Codeforces Round #334 (Div. 2) 604D Modular Arithmetic(數學+快速冪)

D. Moodular Arithmetic

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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:

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

題目鏈接:點擊打開鏈接

給出 k, p, 問滿足f(k * x mod p) == k * f(x) mod p 的函數有多少個.

考慮情況從特殊到一般, k 爲0時, 有 p ^ (p - 1)種函數. k 爲1時, 有p ^ p 種函數. 其他情況時, (k ^ x) mod p會由x增大會變回k, 存在環, 環

的起點有p種可能, 共(p - 1) / ord個環, 故當前情況方案數爲 p ^ ((p - 1) / ord).

AC代碼:

#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;
}