All 6 sides of a cube are to becoated with paint. Each side is is coated uniformly with one color. When a selectionof n different colors of paint is available, how many different cubes can youmake? Note that any two cubes are onlyto be cal

All 6 sides of a cube are to becoated with paint. Each side is is coated uniformly with one color. When a selectionof n different colors of paint is available, how many different cubes can youmake?

Note that any two cubes are onlyto be called "different" if it is not possible to rotate the one intosuch a position that it appears with the same coloring as the other.

Input

Each line of the input filecontains a single integer n(0denoting the number of different colors. Input is terminated by a line wherethe value ofn=0. This line shouldnot be processed.

Output

For each line of input produce oneline of output. This line should contain the number of different cubes that canbe made by using the according number of colors.

SampleInput Outputfor Sample Input

Problem setter: EricSchmidt

Special Thanks: DerekKisman, EPS

题意：求用n中颜色涂立方体的不同种数，能旋转到的算一种

题意：和上一题UVA - 10601 Cubes （组合+置换） 的立方体旋转考虑的分类是一样的，不过这里我们考虑的是涂面的情况

1.不变置换(1)(2)(3)(4)(5)(6), 共1个;

2.沿对面中心轴旋转 90度, 270度 (1)(2345)(6), (1)(5432)(6) 同类共 6个;

3.沿对面中心轴旋转 180度 (1)(24)(35)(6), 同类共 3个;

4.沿对角线轴旋转 120度, 240度 (152)(346), (251)(643) 同类共 8个;

5.沿对边中点轴旋转 180度 (16)(25)(43) 同类共 6个;

<strong>[cpp]</strong> view plaincopy

1. #include <iostream></iostream>

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2. #include <cstdio></cstdio>

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3. #include <cstring></cstring>

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4. #include <cmath></cmath>

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5. #include <algorithm></algorithm>

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6. typedef long long ll;

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7. using namespace std;

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8. ll n;

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9. ll still() {

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10. return n * n * n * n * n * n;

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11. }

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12. ll point() {

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13. return 4 * 2 * n * n;

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14. }

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15. ll edge() {

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16. return 6 * n * n * n;

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17. }

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18. ll plane() {

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19. return 3 * 2 * n * n * n + 3 * n * n * n * n;

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20. }

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21. ll polya() {

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22. ll ans = 0;

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23. ans += still();

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24. ans += point();

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25. ans += edge();

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26. ans += plane();

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27. return ans / 24;

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28. }

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29. int main() {

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30. while (scanf("%lld", &n) != EOF && n) {

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31. printf("%lld\n", polya());

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32. }

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33. return 0;

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34. }

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