Chef and Price Control CodeChef Solution

Problem – Chef and Price Control CodeChef Solution

Chef has N items in his shop (numbered 1 through N); for each valid i, the price of the i-th item is Pi​. Since Chef has very loyal customers, all N items are guaranteed to be sold regardless of their price.

However, the government introduced a price ceiling K, which means that for each item i such that Pi​>K, its price should be reduced from Pi​ to K.

Chef’s revenue is the sum of prices of all the items he sells. Find the amount of revenue which Chef loses because of this price ceiling.

Input

  • The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
  • The first line of each test case contains two space-separated integers N and K.
  • The second line contains N space-separated integers P1​,P2​,…,PN​.

Output

For each test case, print a single line containing one integer ― the amount of lost revenue.

Constraints

  • 1≤T≤100
  • 1≤N≤10,000
  • 1≤Pi​≤1,000 for each valid i
  • 1≤K≤1,000

Subtasks

Subtask #1 (100 points): original constraints

Sample 1:

Input:
3
5 4
10 2 3 4 5
7 15
1 2 3 4 5 6 7
5 5
10 9 8 7 6
Output:
7
0
15

Explanation:

Test Case 1: The initial revenue is 10+2+3+4+5=24. Because of the price ceiling, P1​ decreases from 10 to 4 and P5​ decreases from 5 to 4. The revenue afterwards is 4+2+3+4+4=17 and the lost revenue is 24−17=7.

Test Case 2: The initial revenue is 1+2+3+4+5+6+7=28. For each valid iPi​≤15, so there are no changes, the revenue after introduction of the price ceiling is the same and there is zero lost revenue.

Test Case 3: The initial revenue is 10+9+8+7+6=40. Since Pi​>5 for each valid i, the prices of all items decrease to 5. The revenue afterwards is 5⋅5=25 and the lost revenue is 40−25=15.

Chef and Price Control CodeChef Solution in C++17

#include <iostream>
using namespace std;

int main() {
	// your code goes here
	int T;
	std::cin >> T;
	while(T--){
	    int N,K;
	    cin>>N>>K;
	    int P[N];
	    int PRE=0,F=0;
	    for(int i=0; i<N; i++){
	       cin>>P[i];
	       PRE+=P[i];
	       if(P[i]>K){
	           P[i]=K;
	       }
	       F+=P[i];
	    }
	    std::cout << PRE-F << std::endl;
	}
	return 0;
}

Chef and Price Control CodeChef Solution in Pyth 3

t = int(input())
for tc in range(t):
    n,k=map(int,input().split())
    p=list(map(int,input().split()))
    q=sum(p)
    for i in range(len(p)):
        if p[i]>=k:
            p[i]=k
        else:
            pass
    r=sum(p) 
    print(q-r)

Chef and Price Control 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
	{
		// your code goes here
		Scanner in=new Scanner(System.in);
		int t=in.nextInt();
		while(t-->0){
		    int n=in.nextInt();
		    int k=in.nextInt();
		    int[] arr=new int[n];
		    for(int i=0;i<n;i++){
		        arr[i]=in.nextInt();
		    }
		    int count=0;
		    for(int i=0;i<n;i++){
		        if(arr[i]>k){
		            count+=(arr[i]-k);
		        }
		    }
		    System.out.println(count);
		}
	}
}
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