URI Problem 2623 Solution Categories with Various Products

URI Problem 2623 Solution Categories with Various Products - URI Online Judge Solution

URI Online Judge Solution 2623 Categories with Various Products Using PostgreSQL Query Language.

The sales industry needs a report to know what products are left in stock. To help the sales industry, display the product name and category name for products whose amount is greater than 100 and the category ID is 1,2,3,6 or 9. Show the results in ascendent order by category ID.

Schema

products
Column	Type
id (PK)	numeric
name	character varying (255)
amount	numeric
price	numeric
id_categories (FK)	numeric

categories
Column	Type
id (PK)	numeric
name	character varying (255)

Tables

products
id	name	amount	price	id_categories
1	Blue Chair	30	300.00	9
2	Red Chair	200	2150.00	2
3	Disney Wardrobe	400	829.50	4
4	Blue Toaster	20	9.90	3
5	Solar Panel	30	3000.25	4

categories
id	name
1	Superior
2	Super Luxury
3	Modern
4	Nerd
5	Infantile
6	Robust
9	Wood

Output Sample

name	name
Red Chair	Super Luxury

URI 2623 Solution in SQL:


SELECT products.name, categories.name
FROM products
INNER JOIN categories 
ON products.id_categories = categories.id
WHERE products.amount > 100 
  AND products.id_categories IN (1,2,3,6,9)

Comments

URI Problem 1001 Solution Extremely Basic - - URI Online Judge Solution

URI Online Judge Problem 1001 Extremely Basic Solution using C, Python Programming Language. Read 2 variables, named A and B and make the sum of these two variables, assigning its result to the variable X . Print X as shown below. Print endline after the result otherwise you will get “ Presentation Error ”. Input The input file will contain 2 integer numbers. Output Print the letter X (uppercase) with a blank space before and after the equal signal followed by the value of X, according to the following example. Obs.: don't forget the endline after all. Samples Input Samples Output 10 9 X = 19 -10 4 X = -6 15 -7 X = 8 URI 1001 Solution in Python A = int(input()) B = int(input()) X = A + B print("X =",X); URI 1001 Solution in C: #include int main() { int A,B,X; scanf ("%d %d", &A, &B); X=A+B; printf("X = %d\n",X); return 0; }

URI Problem 1002 Solution Area of a Circle - URI Online Judge Solution

URI Online Judge Problem 1002 Solution Area of a Circle using Python Programming Language. The formula to calculate the area of a circumference is defined as A = π . R2. Considering to this problem that π = 3.14159: Calculate the area using the formula given in the problem description. Input The input contains a value of floating point (double precision), that is the variable R. Output Present the message "A=" followed by the value of the variable, as in the example bellow, with four places after the decimal point. Use all double precision variables. Like all the problems, don't forget to print the end of line after the result, otherwise you will receive "Presentation Error". Input Samples Output Samples 2.00 A=12.5664 100.64 A=31819.3103 150.00 A=70685.7750 URI 1002 Solution in Python R = float(input()); R = R*R*3.14159; print("A=%.4f" % R); URI 1002 Solution in C: #include #define A 3.14159 int main() {...

URI Problem 1179 Solution Array Fill IV - URI Online Judge Solution

URI Online Judge Solution 1179 Array Fill IV Using C, Python Programming Language. In this problem you need to read 15 numbers and must put them into two different arrays: par if the number is even or impar if this number is odd. But the size of each of the two arrrays is only 5 positions. So every time you fill one of two arrays, you must print the entire array to be able to use it again for the next numbers that are read. At the end, all remaining numbers of each one of these two arrays must be printed beginning with the odd array. Each array can be filled how many times are necessary. Input The input contains 15 integer numbers. Output Print the output like the following example. Input Sample Output Sample 1 3 4 -4 2 3 8 2 5 -7 54 76 789 23 98 par[0] = 4 par[1] = -4 par[2] = 2 par[3] = 8 par[4] = 2 impar[0] = 1 impar[1] = 3 impar[2] = 3 impar[3] = 5 impar[4] = -7 impar[0] = 789 impar[1] = 23 par[0] = 54 par[1] = 76 par[2] = 98 Solution Using C: #include <stdi...

URI Online Judge Solution

Search This Blog