Algorithm
You are given two integers x and y. You can perform two types of operations:
- Pay a dollars and increase or decrease any of these integers by 1. For example, if x=0 and y=7 there are four possible outcomes after this operation:
- x=0, y=6;
- x=0, y=8;
- x=−1, y=7;
- x=1, y=7.
- Pay b dollars and increase or decrease both integers by 1. For example, if x=0 and y=7 there are two possible outcomes after this operation:
- x=−1, y=6;
- x=1, y=8.
Your goal is to make both given integers equal zero simultaneously, i.e. x=y=0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
The first line contains one integer t (1≤t≤100) — the number of testcases.
The first line of each test case contains two integers x and y (0≤x,y≤109).
The second line of each test case contains two integers a and b (1≤a,b≤109).
For each test case print one integer — the minimum amount of dollars you have to spend.
2 1 3 391 555 0 0 9 4
1337 0
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391+555+391=1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl '\n'
#define debug(n) cout<<(n)<<endl;
const ll INF = 2e18 + 99;
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin>>t;
while(t--){
ll x, y, a, b;
cin>>x>>y>>a>>b;
if(2* a < b){
cout<<(a * x + y * a)<<endl;
continue;
}
else{
cout<<((max(x, y) - min(x, y)) * a + min(x, y) * b)<<endl;
}
}
}Input
1 3
391 555
0 0
9 4
Output
0
Demonstration
Codeforcess solution 1342-A A. Road To Zero, ,C++, Java, Js and Python, 1342-A,Codeforcess solution