Sale Season Codechef Solution

Hello coders, today we are going to solve Sale Season Codechef Solutions whose Problem Code is SALESEASON.

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Sale Season Codechef Solution

Problem

It’s the sale season again and Chef bought items worth a total of X rupees. The sale season offer is as follows:

Screenshot 2022 08 17 at 10.17.20 PM

Find the final amount Chef needs to pay for his shopping.

Input Format

  • The first line of input will contain a single integer T, denoting the number of test cases.
  • Each test case consists of single line of input containing an integer X.

Output Format

For each test case, output on a new line the final amount Chef needs to pay for his shopping.

Constraints

  • 1 ≤ T ≤ 100
    1 ≤ X ≤ 10000

Sample 1:

Input

4
15
70
250
1000

Output

15
70
225
975

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Explanation:

Test case 11: Since X \le 100X≤100, there is no discount.

Test case 33: Here, X = 250X=250. Since 100 \lt 250 \le 1000100<250≤1000, discount is of 2525 rupees. Therefore, Chef needs to pay 250-25 = 225250−25=225 rupees.

Sale Season Codechef Solution in JAVA

/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		for(int i=0; i<T; i++)
		{
		    int N = sc.nextInt();
		    double r = 0;
		    for(int j=0; j<N; j++)
		    {
		        int p = sc.nextInt();
		        int q = sc.nextInt();
		        int d = sc.nextInt();
		        double pp = p+((double)p*d)/100;
		        pp = pp-(pp*d)/100;
		        r=r+(p-pp)*q;
		    }
		    System.out.println(r);
		}
	}
}

Sale Season Codechef Solution in CPP

 #include <iostream>
using namespace std;
int main() {
	int t;
	cin>>t;
	while(t--){
	    int n;
	    cin>>n;
	    double sum=0;
	    for(int i=1;i<=n;i++){
	        double p,x,d;
	        cin>>p>>x>>d;
	        double dis=double(d)/100;
	        double finalp=p+(p*dis);
	        double finalfp=finalp-(finalp*dis);
	        double loss=x*(p-finalfp);
	        sum+=loss;
	    }
	    cout<<fixed<<sum<<endl;
	}
	return 0;
}

Sale Season Codechef Solution in Python

for testcases in range(int(input())):
    a=int(input())
    count=0
    for i in range(a):
        a,b,c=map(int,input().split())
        d=a*b
        a+=(0.01*c*a)
        a-=(0.01*a*c)
        a*=b
        count+=(d-a)
    print(float(count))

Disclaimer: The above Problem (Sale Season ) is generated by CodeChef but the solution is provided by  Chase2learn.This tutorial is only for Educational and Learning purpose.

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