POJ 1142 暴力!

Smith Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8209		Accepted: 2867

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

先开始一直在想算法，不过后来看了帖子才知道暴力也可以，而且还很快。。（擦，什么鸟数据。。）

Source Code

Problem: 1142		User: bingshen
Memory: 204K		Time: 32MS
Language: C++		Result: Accepted

• Source Code

  #include<stdio.h>
  #include<algorithm>
  
  using namespace std;
  
  bool prime(int n)
  {
  	int i;
  	for(i=2;i*i<=n;i++)
  		if(n%i==0)
  			return false;
  	return true;
  }
  
  int getsum(int x)
  {
  	int sum=0;
  	while(x)
  	{
  		sum=sum+x%10;
  		x=x/10;
  	}
  	return sum;
  }
  
  bool judge(int x)
  {
  	int sum1,sum2=0,i,num=0;
  	int p[10000];
  	sum1=getsum(x);
  	for(i=2;i*i<=x;i++)
  	{
  		if(x%i==0)
  		{
  			p[num++]=i;
  	//		printf("%d/n",i);
  			x=x/i;
  			while(x%i==0)
  			{
  	//			printf("%d/n",i);
  				p[num++]=i;
  				x=x/i;
  			}
  		}
  		if(x==1)
  			break;
  	}
  	if(x>1)
  	{
  		p[num++]=x;
  //		printf("%d/n",x);
  	}
  	for(i=0;i<num;i++)
  	{
  		sum2=sum2+getsum(p[i]);
  //		printf("%d ",p[i]);
  	}
  //	printf("/n");
  	if(sum1==sum2)
  		return true;
  	else
  		return false;
  }
  
  int main()
  {
  	int n,i;
  	while(scanf("%d",&n)!=EOF)
  	{
  		if(n==0)
  			break;
  		for(i=n+1;;i++)
  		{
  			if(prime(i))
  				continue;
  			if(judge(i))
  			{
  				printf("%d/n",i);
  				break;
  			}
  		}
  	}
  	return 0;
  }