HDU 1220 Cube（数学，找规律）

传送门：

http://acm.hdu.edu.cn/showproblem.php?pid=1220

Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2627    Accepted Submission(s): 2064

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

 

    分析：

    题意：

    有一个N*N*N的立方体，
    将其分成1*1*1的单位立方体，
    任意两个立方体的交点个数为0,1,2,4个，
    现在要求的是小于等于2个交点的对数有多少对。
    思路：
    总的对数：n*n*n的立方体可以分成n*n*n个单位立方体，
    所以一共有C（n*n*n，2）对 （即：n^3*(n^3-1)/2）。
    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）
    。一个面的共有 n^2列。底面，左面，前面三个方向相同。
    故总数为：3*n^2*(n-1)。

code：

#include <stdio.h>
#include <string.h>
#define  max_v 35
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int result=n*n*n*(n*n*n-1)/2-3*n*n*(n-1);
        printf("%d\n",result);
    }
    return 0;
}