Algorithm
Problem Main Link: https://www.urionlinejudge.com.br/judge/en/problems/view/1010
URI Online Judge Problem Details:
In this problem, the task is to read a code of a product 1, the number of units of product 1, the price for one unit of product 1, the code of a product 2, the number of units of product 2 and the price for one unit of product 2. After this, calculate and show the amount to be paid.
Input
The input file contains two lines of data. In each line there will be 3 values: two integers and a floating value with 2 digits after the decimal point.
Output
The output file must be a message like the following example where "Valor a pagar" means Value to Pay. Remember the space after ":" and after "R$" symbol. The value must be presented with 2 digits after the point.
| Input Samples | Output Samples |
|
12 1 5.30 |
VALOR A PAGAR: R$ 15.50 |
|
13 2 15.30 |
VALOR A PAGAR: R$ 51.40 |
|
1 1 15.10 |
VALOR A PAGAR: R$ 30.20 |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int main()
{
int a, b;
double c, res;
scanf("%d %d %lf", &a, &b, &c);
res = b * c;
scanf("%d %d %lf", &a, &b, &c);
res += b * c;
printf("VALOR A PAGAR: R$ %.2lf\n", res);
return 0;
}Input
16 2 5.10
Output
Demonstration
#include int main() { int p_code[2], p_u[2], i; float price[2], pay; for(i=0; i<2; i++) { scanf("%d %d %f",&p_code[i],&p_u[i],&price[i]); } pay=((price[0]*p_u[0])+(price[1]*p_u[1])); printf("VALOR A PAGAR: R$ %.2f\n", pay); return 0; }
URI Solution 1010 Simple Calculate Code / URI 1010 Simple Calculate solution in CPP:
#include using namespace std; int main(int argc, char const *argv[]) { int a, b; double c, res; scanf("%d %d %lf", &a, &b, &c); res = b * c; scanf("%d %d %lf", &a, &b, &c); res += b * c; printf("VALOR A PAGAR: R$ %.2lf\n", res); return 0; }
URI Solution 1010 Simple Calculate Code / URI 1010 Simple Calculate solution in Java:
import java.util.Scanner; public class Main{ public static void main(String[] args) { int a, b; double c, res; Scanner sc = new Scanner(System.in); a = sc.nextInt(); b = sc.nextInt(); c = sc.nextDouble(); res = b * c; a = sc.nextInt(); b = sc.nextInt(); c = sc.nextDouble(); res += b * c; System.out.printf("VALOR A PAGAR: R$ %.2f\n", res); } }
URI Solution 1010 Code / URI 1010 solution in Python:
linha1 = input().split(" ") linha2 = input().split(" ") cod1, qtde1, valor1 = linha1 cod2, qtde2, valor2 = linha2 total = (int(qtde1) * float(valor1)) + (int(qtde2) * float(valor2)) print("VALOR A PAGAR: R$ %0.2f" %total)
URI Solution 1010 Code / URI 1010 solution in C# (C Sharp):
URI Online Judge Solution 1010 Code Demonstration:
First take 3 values in store it in a result variable
Again take 3 value and now just increment the result variable. And solved, in C that is
scanf("%d %d %lf", &a, &b, &c); res = b * c; scanf("%d %d %lf", &a, &b, &c); res += b * c;