Smith Numbers 素数打表+暴力+求素因子+poj

Smith Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12106		Accepted: 4141

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

解决方案：10^5以内的素数打表+求素因子。数据有点水，暴力可过。还有，结果不能是素数。

code:

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#define MMAX 100000
using namespace std;
vector<int >prime;
vector<int >pfact;
bool vis[MMAX];
long long n;
void init_prime()
{
    prime.clear();
    memset(vis,false,sizeof(vis));

    prime.push_back(2);
    for(int i=3; i<MMAX; i+=2)
        if(!vis[i])
        {

            prime.push_back(i);
            for(int j=i+i; j<MMAX; j+=i)vis[j]=true;
        }
}
bool ju(int x){
     for(int i=2;i*i<=x;i++){
        if(x%i==0) return false;
      }
  return true;
}
int get_sum(int x)
{
    int sum=0;
    while(x)
    {
        sum+=(x%10);
        x/=10;
    }
    return sum;
}
int main()
{
    init_prime();
    while(~scanf("%lld",&n)&&n)
    {
        bool flag=false;
        for(int i=n+1; ; i++)
        {
            int k=i;
            pfact.clear();
            for(int j=0; j<prime.size(); j++)
            {


                if(k%prime[j]==0)
                {

                    while(k%prime[j]==0)
                    {
                        pfact.push_back(prime[j]);
                        k/=prime[j];
                    }
                }


            }
            if(k>1) pfact.push_back(k);
            int st=get_sum(i);
            int en=0;
            for(unsigned int kk=0; kk<pfact.size(); kk++)
            {
                en+=get_sum(pfact[kk]);
            }
            if(st==en&&!ju(i))
            {
                printf("%d\n",i);
                break;
            }


        }

    }
    return 0;
}