hdu1220组合数学，求公共点小于2的面数对

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Cube

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1192 Accepted Submission(s): 938

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

Sample Output

这道题用到组合计数公式，，

Two unit cubes may have no common points or two common points or four common points

两个单位立方体的间的交点数的关系有0个，1个，2个，4个，题目要我们求不大于2个交点的立方体对，很明显求出所有的立方体对 - 有4个交点的立方体对，，

所有的立方体对好求 C(n 2) n为单位立方体的个数，而4个交点的立方体对是两个立方体共面的情况，所以我们只要求出大的立方体一共有多少个单位面积的公共面就可以了，既所有单位立方体的面数6*n^3减去在大立方体表面的面数6*n^2就可以了，最后除以二是因为那样的面公共，两个面就是一个面所以除以二；

所以最后的答案是：

n^3(n^3-1)/2 - (6*n^3-6*n^2)/2;

别人的思路：

解题思路：
这道题挺简单的：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。公共点的数目可能有：0,1,2,4.很明显我们用总的对数减掉有四个公共点的对数就可以了。总的对数：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<iostream>#include<cstdio>#include<cstring>#include<string>#include<cstdlib>#include<cmath>#include<algorithm>#include<iomanip>using namespace std;const int LEN=10005;int main(){    int n;    while( cin >> n )    {        cout<<n*n*n*(n*n*n-1)/2-(6*n*n*n-6*n*n)/2<<endl;    }    return 0;}

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