Carmichael Numbers - PC110702

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原文地址：http://www.milkcu.com/blog/archives/uva10006.html

原创：Carmichael Numbers - PC110702

作者：MilkCu

题目描述

 Carmichael Numbers 

An important topic nowadays in computer science is cryptography. Some people even think that cryptography is the only important field in computer science, and that life would not matter at all without cryptography.
 Alvaro is one of such persons, and is designing a set of cryptographic procedures for cooking paella. Some of the cryptographic algorithms he is implementing make use of big prime numbers. However, checking if a big number is prime is not so easy. An exhaustive
approach can require the division of the number by all the prime numbers smaller or equal than its square root. For big numbers, the amount of time and storage needed for such operations would certainly ruin the paella.

However, some probabilistic tests exist that offer high confidence at low cost. One of them is the Fermat test.

Let a be a random number between 2 and n - 1 (being n the number whose primality we are testing). Then, n is probably prime if the following equation holds:

If a number passes the Fermat test several times then it is prime with a high probability.

Unfortunately, there are bad news. Some numbers that are not prime still pass the Fermat test with every number smaller than themselves. These numbers are called Carmichael numbers.

In this problem you are asked to write a program to test if a given number is a Carmichael number. Hopefully, the teams that fulfill the task will one day be able to taste a delicious portion of encrypted paella.
As a side note, we need to mention that, according to Alvaro, the main advantage of encrypted paella over conventional paella is that nobody but you knows what you are eating.

Input

The input will consist of a series of lines, each containing a small positive number n ( 2
< n < 65000). A number n = 0 will
mark the end of the input, and must not be processed.

Output

For each number in the input, you have to print if it is a Carmichael number or not, as shown in the sample output.

Sample Input

1729
17
561
1109
431
0

Sample Output

The number 1729 is a Carmichael number.
17 is normal.
The number 561 is a Carmichael number.
1109 is normal.
431 is normal.

---

Miguel Revilla 

2000-08-21

解题思路

Carmichael数肯定是个合数，且对于所有a都满足a^n mod n = a。



根据题目，按部就班的做。



在对乘方求模的时候可以使用递归的方法，减少计算时间：

(a mod n) ^ p mod n = ((a mod n) ^ (p / 2) mod n) * ((a mod n) ^ (p / 2) mod n) * ((a mod n) ^ (p % 2) mod n) mod n



注意不要超过整型范围，增加取模次数，使用long long类型。



不超过100000的16个卡迈克数如下：

561，1105，1729，2465，2821，6601，8911，10585，15841，29341，41041，46657，52633，62745，63973，75361。

代码实现

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int isPri(int n) {
	for(int i = 2; i <= sqrt(n); i++) {
		if(n % i == 0) {
			return 0;
		}
	}
	return 1;
}
long long powmod(int a, int p, int n) {
	if(p == 1) {
		return a % n;
	}
	if(p == 0) {
		return 1 % n;
	}
	return (powmod(a, p / 2, n) % n) * (powmod(a, p / 2, n) % n) * (powmod(a, p % 2, n) % n) % n;
}
int isCar(int n) {
	if(isPri(n)) {
		return 0;
	}
	for(long long a = 2; a < n; a++) {
		if(powmod(a, n, n) != a) {
			//cout << a << endl;
			return 0;
		}
	}
	return 1;
}
int main(void) {
	//cout << powmod(747, 1729, 1729) << endl;
	while(1) {
		int n;
		cin >> n;
		if(n == 0) {
			break;
		}
		if(isCar(n)) {
			cout << "The number " << n << " is a Carmichael number." << endl;
		} else {
			cout << n << " is normal." << endl;
		}
	}
	return 0;
}

（全文完）

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