Smith Numbers

Time Limit: 1000ms Memory Limit: 10000KB 64-bit integer IO format: %lld      Java class name:Main 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

          人是讀大學了，可是連小學的東西都快忘了= =  分解質因數，只要不是素數，就能一直分解，這題便是暴力求解。。。

（還有isprime（）的寫法最好也再注意一下，越是簡單的就越容易出錯）

#include <stdio.h>
#include <string.h>

int isprime(int n)
{
    for(int i = 2; i * i <= n; i++ )
        if(n % i == 0)
            return 0;
    return 1;
}

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

int sumyz(int n)
{
    int sum = 0;
    int i = 2;
    while(1){
        if(n % i == 0){
            sum += sumdigits(i);
            if(isprime(n /= i))
                break;
        }
        else
            i++ ;
    }
    sum += sumdigits(n);
    return sum;
}

int main()
{
    int n, k = 1;
    while(scanf("%d", &n) && n != 0){ 
        k = 1;
        for(int i = n + 1; k == 1; i++){
            if(!isprime(i)){
                if(sumdigits(i) == sumyz(i)){
                    printf("%d\n", i);
                    k = 0;
                }
            }
        }
    }
    return 0;
}