poj 2154 c种颜色染n个格子，旋转算一种

 

Description

Beads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of the necklace can be produced. You should know that the necklace might not use up all the N colors, and the repetitions that are produced by rotation around the center of the circular necklace are all neglected.

You only need to output the answer module a given number P.

Input

The first line of the input is an integer X (X <= 3500) representing the number of test cases. The following X lines each contains two numbers N and P (1 <= N <= 1000000000, 1 <= P <= 30000), representing a test case.

Output

For each test case, output one line containing the answer.

Sample Input

5
1 30000
2 30000
3 30000
4 30000
5 30000

Sample Output

1
3
11
70
629

 

 //

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define maxn 36000
int n, mod, ans;
int prim[35000];
bool flag[maxn + 20];

void get_prim() {
    memset(flag, 0, sizeof (flag));
    for (int i = 2; i <= 1000; i++)if (!flag[i])
            for (int j = i * i; j <= maxn; j += i)flag[j] = true;
    for (int i = 2, k = 0; i <= maxn; i++)
        if (!flag[i])prim[k++] = i;
}

int eular(int n) {
    int i = 0, ans = 1;
    for (i = 0; prim[i] * prim[i] <= n; i++) {
        if (n % prim[i] != 0)continue;
        ans *= prim[i] - 1;
        n /= prim[i];
        while (n % prim[i] == 0) {
            ans *= prim[i];
            n /= prim[i];
        }
    }
    if (n > 1)ans *= n - 1;
    return ans % mod;
}

int f(int c, int k, int mod) {
    int ans = 1;
    c = c % mod;
    while (k) {
        if (k & 1)ans = (c * ans) % mod;
        k >>= 1;
        c = (c * c) % mod;
    }
    return ans;
}

int main() {
    get_prim();
    int i, T;
    scanf("%d", &T);
    while (T-- && scanf("%d%d", &n, &mod)) {
        ans = 0;
        for (i = 1; i * i <= n; i++) {
            if (i * i == n)//枚举循环长度l，找出相应的i的个数：gcd(i,n)=n/l.
                ans = (ans + f(n, i - 1, mod) * eular(i)) % mod;
            else if (n % i == 0)//有长度为l的循环，就会有长度为n/l的循环。
                ans = (ans + f(n, n / i - 1, mod) * eular(i) + eular(n / i) * f(n, i - 1, mod)) % mod;
        }
        printf("%d\n", ans);
    }
}