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/*
>>~~ UVa Online Judge ACM Problem Solution ~~<<
ID: 10006
Name: Carmichael Numbers
Problem: https://onlinejudge.org/external/100/10006.pdf
Language: C++
Author: Arash Shakery
Email: arash.shakery@gmail.com
*/
#include <math.h>
#include <stdio.h>
#include <cstring>
#include <iostream>
using namespace std;
#define N 66000
unsigned int prime[N / 64];
#define gP(n) (prime[n>>6]&(1<<((n>>1)&31)))
void sieve() {
memset(prime, -1, sizeof(prime));
unsigned int i, i2, sqrtN = (unsigned int)sqrt((double)N) + 1;
for(i = 3; i<sqrtN; i+=2)
if(gP(i)) {
i2 = i + i;
for(unsigned int j = i*i; j<N; j+= i2)
prime[j>>6] &= ~(1<<((j>>1)&31));
}
}
bool isPrime(int n) {
return n==2 || ((n&1) && gP(n));
}
int powmod(int b, int p, int m) {
if (!p) return 1;
if (p==1) return b%m;
long long int h = powmod(b, p>>1, m);
return ((p&1) ? h*((b*h)%m) : h*h)%m;
}
bool fermat(int n, int a) {
return powmod(a, n, n) == a;
}
bool isC[65001];
bool isCarmichael(int n) {
if (isPrime(n)) return false;
for(int i=2; i<n; i++)
if (!fermat(n, i))
return false;
return true;
}
int main(){
sieve();
/*
int cnt=0;
for (int i=0; cnt<120 && i<65000; i++)
if (isCarmichael(i))
printf("#%d: %d\n", ++cnt, i);
*/
int n;
while(cin>>n && n)
printf(isCarmichael(n) ? "The number %d is a Carmichael number.\n" : "%d is normal.\n", n);
}
// high probablity: sum of digits is zero
/*
#include <stdio.h>
#include <iostream>
using namespace std;
bool isC[65001];
int main(){
int n;
isC[561]=isC[1105]=isC[1729]=isC[2465]=isC[2821]=isC[6601]=
isC[8911]=isC[10585]=isC[15841]=isC[29341]=isC[41041]=
isC[46657]=isC[52633]=isC[62745]=isC[63973]=true;
while(cin>>n && n)
printf(isC[n] ? "The number %d is a Carmichael number.\n" : "%d is normal.\n", n);
}
*/
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