鏈接：http://poj.org/problem?id=1142
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
思路：找到這樣一個數，
1：比n大
2：其各個位置的數字之和與其素因子各個位上的數字之和相等，
3:且這個數本身不是素數
誤區：因爲給出的數字不超過八位數，所以開一個100000000的數組然後構造素數表，並且一一判斷，導致超時。
Memory 160KB time:141ms ,lenth:462B
#include<stdio.h>
#include<math.h>
int call_sum(int n)
{
int s=0;
while(n)
{s+=n%10;
n=n/10;
}
return s;
}
int check(int n)
{
int s1=call_sum(n);
int s2=0;
int m=n;
for(int i=2;i<=int(sqrt(m*1.0)+0.5);i++)
{
while(n%i==0)
{s2+=call_sum(i);
n=n/i;}
}
if(n>1)
s2+=call_sum(n);
if(s1==s2&&m!=n)
return 1;
else
return 0;
}
int main()
{
int s,n;
while(scanf("%d",&n)!=EOF&&n!=0)
{
while(check(++n)==0);
printf("%d\n",n);
}
return 0;
}