1180 beecrowd URI | Code Ease (2024)

Answered on: Friday 15 March, 2024 / Duration: 14 min read

Programming Language: C# , Popularity :9/10

Solution 1:

The 1180 problem on Beecrowd URI is a simple problem that involves finding the smallest number in an array along with its index. Here is the C# code to solve this problem:

c#
using System;
class Program
{
 static void Main(string[] args)
 {
 int n = int.Parse(Console.ReadLine());
 string[] input = Console.ReadLine().Split(); int smallest = int.Parse(input[0]);
 int index = 0;
 for (int i = 1; i < n; i++)
 {
 int current = int.Parse(input[i]);
 if (current < smallest)
 {
 smallest = current;
 index = i;
 }
 }
 Console.WriteLine("Menor valor: " + smallest);
 Console.WriteLine("Posicao: " + index);
 }
}

Sample Input:


5
2 4 6 8 1

Sample Output:


Menor valor: 1
Posicao: 4

In this code, we first read the number of elements in the array (n) and then read the array elements as input. We initialize the smallest number as the first element of the array and then iterate through the array to find the smallest number and its index. Finally, we print the smallest number and its index as output.

Solution 2:

Problem Statement (1180 - Lowest Number and Position)

Given an integer array of size N, you need to find the lowest number in the array and its position in the array.

Input

The input consists of several test cases. The first line of each test case contains the integer N (1 ≤ N ≤ 1000). The second line contains N space-separated integers.

Output

For each test case, print two integers on a single line: the lowest number in the array and its position in the array. If there are multiple occurrences of the lowest number, print the position of the first occurrence.

Code Example in C#

csharp
using System;
class Program
{
 static void Main(string[] args)
 {
 int testCases = int.Parse(Console.ReadLine());
 for (int i = 0; i < testCases; i++)
 {
 int n = int.Parse(Console.ReadLine());
 int[] arr = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
 // Find the lowest number and its position
 int lowest = arr[0];
 int position = 1;
 for (int j = 1; j < n; j++)
 {
 if (arr[j] < lowest)
 {
 lowest = arr[j];
 position = j + 1;
 }
 }
 // Print the lowest number and its position
 Console.WriteLine($"{lowest} {position}");
 }
 }
}

Input Example:


3
5
1 2 3 4 5
4
10 20 30 40
6
100 200 300 400 500 600

Output Example:


1 1
10 1
100 1

Solution 3:

Problem 1180 from Beecrowd URI (also known as the "URI Online Judge") is a problem related to currency conversion. The problem statement is as follows:

A foreigner wants to know how much, in dollars, a product that costs X reais will cost. Write a program that reads the dollar price and the price of the product in reais and calculates the cost of the product in dollars.

Here's a possible solution in C#:

csharp
using System;
class Program
{
 static void Main(string[] args)
 {
 // Read the price of the dollar and the price of the product in reais
 Console.Write("Enter the dollar price: ");
 double dollarPrice = Convert.ToDouble(Console.ReadLine());
 Console.Write("Enter the product price in reais: ");
 double productPriceReais = Convert.ToDouble(Console.ReadLine());
 // Calculate the cost of the product in dollars
 double productPriceDollars = productPriceReais / dollarPrice;
 // Display the result
 Console.WriteLine($"The product price in dollars is: {productPriceDollars:F2}");
 }
}

In this code, we first read the price of the dollar and the price of the product in reais from the standard input using the Console.ReadLine() method. We then convert the strings to double values using the Convert.ToDouble() method.

Next, we calculate the cost of the product in dollars by dividing the price of the product in reais by the price of the dollar. We use the F2 format specifier in the Console.WriteLine() method to display the result with two decimal places.

Here's an example output of the program:


Enter the dollar price: 5.00
Enter the product price in reais: 100.00
The product price in dollars is: 20.00

In this example, the price of the dollar is R$ 5.00 and the price of the product is R$ 100.00. Therefore, the cost of the product in dollars is US$ 20.00.