uva 11255 - Necklace(置换)

题目链接:uva 11255 - Necklace

题目大意：给定3种颜色的珠子个数，要求所有的珠子都用上的情况下有多少种不同的项链，旋转翻转视为同一种。

解题思路：等价类的计数，polya。

• 旋转：有0，1，~ n-1步。
• 翻转：考虑n为奇数偶数，奇数下，有n条对称轴（过一点）偶数时，有n/2条过两点，n/2条不过点。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 40;

int N, col[3], u[3];
ll c[maxn+5][maxn+5];

void init () {
    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];
    }
}

int gcd (int a, int b) {
    return b == 0 ? a : gcd (b, a % b);
}

ll solve (int k) {
    int n = 0;
    ll ret = 1;

    for (int i = 0; i < 3; i++) {
        if (u[i] % k)
            return 0;
        u[i] /= k;
        n += u[i];
    }

    for (int i = 0; i < 3; i++) {
        ret *= c[n][u[i]];
        n -= u[i];
    }
    return ret;
}

ll polya () {
    ll ret = 0;

    for (int i = 0; i < N; i++) {
        memcpy(u, col, sizeof(col));
        ret += solve(N/gcd(i, N));
    }

    if (N&1) {

        for (int i = 0; i < 3; i++) {
            if (col[i]) {
                memcpy(u, col, sizeof(col));
                u[i]--;
                ret += N * solve(2);
            }
        }
    } else {

        memcpy(u, col, sizeof(col));
        ret += N/2 * solve(2);

        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                if (col[i] && col[j]) {
                    memcpy(u, col, sizeof(col));
                    u[i]--; u[j]--;
                    ret += N/2 * solve(2);
                }
            }
        }
    }
    return ret / 2 / N;
}

int main () {
    init();

    int cas;
    scanf("%d", &cas);
    while (cas--) {
        N = 0;
        for (int i = 0; i < 3; i++) {
            scanf("%d", &col[i]);
            N += col[i];
        }
        printf("%lld\n", polya());
    }
    return 0;
}