置换群，burnside引理的题目。

这道题代表了一类题目的优化

裸的算法是 ∑n^(gcd(n,i)) 1<i<=n

复杂度过高，进行优化。

置换群种循环的个数L = n / gcd(n, i)

因为如果L | n, 则有n / L | n

则环的长度L的范围是1 ~ sqrt(L)

令a = gcd(n, i), 设i = at

则只有i与t互质的时候，gcd(i, t) = a

则最后可以优化为 ∑(Φ(i) * n ^ (i)) % p

###Code

POJ 2154
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//By Brickgao
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;
#define out(v) cerr << #v << ": " << (v) << endl
#define SZ(v) ((int)(v).size())
#define LL long long
const int maxint = -1u>>1;
template <class T> bool get_max(T& a, const T &b) {return b > a? a = b, 1: 0;}
template <class T> bool get_min(T& a, const T &b) {return b < a? a = b, 1: 0;}

long long ans;
int t, num, n, p;
int isprime[50001];
int prime[8001];

void getprime() {
    num = 0;
    for(int i = 2; i <= 50000; i ++) {
        if(!isprime[i]) {
            prime[num ++] = i;
            for(int j = 1; j * i <= 50000; j ++) {
                isprime[i * j] = 1;
            }
        }
    }
}

int euler(int x) {
    int res = x;
    for(int i = 0; i < num && prime[i] * prime[i] <= x; i++) {
        if(x % prime[i] == 0) {
            res = res / prime[i] * (prime[i] - 1);
            while(x % prime[i] == 0) {
                x /= prime[i];
            }
        }
    }
    if(x > 1) res = res / x * (x - 1);
    return res;
}

LL quickpow(LL m , LL n , LL k) {
    LL tmp = 1; 
    m %= k;
    while(n) { 
        if(n & 1)
            tmp = (tmp * m) % k; 
        m = (m * m) % k;
        n >>= 1;
    } 
    return tmp;
} 

int main() {
    getprime();
    scanf("%d", &t);
    while(t --) {
        ans = 0;
        scanf("%d%d", &n, &p);
        for(int i = 1; i < sqrt(n); i ++) {
            if(n % i == 0) {
                ans = (ans + euler(i) % p * quickpow(n, n / i - 1, p) + euler(n / i) % p * quickpow(n, i - 1, p)) % p;
                //cout << quickpow(n,n-1,p) << " " << quickpow(n, n / i - 1, p) << " " << euler(i) << " " << euler(n / i) << endl;
            }
        }
        if((int)sqrt(n) * (int)sqrt(n) == n) {
            ans = (ans + quickpow(n, sqrt(n) - 1, p) * (euler(sqrt(n)) % p)) % p;
        }
        ans %= p;
        cout << ans << endl;
    }
    return 0;
}