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 ；