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Copy file name to clipboardExpand all lines: Content/Chapter-6-2-nested-loops-exam-problems/exam-problems/arrow/arrow.md
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From the explanation we see that **the input data** will be read from one input line only, which will contain **an integer** within the range [**3 … 1000**]. This is why we will use **a variable** of **`int`** type.
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### Divide the Figure into Parts
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We can divide the figure into **3 parts** – upper, middle and lower one. **The upper part** contains two sub-parts – a first row and a body of the arrow. We can see from the examples that the count of **the outer dots** on the first row and in the body of the arrow is **`(n - 1) / 2`**. We can write this value in **a variable****`outerDots`**.
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## Printing the Body and the Middle Row of the Arrow
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###Printing the Body and the Middle Row of the Arrow
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In order to draw **the body of the arrow**, we need to create **a loop**, which runs **`n - 2`** times.
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## Printing the Lower Part of the Arrow
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###Printing the Lower Part of the Arrow
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In order to draw **the lower part of the arrow**, we need to assign new values to **the variables****`outerDots`** and **`innerDots`**.
Copy file name to clipboardExpand all lines: Content/Chapter-6-2-nested-loops-exam-problems/exam-problems/axe/axe.md
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### Divide the Figure into Parts
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We divide the figure into 3 parts: upper part, middle part (the handle), down part.
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After we have declared and initialized **the variables**, we can begin drawing the figure by starting with **the upper part**. We can see from the examples what the structure of **the first row** is, and we can create a loop, which runs **`n`** times. At each iteration of the loop **the middle dashes** increase by 1, and **the right dashes** decrease by 1.
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## Printing the Handle
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###Printing the Handle
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We can draw **the handle of the axe** by creating a loop, which runs **`n - 2`** times. We can see in the examples what its structure is.
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## Printing the Lower Part of the Axe
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###Printing the Lower Part of the Axe
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We need to divide **the lower part** of the figure into two sub-parts – **head of the axe** and **the last row of the figure**. We will print **the head of the axe** on the console by creating a loop that runs **`n / 2 - 1`** times. At each iteration **the left dashes** and **the right dashes** decrease by 1, and **the middle dashes** increase by 2.
We can divide the figure into 3 parts – upper wing, body and lower wing. In order to draw the upper wing, we need to divide it into parts – beginning **`*`**, middle part **`\ /`** and end **`*`**. After looking at the examples we find out that the beginning is with size **`n - 2`**.
In order to create **the body of the butterfly** we can use **the variable****`halfRowSize`** again and print on the console exactly **one** row. The body structure begins with **`(white space)`**, middle **`@`** and ends with **`(white space)`**.
Let’s solve the problem step by step: read the input, perform some calculations, print the fort roof, print the fort body, print the fort base.
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### Reading the Input Data
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We can see from the explanation that **the input data** will be only one line which will contain **an integer** within the range [**3 … 1000**]. Therefore, we will use **a variable** of **`int`** type.
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* body
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* base
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## Calculating and Printing the Roof
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###Calculating and Printing the Roof
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We can see from the examples that **the roof** is made of **two towers** and **a middle part**. Each tower has a beginning **`/`**, middle part **`^`** and an end **`\`**.
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**The body of the fort** contains a beginning **`|`**, a middle part **`(white space)`** and an end **`|`**. **The middle part** is a blank space with size of **`2 * n - 2`**. The number of **the rows** used for walls can be found in the given parameters: **`n - 3`**. This code prints the body of the fort:
In order to draw the last but one row, which is a part of the base, we need to print a beginning **`|`**, middle part **`(white space)_(white space)`** and an end **`|`**. In order to do this, we can use the already declared variables **`colSize`** and **`midSize`** again, because we can see from the examples that they are equal to the **`_`** in the roof.
Copy file name to clipboardExpand all lines: Content/Chapter-6-2-nested-loops-exam-problems/exam-problems/stop/stop.md
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### Divide the Figure into Parts
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We can **divide** the figure into **3 parts** – upper, middle and lower. **The upper part** contains two sub-parts – first row and rows in which the sign widens. **The first row** is made of a beginning **`.`**, middle part **`_`** and an end **`.`**. After looking at the examples we can say that the beginning is **`n + 1`** columns wide, so it is good to write this **value** in a separate **variable**.
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## Printing the Upper Part of the Sign
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###Printing the Upper Part of the Sign
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In order to draw the rows in which the sign is getting **"wider"**, we need to create **a loop**, which runs **`n`** times. The structure of a row contains a beginning **`.`**, **`//`** + middle part **`_`** + **`\\`** and an end **`.`**. In order to reuse the already created **variables**, we need to decrease **`dots`** by 1 and **`underscores`** by 2, because we have already **printed** the first row, and the dots and underscores in the upper part of the figure are **decreasing**.
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## Printing the Middle Row and the Lower Part of the Sign
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###Printing the Middle Row and the Lower Part of the Sign
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**The middle part** of the figure begins with **`//`** + **`_`**, middle part **`STOP!`** and an end **`_`** + **`\\`**. The count of the underscores **`_`** is **`(underscores - 5) / 2`**.
Let's declare and initialize the **variables** that will store the **characters** of the password: for the characters of **digit** type – **`int`** – **`d1`**, **`d2`**, **`d3`**, and for the **letters** – of **`char`** type – **`l1`**, **`l2`**. To make it easier we will skip explicit specification of the type by replacing it with the keyword **`var`**.
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## Processing the Input Data and Printing Output
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###Processing the Input Data and Printing Output
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We need to create **five****`for`** nested loops, one for each variable. To ensure that the last digit **d3** is **greater** than the first two, we will use the built-in function **`Math.Max (...)`**.
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