Codeforces Round #334 D. Moodular Arithmetic（置换）

题意:

给定3≤p≤106的prime,0≤k≤p−1,求满足
f(kx mod p)≡kf(x) mod p
f:{0,1,2,...,p−1}→{0,1,2,...,p−1}
这样的函数有多少,答案mod (109+7)

分析:

这题−−看cf题解吧是群论的只是,不懂,看qscqesze的似懂非懂,然后题解看懂了k=0,k=1的特判
If k = 0, then the functional equation is equivalent to f(0) = 0. Therefore, pp − 1 functions satisfy this
because the values f(1), f(2), ..., f(p − 1) can be anything in {0, 1, 2, ..., p − 1}.

If k = 1, then the equation is just f(x) = f(x). Therefore pp functions satisfy this
because the values f(0), f(1), f(2), ..., f(p − 1) can be anything in {0, 1, 2, ..., p − 1}.
剩下的就是置换的环的个数−−pcnt,0不算

代码:

////  Created by TaoSama on 2015-12-02//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;int p, k;bool vis[N];LL ksm(LL x, int n) {    LL ret = 1;    for(; n; n >>= 1, (x *= x) %= MOD)        if(n & 1)(ret *= x) %= MOD;    return ret;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d%d", &p, &k) == 2) {        if(k == 0) printf("%lld\n", ksm(p, p - 1));        else if(k == 1) printf("%lld\n", ksm(p, p));        else {            int cnt = 0;            for(int i = 1; i < p; ++i) {                if(vis[i]) continue;                for(LL j = i; !vis[j]; j = k * j % p) vis[j] = true;                ++cnt;            }            printf("%lld\n", ksm(p, cnt));        }    }    return 0;}