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hdu 1220 Cube

2014-08-01 15:22 369 查看

Cube

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1339 Accepted Submission(s): 1057

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

HintHint 
The results will not exceed int type.


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)。

#include<iostream>
using namespace std; 
int main() 
{
     int a;
     while(scanf("%d",&a)!=EOF)
          printf("%d\n",a*a*a*(a*a*a-1)/2-3*a*a*(a-1));
     return 0;
}

                                            

                                        

                        
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