UVA10294 Arif in Dhaka

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)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath> 
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(long long &a, long long &b)
{
    long long tmp = a;a = b;b = tmp;
}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const long long INF = 0x3f3f3f3f;

long long n,t;

long long pow(long long a, long long b)
{
    long long r = 1, base = a;
    for(;b;b >>= 1)
    {
        if(b & 1) r *= base;
        base *= base;
    }
    return r;
}

long long gcd(long long a, long long b)
{
    return !b ? a : gcd(b, a % b);
}

int main()
{
    while(scanf("%lld %lld", &n, &t) != EOF)
    {
        long long a = 0,b = 0; 
        for(register long long i = 1;i <= n;++ i) a += pow(t, gcd(i, n));
        if(n&1) b = n * pow(t, (n + 1) >> 1); 
        else b = n/2 * (pow(t, n/2) + pow(t, n/2 + 1));
        printf("%lld %lld
", a/n, (a + b) / (n <<  1));
    }
    return 0;
}