HDU 1220 Cube

Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1772    Accepted Submission(s): 1406

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

123

 

Sample Output

016297
Hint
Hint
 The results will not exceed int type.

 

Author

Gao Bo

 

Source

杭州电子科技大学第三届程序设计大赛

 

Recommend

Ignatius.L   |   We have carefully selected several similar problems for you:  1221 1219 1225 1222 1224 

题意：一个n*n*n的立方体，被切割成1*1*1的小块。两两配对，问两块公共点不超过2的对数。先算出总对数，减去共面的对数。顶点8个，每个共面3对，棱12条，每条有n-2个方块，每个方块共面4对，非顶点非楞的面上小方块，每个共面5对，一共6个面，每个面有(n-2)^2个这样的小方块。剩下的内部小方块每个共面6对。最后总共面数要除以2，因为每对算了2次。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int init(int n){int a,b,c,d;a=8;b=(n-2)*12;c=(n-2)*(n-2)*6;d=n*n*n-a-b-c;return a*3+b*4+c*5+d*6;}int main(){int n,sum;while(~scanf("%d",&n)){if(n==1){printf("0\n");continue;}else{sum=n*n*n;sum=sum*(sum-1)/2;printf("%d\n",sum-init(n)/2);}}return 0;}