Codeforces Round #334 (Div. 2) 604D Modular Arithmetic(数学+快速幂)

本文深入探讨了一个在整数范围内关于特定模运算的函数方程式，着重于两个固定整数k和p（其中p为奇质数），通过解析函数特性与模运算的关系，揭示了满足该方程的不同函数数量，并特别关注了k的不同取值情况下的解决方案数量。通过计算和分析，得出了解决方案的数量在不同条件下变化的规律。

As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that

$\text{[math]}$

for some function $\text{[math]}$ . (This equation should hold for any integer x in the range 0 top - 1, inclusive.)

It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.

Input

The input consists of two space-separated integers p and k (3 ≤ p ≤ 1 000 000, 0 ≤ k ≤ p - 1) on a single line. It is guaranteed that p is an odd prime number.

Output

Print a single integer, the number of distinct functions f modulo 109 + 7.

   Sample test(s)
 
     input
   
3 2

     output
   
3

     input
   
5 4

     output
   
25

Note

In the first sample, p = 3 and k = 2. The following functions work:

f(0) = 0, f(1) = 1, f(2) = 2.
f(0) = 0, f(1) = 2, f(2) = 1.
f(0) = f(1) = f(2) = 0.

题目链接:点击打开链接

给出 k, p, 问满足f(k * x mod p) == k * f(x) mod p 的函数有多少个.

考虑情况从特殊到一般, k 为0时, 有 p ^ (p - 1)种函数. k 为1时, 有p ^ p 种函数. 其他情况时, (k ^ x) mod p会由x增大会变回k, 存在环, 环

的起点有p种可能, 共(p - 1) / ord个环, 故当前情况方案数为 p ^ ((p - 1) / ord).

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
ll p, k;
ll power_mod(ll a, ll b)
{
    ll res = 1;
    while(b) {
        if(b & 1) {
            res = res * a % MOD;
            p--;
        }
        b >>= 1;
        a = a * a % MOD;
    }
    return res;
}
int main(int argc, char const *argv[])
{
    scanf("%lld%lld", &p, &k);
    if(k == 0) printf("%lld\n", power_mod(p, p - 1));
    else if(k == 1) printf("%lld\n", power_mod(p, p));
    else {
        ll ord = 1, cur = k;
        for(; cur != 1; ++ord) 
            cur = cur * k % p;
        printf("%lld\n", power_mod(p, (p - 1) / ord));
    }
    return 0;
}