/en.md
new file mode 100644
index 00000000..870accdf
--- /dev/null
+++ b/dimik-square-number/en.md
@@ -0,0 +1,26 @@
+# Dimik - Summation
+
+In this problem, you will be given `T` testcases. Each line of the testcase consists of a 5 digit integer `n`. We just have to print the summation of leftmost and rightmost digit of an integer `n`.
+
+### Solution
+We can find the solution by square rooting the value of `n` using `sqrt` function and multiply against itself through the use of `floor` function. Because the value returned by `sqrt` function is `double` and to compare with the `integer` value of `n`, the datatype `double` is converted to `int` using `floor` function.
+
+### C++
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+int main()
+{
+    int t;
+    cin >> t;
+    for (int k = 1; k <= t; k++)
+    {
+        int n;
+        cin >> n;
+        if (floor(sqrt(n)) * floor(sqrt(n)) == n)
+            cout << "YES" << endl;
+        else
+            cout << "NO" << endl;
+    }
+}
+```

From 909e1d7af1fb7830f6712e1317f2f8624bd3fc08 Mon Sep 17 00:00:00 2001
From: Nafis Fuad Pranta <nfpranta@gmail.com>
Date: Mon, 31 Jul 2023 16:50:42 +0600
Subject: [PATCH 2/3] Update en.md

---
 dimik-square-number/en.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/dimik-square-number/en.md b/dimik-square-number/en.md
index 870accdf..51acd442 100644
--- a/dimik-square-number/en.md
+++ b/dimik-square-number/en.md
@@ -1,6 +1,6 @@
-# Dimik - Summation
+# Dimik - Square Number
 
-In this problem, you will be given `T` testcases. Each line of the testcase consists of a 5 digit integer `n`. We just have to print the summation of leftmost and rightmost digit of an integer `n`.
+In this problem, you will be given `T` testcases. Each line of the testcase consists of an integer `n`. We just identify is the value of `n` is a square number or not.
 
 ### Solution
 We can find the solution by square rooting the value of `n` using `sqrt` function and multiply against itself through the use of `floor` function. Because the value returned by `sqrt` function is `double` and to compare with the `integer` value of `n`, the datatype `double` is converted to `int` using `floor` function.

From 97a66e3b146a70514fb9a320eb45da576acd898f Mon Sep 17 00:00:00 2001
From: Nafis Fuad Pranta <nfpranta@gmail.com>
Date: Mon, 31 Jul 2023 16:53:17 +0600
Subject: [PATCH 3/3] Update en.md

---
 dimik-square-number/en.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/dimik-square-number/en.md b/dimik-square-number/en.md
index 51acd442..6a6a0786 100644
--- a/dimik-square-number/en.md
+++ b/dimik-square-number/en.md
@@ -1,6 +1,6 @@
 # Dimik - Square Number
 
-In this problem, you will be given `T` testcases. Each line of the testcase consists of an integer `n`. We just identify is the value of `n` is a square number or not.
+In this problem, you will be given `T` testcases. Each line of the testcase consists of an integer `n`. We just have to identify if the value of `n` is a square number or not.
 
 ### Solution
 We can find the solution by square rooting the value of `n` using `sqrt` function and multiply against itself through the use of `floor` function. Because the value returned by `sqrt` function is `double` and to compare with the `integer` value of `n`, the datatype `double` is converted to `int` using `floor` function.