Chef in Vaccination Queue codechef solution in c++

 

Chef in Vaccination Queue Problem Code: VACCINQSubmit

There are N$N$ people in the vaccination queue, Chef is standing on the Pth$P^{th}$ position from the front of the queue. It takes X$X$ minutes to vaccinate a child and Y$Y$ minutes to vaccinate an elderly person. Assume Chef is an elderly person.

You are given a binary array A1,A2,…,AN$A_1,A_2,\dots ,A_N$ of length N$N$, where Ai=1$A_i=1$ denotes there is an elderly person standing on the ith$i^{th}$ position of the queue, Ai=0$A_i=0$ denotes there is a child standing on the ith$i^{th}$ position of the queue. You are also given the three integers P,X,Y$P,X,Y$. Find the number of minutes after which Chef's vaccination will be completed.

Input Format

• First line will contain T$T$, number of testcases. Then the testcases follow.
• The first line of each test case contains four space-separated integers N,P,X,Y$N,P,X,Y$.
• The second line of each test case contains N$N$ space-separated integer A1,A2,…,AN$A_1,A_2,\dots ,A_N$.

Output Format

For each testcase, output in a single line the number of minutes after which Chef's vaccination will be completed.

Constraints

• 1≤T≤100$1\le T\le 100$
• 1≤N≤100$1\le N\le 100$
• 1≤P≤N$1\le P\le N$
• 1≤X,Y≤10$1\le X,Y\le 10$
• 0≤Ai≤1$0\le A_i\le 1$
• AP=1$A_P=1$

Sample Input 1 

3
4 2 3 2
0 1 0 1
3 1 2 3
1 0 1
3 3 2 2
1 1 1

Sample Output 1 

5
3
6

Explanation

Test case 1$1$: The person standing at the front of the queue is a child and the next person is Chef. So it takes a total of 3+2=5$3+2=5$ minutes to complete Chef's vaccination.

Test case 2$2$: Chef is standing at the front of the queue. So his vaccination is completed after 3$3$ minutes.

Test case 3$3$: Chef is standing at the rear of the queue. So it takes a total of 2+2+2=6$2+2+2=6$ minutes to complete Chef's vaccination.

solution in c++

#include<iostream>
#include<algorithm>
using namespace std;

class solution
{
public:
	void solve()
	{
		int n, p, x, y;
		cin >> n >> p >> x >> y;
		int* arr = new int[n];
		for (int i = 0; i < n; i++)
		{
			cin >> arr[i];
		}
		int total = 0;
		for (int i = 0; i < p; i++)
		{
			if (arr[i] == 1)
				total += y;
			else
				total += x;
		}
		cout << total << "\n";
	}
};
int main()
{
	solution ss;

	int t;
	cin >> t;
	while (t--)
    {
		ss.solve();
	}

	return 0;
}