POJ

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

Hint

题意

smith数
一个数的质因子的各位之和等于它本身个位数之和
要求找到大于n的最小smith数

题解：

首先这个数一定是合数 只有合数才能够分解质因子
而且合数的分布较密 不断往后枚举判断就可以
看了别人的思路 使用了分治的思想 （是吧？）
就是不断分解它的因子到质因子 然后计算

AC代码

#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;bool isprime(int n){    int t = sqrt(n+0.5);    for (int i = 2; i <= t; ++i){        if (n%i==0) return false;       }    return true;}int sum(int n){    int s = 0;    int t;    while (n){        t=n%10;        s+=t;        n/=10;     }    return s;}int cut(int n){    if (isprime(n)) return sum(n);    for (int i = (int)sqrt(n+0.5); i > 1; i--){        if (n%i==0){            return cut(i)+cut(n/i);        }    }}int main(){    int n;    while (scanf("%d",&n),n){        while (n++){            if (!isprime(n)&&sum(n)==cut(n)){                break;            }        }        printf("%d\n",n);     }    return 0;}