uva 10294 - Arif in Dhaka (First Love Part 2)(置换)
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题目链接:uva 10294 - Arif in Dhaka (First Love Part 2)
题目大意:项链和手镯都是由若珠子穿成的环形首饰,区别在于手镯可以翻转,但是项链不行。给定n和t,表示用t种颜色的n个珠子能制作的项链和手镯的个数。
解题思路:等价类计数,一共两种置换,旋转或者翻转。
- 旋转:枚举间距0,1,2,3…,n−1,所以不动点a=∑i=0n−1tgcd(n,i)
- 翻转:当n为奇数时,对称轴有n条,每条对称轴形成n−12个长度为2的循环和一个长度为1的循环,所以不动点b1=ntn+1/2;当n为偶数时,有两种对称轴,一种是穿过珠子的,一种是不穿过珠子的,都有n2条,但是形成的循环分别为n2−1个长度为2和两个长度为1,n2个长度为2,所以b2=n∗(tn/2+1+tn/2)2
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef unsigned long long ll;const int maxn = 50;int gcd (int a, int b) { return b == 0 ? a : gcd(b, a % b);}int main () { int n, t; ll p[maxn]; while (scanf("%d%d", &n, &t) == 2) { p[0] = 1; for (int i = 1; i <= n; i++) { p[i] = p[i-1] * t; printf("%lld\n", p[i]); } ll a = 0, b = 0; for (int i = 0; i < n; i++) a += p[gcd(n, i)]; if (n&1) b = n * p[(n+1)/2]; else b = n / 2 * (p[n/2+1] + p[n/2]); printf("%lld %lld\n", a / n, (a + b) / 2 / n); } return 0;}
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