POJ1142——Smith Numbers

Smith Numbers

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 12388 Accepted: 4244

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

Source

Mid-Central European Regional Contest 2000

简单的质因数分解题

#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int a[10010];int b[10010];int tot;bool is_prime(int n){int temp = (int)sqrt((double)n + 1);for(int i = 2; i <= temp; i ++)if( n % i == 0)return false;return true;}int digit_sum(int n){int sum = 0;while(n){sum += n % 10;n /= 10;}return sum;}void prime_factor(int n){int temp = (int)sqrt((double)n + 1);int cur = n;for(int i = 2;i <= temp; i ++){if(cur % i == 0){a[++ tot] = i;b[tot] = 0;while(cur % i == 0){b[tot] ++;cur /= i;} }}if(cur != 1){a[++ tot] = cur;b[tot] = 1;}}int main(){int n;while(~scanf("%d", &n), n){int cur = n;int sum1, sum2;while(1){cur ++;if( is_prime(cur) )continue;sum1 = digit_sum(cur);sum2 = 0;tot = 0;prime_factor(cur);for(int i = 1; i <= tot; i ++){//printf("%d %d\n", a[i], b[i]);sum2 += digit_sum(a[i]) * (b[i]);}if(sum1 == sum2)break;}printf("%d\n", cur);}return 0;}