Skip to main content

URI Problem 2617 Solution Provider Ajax SA - URI Online Judge Solution


URI Online Judge Solution 2617 Provider Ajax SA Using PostgreSQL Query Language.

The financial sector has encountered some problems in the delivery of one of our providers, the delivery of the products does not match the invoice.
Your job is to display the name of the products and the name of the provider, for the products supplied by the provider 'Ajax SA'.

Schema

providers
ColumnType
id (PK)numeric
namecharacter varying (255)
streetcharacter varying (255)
citycharacter varying (255)
statechar (2)
 
products
ColumnType
id (PK)numeric
namecharacter varying (255)
amountnumeric
pricenumeric
id_providers (FK)numeric

Tables

providers
idnamestreetcitystate
1Ajax SAPresidente Castelo BrancoPorto AlegreRS
2Sansul SAAv BrasilRio de JaneiroRJ
3South ChairsAv MoinhoSanta MariaRS
4Elon ElectroApoloSão PauloSP
5Mike ElectroPedro da CunhaCuritibaPR
 
products
idnameamountvalueid_providers
1Blue Chair30300.005
2Red Chair502150.001
3Disney Wardrobe400829.504
4Blue Toaster209.903
5Solar Panel303000.254

Output Sample

namename
Red ChairAjax SA

URI 2617 Solution in SQL:


SELECT products.name, providers.name
FROM products
INNER JOIN providers ON products.id_providers = providers.id
WHERE providers.name = 'Ajax SA'

Comments

Popular posts from this blog

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...