UVA 11255 - Necklace(Ploya)

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

;

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;
}