Ultraviolet A 10601-cubes (replacement)

Link to the question: Ultraviolet A 10601-cubes

There are 12 equi-long wooden sticks, and then each wooden stick enters the color number of each wooden stick. Your task is to figure out how many different cubes are spelled out with them, the identical cubes after rotation are determined to be the same.

Solution: polyA, and then the corresponding cube has 24 kinds of rotation:

• Do not rotate (still): 1 type, cycle length is 12
• Take the pair of vertices as the axis (rot_point): 4 groups, cycle length is 3
• Take the center of the opposite side as the axis (rot_plane): three groups with 90,180,270 degrees of rotation respectively, corresponding to the cycle length 3, 2, 3
• Take the opposite side as the axis (rot_edge): 6 groups, except that the cycle length of the two sides is 1, and the other is 2.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 12;int u[maxn+5], rod[maxn+5];ll C[maxn+5][maxn+5];void init () {    memset(C, 0, sizeof(C));    for (int i = 0; i <= maxn; i++) {        C[i][0] = C[i][i] = 1;        for (int j = 1; j < i; j++)            C[i][j] = C[i-1][j-1] + C[i-1][j];    }}ll solve (ll k) {    int n = 0;    ll ret = 1;    for (int i = 0; i < 6; i++) {        if (u[i] % k)            return 0;        u[i] /= k;        n += u[i];    }    for (int i = 0; i < 6; i++) {        ret *= C[n][u[i]];        n -= u[i];    }    //printf("%lld %lld!!\n", k, ret);    return ret;}ll still () {    memcpy(u, rod, sizeof(rod));    return solve(1);}ll rot_point () {    memcpy(u, rod, sizeof(rod));    return 4 * 2 * solve(3);}ll rot_edge () {    ll ret = 0;    for (int i = 0; i < 6; i++) {        for (int j = 0; j < 6; j++) {            if (rod[i] && rod[j]) {                memcpy(u, rod, sizeof(rod));                u[i]--; u[j]--;                ret += 6 * solve(2);            }        }    }    return ret;}ll rot_plane () {    ll ret = 0;    memcpy(u, rod, sizeof(rod));    ret += solve(4) * 2 * 3;    memcpy(u, rod, sizeof(rod));    ret += solve(2) * 3;    return ret;}inline ll polya () {    return still() + rot_point() + rot_edge() + rot_plane();}int main () {    init();    int cas, x;    scanf("%d", &cas);    while (cas--) {        memset(rod, 0, sizeof(rod));        for (int i = 0; i < maxn; i++) {            scanf("%d", &x);            rod[x-1]++;        }        printf("%lld\n", polya() / 24);    }    return 0;}