POJ2154 Color(Polya定理)

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

Source

POJ Monthly,Lou Tiancheng

Polya定理：

假设$G$是$p$个对象的一个置换群，用$m$种颜色涂染$p$个对象，则不同颜色的方案数为

$L = \frac{1}{|G|}\sum_{g_i \in G}m^{c(g_i)}$

$G = \{g_1, g_2, \dots g_s \}$，$c(g_i)$为置换$g_i$的循环节数

本题而言第$i$种置换的循环节数为$gcd(n, i)$

因此答案为$L = \frac{1}{n}\sum_{i = 1}^n n^{gcd(i, n}$

枚举约数，用欧拉函数计算，时间复杂度$O(T\sqrt(N) f(n))$，$f(n)$表示小于$\sqrt(n)$的质因子的个数

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
#define LL long long
const int MAXN = 1e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();}
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
    return x * f;
}
int T, N, mod;
int fastpow(int a, int p, int mod) {
    int base = 1; a %= mod;
    while(p) {
        if(p & 1) base = (base * a) % mod;
        a = (a * a) % mod; p >>= 1;
    }
    return base % mod;
}
int prime[MAXN], tot, vis[MAXN];
void Prime() {
    for(int i = 2; i <= MAXN - 10; i++) {
        if(!vis[i]) prime[++tot] = i;
        for(int j = 1; j <= tot && prime[j] * i <= MAXN - 10; j++) {
            vis[i * prime[j]] = 1;
            if(!i % prime[j]) break;
        }
    }
}
int phi(int x, int mod) {
    int limit , ans = x;
    for(int i = 1; i <= tot && prime[i] * prime[i] <= x; i++) {
        if(!(x % prime[i])) {
            ans = ans - ans / prime[i];
            while((x % prime[i]) == 0) x /= prime[i];
        }
    }
    if(x > 1) ans = ans - ans / x;
//    printf("%d", ans % mod);
    return ans % mod;
}
main() {
    T = read();
    Prime();
    while(T--) {
        N = read(); mod = read();
        int ans = 0, now = N;
        for(int d = 1; d * d<= N; d++) {
            if(d * d == N)
                ans = (ans + fastpow(N, d - 1, mod) % mod * phi(N / d, mod) % mod) % mod;
            else if( (N % d) == 0) {
                ans = (ans + fastpow(N, d - 1, mod) * phi(N / d, mod) + fastpow(N, N / d - 1, mod) * phi(d, mod)) % mod;
            }

            //printf("%d\n", ans);
        }
        //if(now > 0) ans += fastpow(N, now - 1, mod) * phi(N / now, mod);
        printf("%d\n", ans % mod);
    }
}

原文地址：https://www.cnblogs.com/zwfymqz/p/9293465.html