A. Golden Plate

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最新推荐文章于 2020-08-20 19:13:48 发布

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本文介绍了一种算法，用于计算在一个矩形盘子上，按照特定规则镀金的单元格总数。盘子被划分为w×h个单元格，需要镀k层金环，每层金环距离边缘递增2个单元格，宽度为1个单元格。文章提供了输入输出样例及代码实现。

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You have a plate and you want to add some gilding to it. The plate is a rectangle that we split into w×hw×h cells. There should be kk gilded rings, the first one should go along the edge of the plate, the second one — 22 cells away from the edge and so on. Each ring has a width of 11 cell. Formally, the ii-th of these rings should consist of all bordering cells on the inner rectangle of size (w−4(i−1))×(h−4(i−1))(w−4(i−1))×(h−4(i−1)).

The picture corresponds to the third example.
Your task is to compute the number of cells to be gilded.

Input
The only line contains three integers ww, hh and kk (3≤w,h≤1003≤w,h≤100, 1≤k≤⌊min(n,m)+14⌋1≤k≤⌊min(n,m)+14⌋, where ⌊x⌋⌊x⌋ denotes the number xx rounded down) — the number of rows, columns and the number of rings, respectively.

Output
Print a single positive integer — the number of cells to be gilded.

Examples
inputCopy
3 3 1
outputCopy
8
inputCopy
7 9 1
outputCopy
28
inputCopy
7 9 2
outputCopy
40
Note
The first example is shown on the picture below.
The second example is shown on the picture below.
The third example is shown in the problem description.

#include<iostream>
using namespace std;

int main()
{
    int w, h, k;
    while(cin >> w >> h >> k)
    {
        int sum = 0;
        for(int i = 0; i < k; ++i)
        {
            sum += 2 * (w + h) - 4;
            w -= 4;
            h -= 4;
        }
        cout << sum << '\n';
    }
    return 0;
}