hdu 1220 容斥

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): 2260    Accepted Submission(s): 1819

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.

     

       对于一个N*N*N的立方体，问有几对小立方体满足公共点个数<=2,由于题目中说了这个数目只有三种情况0/2/4，我们可以让总组合数减去交点是4的对数就是答案，全部方案个数就是C(N*N*N,2)

有四个公共顶点也就是说有公共面，可能位于水平线上或者垂线上或者纵线上，类似于xyz轴，显然对于每一行/列/垂线上都有(N-1)对，这个结果就是3*(N-1)*N*N,因为有N*N个x/y/z。

  

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
#define LL long long
LL qpow(LL a,LL b){LL r=1;for(;b;b>>=1,a=a*a)if(b&1)r=r*a;return r;}
int main()
{
    LL n;
    while(cin>>n)
        cout<<((n*n*n)*(n*n*n-1)/2-3*(n-1)*n*n)<<endl;
    return 0;
}