分解质因数——Poj 1142 Smith Numbers

Smith Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10627		Accepted: 3723

Description

While skimming his phone directory in 1982, Albert Wilansky, a mathematician of Lehigh University,noticed that the telephone number of his brother-in-law H. Smith had the following peculiar property: The sum of the digits of that number was equal to the sum of the digits of the prime factors of that number. Got it? Smith's telephone number was 493-7775. This number can be written as the product of its prime factors in the following way: 
4937775= 3*5*5*65837
The sum of all digits of the telephone number is 4+9+3+7+7+7+5= 42,and the sum of the digits of its prime factors is equally 3+5+5+6+5+8+3+7=42. Wilansky was so amazed by his discovery that he named this kind of numbers after his brother-in-law: Smith numbers. 
As this observation is also true for every prime number, Wilansky decided later that a (simple and unsophisticated) prime number is not worth being a Smith number, so he excluded them from the definition. 
Wilansky published an article about Smith numbers in the Two Year College Mathematics Journal and was able to present a whole collection of different Smith numbers: For example, 9985 is a Smith number and so is 6036. However,Wilansky was not able to find a Smith number that was larger than the telephone number of his brother-in-law. It is your task to find Smith numbers that are larger than 4937775!

Input

The input file consists of a sequence of positive integers, one integer per line. Each integer will have at most 8 digits. The input is terminated by a line containing the number 0.

Output

For every number n > 0 in the input, you are to compute the smallest Smith number which is larger than n,and print it on a line by itself. You can assume that such a number exists.

Sample Input

4937774
0

Sample Output

4937775

思路：

枚举大于给定n的所有合数（题目中说： a prime number is not worth being a Smith number）是否为Smith Number。

根据质因数分解定理，分解质因数时，可以这样做：

设要分解的合数为N，M为试除的质因数，初值为2；

1、如果N被M整除，则M为N的质因数，令N = N / M，M不变；

2、若不能整除，则N不变，M++；

重复上面两步，直到N是一个素数为止。

需要注意的是：

不需要判断当前M是否为素数，因为若为合数则必然不能整除。

#include <iostream>
#include <cstdio>

using namespace std;

bool IsPrime(int n)
{
	int i;
	for (i = 2; i*i <= n; i++)
	{
		if (n % i == 0)
		{
			return false;
		}
	}
	return true;
}

int SumOfDigits(int n)
{
	int sum = 0;
	while (n != 0)
	{
		sum += n%10;
		n /= 10;
	}
	return sum;
}

int SumOfPrimeFactors(int n)
{
	int sum = 0;
	int i = 2;
	while (true) //合数是一定可以被分解质因数的，所以一直循环好了
	{
		if (n % i == 0)
		{
			sum += SumOfDigits(i);
			if (IsPrime(n /= i))
			{
				break;
			}
		}
		else
		{
			i++;
		}
	}
	return sum += SumOfDigits(n);
}

int main(void)
{
	int n;
	while (scanf("%d", &n) != EOF && n != 0)
	{
		bool finished = false;
		int i;
		for (i = n+1; !finished; i++)
		{
			if (!IsPrime(i)) //素数不可能是Smith Number
			{
				if (SumOfDigits(i) == SumOfPrimeFactors(i))
				{
					printf("%d\n", i);
					finished = true;
				}
			}
		}
	}
	return 0;
}