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 << endl
const int N = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

typedef long long LL;
int p, k;
bool vis
;

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;
}