Uva 10006-Carmichael Numbers（快速幂）

题目链接：点击打开链接

题意：给一个数n，问这个数是不是Carmichael Numbers，Carmichael Numbers的定义为：一个数n如果不是素数且对于对于任意的    2=<a<n 都满足 a^n%n=a,那么这个数就是Carmichael Numbers，否则不是。快速幂暴力解决。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cctype>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define maxn 65002
#define _ll __int64
#define ll long long
#define INF 0x3f3f3f3f
#define Mod 10000007
#define pp pair<int,int>
#define ull unsigned long long
using namespace std;
ll n;
bool pri[maxn];
void init()
{
	memset(pri, 1, sizeof(pri));
	pri[0] = 0;
	pri[1] = 0;

	for (int i = 2; i * i <= maxn; i++) {
		if (pri[i]) {
			for (int j = i * i; j <= maxn; j += i) {
				pri[j] = 0;
			}
		}
	}
}
ll pow_mod(ll a, ll n, ll p)
{
	if (n == 0) {
		return 1;
	}

	ll ans = pow_mod(a, n / 2, p);
	ans = ans * ans % p;

	if (n & 1) {
		ans = ans * a % p;
	}

	return ans;
}
void solve()
{
	if (pri[n]) {
		printf("%d is normal.\n", n);
		return ;
	}

	for (ll i = 2; i < n; i++) {
		if (pow_mod(i, n, n) != i) {
			printf("%d is normal.\n", n);
			return ;
		}
	}

	printf("The number %d is a Carmichael number.\n", n);
}
int main()
{
	init();

	while (~scanf("%lld", &n) && n) {
		solve();
	}

	return 0;
}