A. Road To Zero
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
Input
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).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
input
2
1 3
391 555
0 0
9 4
output
1337
0
Note
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.
My Answer code:
/*
Author:Albert Tesla Wizard
Time:2020/12/9 11:09
*/
#include<bits/stdc++.h>
using namespace std;
using ull=unsigned long long;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
vector<vector<ull>>a(t+1,vector<ull>(2));
vector<vector<ull>>b(t+1,vector<ull>(2));
vector<ull>ans(t+1,0);
for(int i=1;i<=t;i++)cin>>a[i][0]>>a[i][1]>>b[i][0]>>b[i][1];
for(int i=1;i<=t;i++)
{
ull total1=(a[i][0]+a[i][1])*b[i][0],total2;
ull Min=min(a[i][0],a[i][1]);
ull Max=max(a[i][0],a[i][1]);
total2=Min*b[i][1]+(Max-Min)*b[i][0];
ans[i]=min(total1,total2);
}
for(int i=1;i<=t;i++)cout<<ans[i]<<'\n';
return 0;
}
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