Total Expenses | CodeChef Solution

Hello coders, today we are going to solve Total Expenses CodeChef Solution whose Problem Codes is FLOW009.

Task

While purchasing certain items, a discount of 10% is offered if the quantity purchased is more than 1000.
If the quantity and price per item are input, write a program to calculate the total expenses.

Input Format

The first line contains an integer T, total number of test cases. Then follow T lines, each line contains integers quantity and price.

Output Format

For each test case, output the total expenses while purchasing items, in a new line.

Constraints

• 1 ≤ T ≤ 1000
• 1 ≤ quantity,price ≤ 100000

Example

Sample Input

3 
100 120
10 20
1200 20

Sample Output

12000.000000
200.000000
21600.000000

Solution – Total Expenses | CodeChef Solution

C++

#include <iostream>
using namespace std;

int main() {
	// your code goes here
	int t;
	double e,f,p,d;
	cin>>t;
	while(t--){
	    cin>>e>>f;
	    p = e*f;
	    if(e>1000){
	        d=p/10;
	        p-=d;
	        printf("%f\n",p);
	    }
	    else{
	          printf("%f\n",p);
	    }
	}
	return 0;
}

Python

#Solution Provided by CodingBroz
for i in range(int(input())):
    total = 0
    m, n = map(int, input().split())
    total += (m * n)
    if m > 1000:
        discount = total - (total * 10/100)
        print(discount)
    else:
        print(total)

Java

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
	     Scanner sc=new Scanner(System.in);
	     int TEST = sc.nextInt();
	     while(TEST-->0)
	     {
	         int a=sc.nextInt();
	         int b=sc.nextInt();
	         float dis=a*b;
	         if(a>1000)
	         {
	             dis=dis-(dis/10);
	             System.out.println(dis);
	         }
	         else
	         {
	             System.out.println(dis);
	         }
	     }
	}
}

Disclaimer: The above Problem (Total Expenses) is generated by CodeChef but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.