杭电1220--Cube

Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1562    Accepted Submission(s): 1243

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

 

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。

 

分析：

公共点的数目只可能有：0,1,2,4. 用总的对数减掉有四个公共点的对数为结果;

总的公共点对数：n^3*(n^3-1)/2(一共有

n^3块小方块，从中选出2块

)

 ;

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，底面和左面，前面三个方向相同，同理可得，故总数为：3*n^2(n-1) ;
所以结果为：n^3 * (n^3-1) - 3*n^2(n-1);

#include <stdio.h>
int main()
{
    int n ;
    while(~scanf("%d", &n))
    {
        int num = n*n*n*(n*n*n-1)/2 - n*n*(n-1)*3 ;
        printf("%d
", num) ;
    }
    return 0 ;
}