C - Smith Numbers POJ - 1142

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

题意：当一个数分解为几个质因子相乘时，这几个质因子的数位之和等于这个数本身数位之和，那么称这个数 为smith number,题目告诉我们质数不算smith number;让我们找出大于n的最小的smith number;

思路：很明显直接分解质因数求和检验就可以了；

下面附上代码：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
int a[10005],b[10005];
int t,m;
int IS(int m)
{
	int sum=0;
	while(m)
	{
		sum+=m%10;
		m/=10;
	}
	return sum;
}
void DIV(int n)
{
	t=0;
	m=(int)sqrt(n*1.0);
	for(int i=2;i<=m;i++)
	{
		if(n%i==0)
		{
			a[t]=i;
			b[t]=0;
			while(n%i==0)
			{
				b[t]++;
				n/=i;
			}
			++t;
		}
	}
	if(n>1)
	{
		a[t]=n;
		b[t++]=1;
	}
}
int check(int n)
{
	int x=IS(n);
	int y=0;
	for(int i=0;i<t;i++)
		y+=IS(a[i])*b[i];
	if(x==y) return 1;
	return 0;
}
int main()
{
	int n,i;
	while(cin>>n&&n)
	{
		for(i=n+1;;i++)
		{
			DIV(i);
			if(a[0]!=i&&check(i)) break;
		}
		printf("%d\n",i);	
	}
	return 0;
}