Chef and Price Control||codechef||beginner solution ||programming info

Chef and Price Control

Chef has

N

items in his shop (numbered

1

through

N

); for each valid

i

, the price of the

i

-th item is

P_{i}

. 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 $P_{i} > K$ its price should be reduced from $P_{i}$ 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 $P_{1}, P_{2}, \dots, P_{N}$

Output

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

Constraints

$1 \leq T \leq 100$
$1 \leq N \leq 10, 000$
$1 \leq P_{i} \leq 1, 000$ for each valid $i$
$1 \leq K \leq 1, 000$

Subtasks

Subtask #1 (100 points): original constraints

Example 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

Example Output

7
0
15

Explanation

Test Case 1: The initial revenue is $10 + 2 + 3 + 4 + 5 = 24$ . Because of the price ceiling, $P_{1}$ decreases from $10$ to $4$ and $P_{5}$ 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 $i$ , $P_{i} \leq 15$ , 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 $P_{i} > 5$ for each valid $i$ , the prices of all items decrease to $5$ . The revenue afterwards is $5 \cdot 5 = 25$ and the lost revenue is $40 - 25 = 15$ .

SOLUTION USING C LANGUAGE:-

#include<stdio.h>
int main()
{
	int t,i;
	scanf("%d",&t);
	int n,k,p[10000];
	while(t--)
	{
	    int s=0,l=0;
		scanf("%d %d",&n,&k);
	for(i=0;i<n;i++)
	{
		scanf("%d\t",&p[i]);
	}
	for(i=0;i<n;i++)
	{
		s=s+p[i];
	}
	for(i=0;i<n;i++)
	{
		if(p[i]>k)
		p[i]=k;
	}
	for(i=0;i<n;i++)
	{
	    l=l+p[i];
	}
		printf("%d\n",(s-l));
	
		}
}