UVA10294 Arif in Dhaka

Posted 嘒彼小星

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题目大意:给定n颗珠子,用k种颜色对其进行染色,问染成项链(旋转相同,反转不同)和珠子(旋转相同,翻转相同)分别有多少种不同方法

 

旋转1..n颗数字,有n种置换。旋转i颗,有gcd(n,i)个循环, 该置换不动点数目为t^gcd(n,i)
n为奇数时,有n条对称轴,即n种置换,每种置换分为循环节为(n - 1) / 2的循环 和 循环节为1的循环

每种置换不动点的数目为t^((n + 1)/2)

n为偶数时,有n/2条对称轴不穿珠子分为(n/2个)循环节为2的置换,有n/2条对称轴穿过珠子分为((n-2)/2个)循环节为2的循环和(2个)循环为1的循环
每种置换不动点数目为t^(n/2) + t^(n/2 + 1)

 

技术分享图片
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <vector>
 8 #include <cmath> 
 9 #define min(a, b) ((a) < (b) ? (a) : (b))
10 #define max(a, b) ((a) > (b) ? (a) : (b))
11 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
12 inline void swap(long long &a, long long &b)
13 {
14     long long tmp = a;a = b;b = tmp;
15 }
16 inline void read(long long &x)
17 {
18     x = 0;char ch = getchar(), c = ch;
19     while(ch < 0 || ch > 9) c = ch, ch = getchar();
20     while(ch <= 9 && ch >= 0) x = x * 10 + ch - 0, ch = getchar();
21     if(c == -) x = -x;
22 }
23 
24 const long long INF = 0x3f3f3f3f;
25 
26 long long n,t;
27 
28 long long pow(long long a, long long b)
29 {
30     long long r = 1, base = a;
31     for(;b;b >>= 1)
32     {
33         if(b & 1) r *= base;
34         base *= base;
35     }
36     return r;
37 }
38 
39 long long gcd(long long a, long long b)
40 {
41     return !b ? a : gcd(b, a % b);
42 }
43 
44 int main()
45 {
46     while(scanf("%lld %lld", &n, &t) != EOF)
47     {
48         long long a = 0,b = 0; 
49         for(register long long i = 1;i <= n;++ i) a += pow(t, gcd(i, n));
50         if(n&1) b = n * pow(t, (n + 1) >> 1); 
51         else b = n/2 * (pow(t, n/2) + pow(t, n/2 + 1));
52         printf("%lld %lld\n", a/n, (a + b) / (n <<  1));
53     }
54     return 0;
55 }
UVA10294

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