Cube（规律）

Cube

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1722 Accepted Submission(s): 1367

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, you should output the number of pairs that was described above in one line.

[align=left]Sample Input[/align]

1
2
3

[align=left]Sample Output[/align]

0
16
297

Hint

Hint

The results will not exceed int type.

题解：好菜，自己竟然没推出来。。。

总对数n^3*(n^3-1)/2;

点数为4的对数：一行有n-1对，一个面有总共n*n行，三个面面要乘以3；

总的减去4的就是<=2的了；

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
const int INF=0x3f3f3f3f;
int main(){
int n;
while(~scanf("%d",&n)){
printf("%d\n",n*n*n*(n*n*n-1)/2-n*n*(n-1)*3);
}
return 0;
}