【杭电oj】1220 - Cube（数学水题）

Cube

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1768 Accepted Submission(s): 1402

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

   1
2
3

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

观察就行了，数学题，不难。

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
	int n;
	long long oppo;		//相邻的立方体数 
	long long sum;		//总方案数 
	long long ans;
	while (~scanf ("%d",&n))
	{
		if (n == 1)
		{
			printf ("0\n");
			continue;
		}
		else if (n == 2)
		{
			printf ("16\n");
			continue;
		}
		oppo = 0;
		oppo += 2 * (3 * 4 + 4 * (n-2) * 4 + (n-2) * (n-2) * 5);
		oppo += (n-2) * (4 * 4 + 5 * (n-2) * 4 + (n-2) * (n-2) * 6);
		oppo /= 2;
		n = n * n * n;
		sum = (n-1) * n / 2;
		ans = sum - oppo;
		printf ("%ld\n",ans);
	}
	return 0;
}