MRKT - EDITORIAL

general
,
1

PROBLEM LINK:

Contest

Practice

Author: Aditya Chandran

Tester: Aishwarya

Editorialist: Aishwarya

DIFFICULTY:

CAKEWALK.

PREREQUISITES:

Greedy, Math.

PROBLEM:

You are given an array of size 3, having weights of vegetables needed.

Next n lines are arrays prices per kilo of the respective vegetables.

Find minimum cost required to but the vegetables.

EXPLANATION:

Simply multiplying the prices of the vegetables to the weight required and adding this cost for all vegetables will give you money required to be sent in that particular market.

Whichever market gives you the minimum cost, is the correct answer.

cost = \min_{k=i}^{n}(cost,(w1*p_i1 + w2*p_i2 + w3*p_i3))

TIME COMPLEXITY:

O(N).

SOLUTIONS:

Setter's Solution 

import math

n=int(input())

w1,w2,w3=list(map(int,input().split()))

c=1e18

for i in range(n):

    p1,p2,p3=list(map(int,input().split()))

    cost=p1*w1+p2*w2+p3*w3

    c=min(c,cost)

print(c)
Tester's Solution

#include <bits/stdc++.h>

#include <iostream>

using namespace std;

int main() {

    // your code goes here

    int n;

    cin >> n;

    int w1,w2,w3;

    cin >> w1 >> w2 >> w3;

    int cost = INT_MAX;

    for(int i=0;i<n;i++)

    {

        int p1,p2,p3;

        cin >> p1 >> p2 >>p3;

        cost = min(cost,(w1*p1 + w2*p2 + w3*p3));

    }

    cout << cost << endl;

    return 0;

}