本文主要是介绍UVA 11255 - Necklace(Ploya)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
UVA 11255 - Necklace
题目链接
题意：一个链子，由三种颜色的珠子构成，现在给定三种颜色的珠子个数，求能组成多少种（旋转，翻转算同一种）
思路：利用ploya定理，然后分类讨论即可
代码：
#include <cstdio>
#include <cstring>typedef long long ll;
const int N = 45;int t, a, b, c, n;
ll C[N][N];void getC() {for (int i = 0; i <= 40; 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) {while (b) {int tmp = b;b = a % b;a = tmp;}return a;
}ll cal(int a, int b, int c, int d, int len) {if (a < 0 || b < 0 || c < 0) return 0;if (a % len || b % len || c % len) return 0;int az = a / len, bz = b / len;return C[d][az] * C[d - az][bz];
}int main() {getC();scanf("%d", &t);while (t--) {scanf("%d%d%d", &a, &b, &c);n = a + b + c;ll ans = 0;for (int i = 0; i < n; i++) {int d = gcd(i, n);int len = n / d;ans += cal(a, b, c, d, len);}if (n&1) {ans += cal(a - 1, b, c, n / 2, 2) * n;ans += cal(a, b - 1, c, n / 2, 2) * n;ans += cal(a, b, c - 1, n / 2, 2) * n;}else {ans += cal(a, b, c, n / 2, 2) * (n / 2);ans += cal(a - 1, b - 1, c, n / 2 - 1, 2) * n;ans += cal(a - 1, b, c - 1, n / 2 - 1, 2) * n;ans += cal(a, b - 1, c - 1, n / 2 - 1, 2) * n;ans += cal(a - 2, b, c, n / 2 - 1, 2) * (n / 2);ans += cal(a, b - 2, c, n / 2 - 1, 2) * (n / 2);ans += cal(a, b, c - 2, n / 2 - 1, 2) * (n / 2);}ans /= 2 * n;printf("%lld\n", ans);}return 0;
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