HDU 1220 Cube (组合数学)

最新推荐文章于 2018-05-31 19:31:01 发布

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最新推荐文章于 2018-05-31 19:31:01 发布

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分类专栏： ACM_组合数学 HDUOJ 文章标签： HDU 1220 Cube 组合数学

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针对一个给定边长的正方体，将其切割成单位体积的小正方体，并计算所有公共顶点数不超过2的小正方体对数。通过组合数学原理推导公式并给出AC代码。

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Cube

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1853 Accepted Submission(s): 1482

Problem Description

Cowl is good at solving math problems. One day a friend asked him such a question: You are given a cube whose edge length is N, it is cut by the planes that was paralleled to its side planes into N * N * N unit cubes. Two unit cubes may have no common points or two common points or four common points. Your job is to calculate how many pairs of unit cubes that have no more than two common points.

Process to the end of file.

Input

There will be many test cases. Each test case will only give the edge length N of a cube in one line. N is a positive integer(1<=N<=30).

Output

For each test case, you should output the number of pairs that was described above in one line.

Sample Input

Sample Output

  
   0
16
297


   
    
     Hint
    
    Hint
   
    
The results will not exceed int type.

Author

Gao Bo

Source

杭州电子科技大学第三届程序设计大赛

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题意：给你一个正方体，切割成单位体积的小正方体，求所有公共顶点数<=2的小正方体的对数。

题解：因为公共点的数目只可能有：0,1,2,4.
很明显我们用总的对数减掉有四个公共点的对数就可以了。
总的公共点对数：C（2，n^3）= n^3*(n^3-1)/2(一共有n^3块小方块，从中选出2块)。
因为（只有两个小方块之间才存在公共点，我们从所有的小方块中任意选出两个，自然就确定了这两个小方块的公共点的对数，从所有小方块中任意选取两个，所以总的选取方法数就是所有种类对数数目的总和！）
公共点为4的对数：一列有n-1对（n个小方块，相邻的两个为一对符合要求），
一个面的共有 n^2列，选上面和左面，前面三个方向，同理可得，故总数为：3*n^2(n-1)。

综上：
所以公数学式为：(n^3*(n^3-1))/2 - 3*n^2(n-1)。

AC代码:

#include<stdio.h>
#include<iostream>
int main()
{
	int n;
	while(std::cin>>n)
	{
		std::cout<<(n*n*n*(n*n*n-1))/2-3*(n*n)*(n-1)<<std::endl;
	}
	return 0;
}