标签：hdu1220
Cube
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1372 Accepted Submission(s): 1085
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
0
16
297
The results will not exceed int type.
Author
Gao Bo
由于路计算了两个方向，所以结果需要/2
#include <stdio.h>
int main()
{
int n, a, arr[] = {0, 0, 16};
while(scanf("%d", &n) == 1){
if(n < 3){
printf("%d\n", arr[n]);
continue;
}
a = (n*n*n - 4) * 8;
a += (n*n*n - 5)*12*(n-2);
a += (n-2)*(n-2)*6*(n*n*n - 6);
a += (n-2)*(n-2)*(n-2)*(n*n*n-7);
printf("%d\n", a >> 1);
}
return 0;
}
HDU1220 Cube
标签：hdu1220
原文地址：http://blog.csdn.net/chang_mu/article/details/38893387
