【POJ】1142 - Smith Numbers（容斥原理）

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Smith Numbers

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

Source

Mid-Central European Regional Contest 2000

利用容斥原理筛选质因数，枚举过去就可以了。

代码如下：

#include <cstdio>
int n;
int dis(int x)
{
	int sum = 0;
	while (x)
	{
		sum += x % 10;
		x /= 10;
	}
	return sum;
}
int pr(int x)		//分解质因数 
{
	int sum = 0;
	int t = x;
	for (int i = 2 ; i * i <= x ; i++)
	{
		if (t % i == 0)
		{
			while (t % i == 0)
			{
				sum += dis(i);
				t /= i;
			}
		}
	}
	if (t == x)		//本身是素数 
		return -1;
	if (t > 1)
		return sum + dis(t);
	else
		return sum;
}
int main()
{
	int t;
	while (~scanf ("%d",&n) && n)
	{
		for (int i = n + 1 ; ; i++)
		{
			t = pr(i);
			if (t != -1 && t == dis(i))
			{
				printf ("%d\n",i);
				break;
			}
		}
	}
	return 0;
}