Cube HDU - 1220(数学)

最新推荐文章于 2021-07-29 09:54:18 发布

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本文介绍了一个数学问题的解决方法，该问题是计算给定边长为N的大立方体被切割成N*N*N个小立方体后，有多少对小立方体最多只共有点的情况。通过组合数学的方法减去所有共面的小立方体对，最终得出答案。

Cube

HDU - 1220

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

1
2
3

Sample Output

0
16
297


        
  
The results will not exceed int type.

Hint

Hint
        
 思路：在所有的立方体中任选2个是所有的情况，减去四个公共公共点的情况

总的种类数是C（n*n*n，2） = n*n*n*（n*n*n-1）/2

四个公共点即有一个公共面，那把立方体的所有面数减去外表面的面数，6*n*n*n-6*n*n就是有公共面的立方体数，然后除以2就是有四个公共点的对数（6*n*n*n-6*n*n）/2

然后两个式子相减就是答案

code：

#include <iostream>
using namespace std;
int main(){
    int n,m;
	while(~scanf("%d",&n)){
		m = n*n*n*(n*n*n-1)/2 - (6*n*n*n-6*n*n)/2;
		printf("%d\n",m);
	}	
	return 0;
}