Algorithm
Problem Name: beecrowd | 1985
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/1985
MacPRONALTS
By Victor Jatobá, UNIME 
Brazil
Timelimit: 1
The MacPRONALTS is with a super promotion, exclusive to the contestants of the first Selective MaratonaTEC. But they had a problem, all runners were trying to buy at the same time, so this generated a very long queue. The worst is that the cashier girl no have calculator or a program to help her to calculate more quickly. You are the person that will help the girl and the MacPRONALTS increase their sells. Bellow has a menu day, that contains the product number and its value.
1001 | R$ 1.50
1002 | R$ 2.50
1003 | R$ 3.50
1004 | R$ 4.50
1005 | R$ 5.50
Input
The first entry is reported the number of purchased products (1 <= p <= 5). For each product follows the quantity (1 <= q <= 500) that the customer purchased.
Obs .: the product number will not be repeated.
Output
You must print the purchase amount with two decimal places.
| Input Sample | Output Sample |
|
2 1001 2 1005 3 |
19.50 |
|
1 1003 500 |
1750.00 |
|
5 1001 500 1005 300 1003 23 1002 52 1004 44 |
2808.50 |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include<stdio.h>
int main()
{
int n,i,b;
float a,s=0.0;
scanf("%d",&n);
for (i = 1; i < = n; i++)
{
scanf ("%f %d",&a,&b);
if (a == 1001)
a = 1.50;
if (a == 1002)
a = 2.50;
if (a == 1003)
a = 3.50;
if (a == 1004)
a = 4.50;
if (a == 1005)
a = 5.50;
s=s+a*b;
}
printf ("%0.2f\n",s);
return 0;
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
using namespace std;
int main()
{
float n , p ,q , total = 0;
cin >> n;
for(int i = 0 ; i < n ; i++){
cin >> p >> q;
if(p == 1001){
total+= 1.5*q;
}
else if(p == 1002){
total+= 2.5*q;
}
else if(p == 1003){
total+= 3.5*q;
}
else if(p == 1004){
total+= 4.5*q;
}
else if(p == 1005){
total+= 5.5*q;
}
}
printf("%.2f\n" , total);
return 0;
}
Input
Output