codeforces 603B Moodular Arithmetic

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本文探讨了一个有趣的数学问题：对于固定的整数k和奇质数p，如何计算满足f(kx%p) = kf(x)%p的所有可能函数f的数量，并提供了解决方案及代码实现。

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http://www.elijahqi.win/archives/2942
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

for some function . (This equation should hold for any integer x in the range 0 to p - 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.

Examples
Input

Copy
3 2
Output
3
Input

Copy
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.
非常有趣的一个数学题感觉非常好玩就来研究了下

f(kx%p)=kf(x)%p给出这样一个方程求集合A—>B的映射有几种情况两个集合都是0~p-1的

那么可以知道k=0的时候除了为0的情况其他都是随便对应p^(p-1)

k=1是全部都可以随便对应p^p

反之我们一定可以找一个循环节出来k^m一定会随着m增大在%p意义下重新变成相同的一个序列那么根据费马小定理可知道x^(p-1)≡1(mod p)p为质数这个m一定是p-1的因子

每个环起始位置p种可能一共(p-1)/m个环所以方案数是p^((p-1)/m)

好啦不玩了辣鸡蒟蒻elijahqi该睡觉了..

#include<cstdio>
#include<cctype>
#include<algorithm>
#define ll long long
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=gc();}
    while(isdigit(ch)) x=x*10+ch-'0',ch=gc();
    return x*f;
}
const int mod=1e9+7;
int p,k;
inline int ksm(ll b,int t){
    ll tmp=1;for (;t;b=b*b%mod,t>>=1) if (t&1) tmp=tmp*b%mod;return tmp;
}
int main(){
    freopen("cf603b.in","r",stdin);
    p=read();k=read();
    if (!k) {printf("%d\n",ksm(p,p-1));return 0;}
    if (k==1) {printf("%d\n",ksm(p,p));return 0;}
    ll tmp=1;int cnt=1;
    for (int i=1;i<=p;++i) {tmp*=k,tmp%=p;if (tmp==1) break;++cnt;}
    printf("%d\n",ksm(p,(p-1)/cnt));
    return 0;
}