Commentary Boxes CodeForces - 990A（水题）

最新推荐文章于 2019-08-09 19:23:31 发布

Levi——

最新推荐文章于 2019-08-09 19:23:31 发布

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解决足球赛事中评论箱分配问题，确保每个代表团获得相同数量的评论箱。通过新建或拆除评论箱来达到总数能被代表团数量整除的目标。

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Berland Football Cup starts really soon! Commentators from all over the world come to the event.

Organizers have already built $n$ commentary boxes. $m$ regional delegations will come to the Cup. Every delegation should get the same number of the commentary boxes. If any box is left unoccupied then the delegations will be upset. So each box should be occupied by exactly one delegation.

If $n$ is not divisible by $m$ , it is impossible to distribute the boxes to the delegations at the moment.

Organizers can build a new commentary box paying $a$ burles and demolish a commentary box paying $b$ burles. They can both build and demolish boxes arbitrary number of times (each time paying a corresponding fee). It is allowed to demolish all the existing boxes.

What is the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$ )?

Input

The only line contains four integer numbers $n$ , $m$ , $a$ and $b$ ( $1 \leq n, m \leq 10^{12}$ , $1 \leq a, b \leq 100$ ), where $n$ is the initial number of the commentary boxes, $m$ is the number of delegations to come, $a$ is the fee to build a box and $b$ is the fee to demolish a box.

Output

Output the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$ ). It is allowed that the final number of the boxes is equal to $0$ .

Examples
inputCopy
9 7 3 8
outputCopy
15
inputCopy
2 7 3 7
outputCopy
14
inputCopy
30 6 17 19
outputCopy
0

Note

In the first example organizers can build $5$ boxes to make the total of $14$ paying $3$ burles for the each of them.

In the second example organizers can demolish $2$ boxes to make the total of $0$ paying $7$ burles for the each of them.

In the third example organizers are already able to distribute all the boxes equally among the delegations, each one get $5$

boxes

题意：给出n，m，a，b。表示有n个休息室，m个代表团，每个代表团必须有同样的休息室，如果m不能被n整除的话，你可以拆除，或新建一些休息室，新建花费a，拆除花费b，问怎么使花费最少。

思路：比较第一个比m小的m的倍数和第一个比m大的倍数的花费即可。

#include "iostream"
#include "algorithm"
using namespace std;
typedef long long ll;
int main()
{
    ios::sync_with_stdio(false);
    ll n,m,a,b;
    cin>>n>>m>>a>>b;
    if(n%m==0)
        cout<<"0"<<endl;
    else{
        ll x=n/m;
        ll y=(n/m)+1;
        cout<<min((y*m-n)*a,(n-m*x)*b)<<endl;
    }
    return 0;
}