Sale Season Codechef Solution

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

Sale Season Codechef Solution
Sale Season Codechef Solution

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:

Sale Season Codechef Solution

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

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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