Smith Numbers（分解质因数）

Smith Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 14173		Accepted: 4838

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

 

AC代码:

#include<iostream>

using namespace std;

int CalDigitsSum(int num)
{
    int sum = 0;
    while(num)
    {
        sum += num % 10;
        num /= 10;
    }
    return sum;
}

int PrimaryCal(int num)
{
    int total = 0;
    int tempNum = num;
    for(int i = 2; i * i <= num; i++)
    {
        int temp;
        if(num % i == 0)
            temp = CalDigitsSum(i);
        while(num % i == 0)
        {
            total += temp;
            num /= i;
        }
    }
    if(tempNum == num)
        return -1;
    if(num != 1)
        total += CalDigitsSum(num);
    return total;
}

int main()
{
    int n;
    while(1)
    {
        cin >> n;
        if(n == 0)
            break;
        for(int i = n + 1; ; i++)
        {
            if(CalDigitsSum(i) == PrimaryCal(i))
            {
                cout << i << endl;
                break;
            }
        }
    }
    return 0;
}