POJ 1142 Smith Numbers

 

Smith Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7141		Accepted: 2505

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

 

/* http://acm.pku.edu.cn/JudgeOnline/problem?id=1142 map 哈希 + 求质因数，注意质数不是合法结果 */ #include <iostream> #include <map> using namespace std; int input; map<int, int>::iterator itr; int getDigitSum(int a) { int res = 0; while(a != 0) { res += a % 10; a /= 10; } return res; } bool check(int inNum) { map<int, int> countm; int curN, input = inNum; for(curN = 2; curN * curN <= input; curN++) { //cout<<curN<<endl; if(input % curN == 0) { while(input % curN == 0) { itr = countm.find(curN); if(itr == countm.end()) countm[curN] = 1; else countm[curN]++; input /= curN; //cout<<input<<endl; } } } if(input == inNum) return false; if(input != 1) countm[input] = 1; int total = 0; //cout<<countm.size()<<endl; for(itr = countm.begin(); itr != countm.end(); itr++) { int digit = itr->first; int cn = itr->second; //cout<<digit<<" "<<cn<<endl; total += getDigitSum(digit) * cn; //cout<<total<<endl; } //cout<<getDigitSum(inNum)<<endl; if(total == getDigitSum(inNum)) return true; else return false; } int main() { while(cin>>input && input != 0) { while(true) { input++; //cout<<input<<endl; if(check(input)) { cout<<input<<endl; break; } } } return 0; }