poj 1142 Smith Numbers

Smith Numbers

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 12967 Accepted: 4417

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

49377740

Sample Output

4937775

嗯，题目大意就是说，求比所给数大的Smith数，离所给数最近的那个数。Smith数就是它本身各个数位数字之和等于它的每个因子的各个数位数字之和的数。

#include<cstdio>#include<cstring>#define max 100000010int prime(int n){if(n==1||n==0)return 1;if(n==2)return 0;for(int i=2;i*i<=n;++i){if(n%i==0)return 1;}return 0;}int sum(int n){int s=0;while(n){s+=n%10;n/=10;}return s;}int get(int n){int s=0;for(int i=2;;){if(!prime(i)&&n%i==0&&n){n/=i;s+=sum(i);if(!prime(n)){s+=sum(n);break;}}else i++;}return s;}int main(){int n,m,s;while(scanf("%d",&n)&&n){for(int i=n+1;;++i){if(prime(i)){m=i;m=sum(m);s=get(i);if(m==s){printf("%d\n",i);break;}}}}return 0;}