HDOJ题目1220 Cube（组合数学）

Cube

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

Total Submission(s): 1351    Accepted Submission(s): 1069


[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

HintHint
The results will not exceed int type.

 

[align=left]Author[/align]
Gao Bo
 

[align=left]Source[/align]
杭州电子科技大学第三届程序设计大赛

 

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思路：

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，，

ac代码

#include<stdio.h>
#include<math.h>
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
__int64
sum=pow(n,3)*(pow(n,3)-1)/2-(6*pow(n,3)-6*n*n)/2;
printf("%I64d\n",sum);
}
}