【POJ】1142 - Smith Numbers（容斥原理）

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Smith Numbers

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13488 Accepted: 4593

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

利用容斥原理筛选质因数，枚举过去就可以了。

代码如下：

#include <cstdio>int n;int dis(int x){int sum = 0;while (x){sum += x % 10;x /= 10;}return sum;}int pr(int x)//分解质因数 {int sum = 0;int t = x;for (int i = 2 ; i * i <= x ; i++){if (t % i == 0){while (t % i == 0){sum += dis(i);t /= i;}}}if (t == x)//本身是素数 return -1;if (t > 1)return sum + dis(t);elsereturn sum;}int main(){int t;while (~scanf ("%d",&n) && n){for (int i = n + 1 ; ; i++){t = pr(i);if (t != -1 && t == dis(i)){printf ("%d\n",i);break;}}}return 0;}