poj1142.Smith Number（数学推导）

Smith Number

Time Limit: 1 Sec  Memory Limit: 64 MB Submit: 825  Solved: 366

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

 #include<stdio.h>
 int n ;

 int main ()
 {
     //freopen ("a.txt" , "r" , stdin ) ;
     while (~ scanf ("%d" , &n) ) {
         if (n == )
             break ;
         while () {
             n ++ ;
             int x ;
             int sum =  ;
             int m = n ;
             while (m) {
                 sum += (m % ) ;
                 m /=  ;
             }
             int sum2 =  ;
             m = n ;
             for (int i =  ; i * i <=  n ; i++) {
                 if (m % i == ) {
                     int y = i ;
                     int k =  ;
                     while (y) {
                         k += y %  ;
                         y /=  ;
                     }
                     while (m % i == ) {
                         sum2 += k ;
                         m /= i ;
                     }
                 }
             }
             if (m == n)
                 continue ;
             if (m != ) {
                     int y = m ;
                     int k =  ;
                     while (y) {
                     k += y %  ;
                     y /=  ;
                 }
                 sum2 += k ;
             }

             if (sum2 == sum) {
                 break ;
             }
         }
         printf ("%d\n" , n ) ;
     }
 }

在O（sqrt（N））时间内求得n的所有质因数。
注意1.若跳出循环后n除尽，则cnt 不加1 ； 反之 加一。

2.跳出若为本身，则cnt = 0 ；