poj1142(质因数分解)

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http://poj.org/problem?id=1142

Smith Numbers

Time Limit: 1000MSMemory Limit: 10000KTotal Submissions: 10279Accepted: 3610

Description

While skimming hisphone directory in 1982, Albert Wilansky, a mathematician of LehighUniversity,noticed that the telephone number of his brother-in-lawH. Smith had the following peculiar property: The sum of the digitsof that number was equal to the sum of the digits of the primefactors of that number. Got it? Smith's telephone number was493-7775. This number can be written as the product of its primefactors 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 equally3+5+5+6+5+8+3+7=42. Wilansky was so amazed by his discovery that henamed this kind of numbers after his brother-in-law: Smithnumbers.
As this observation is also true for every prime number, Wilanskydecided later that a (simple and unsophisticated) prime number isnot worth being a Smith number, so he excluded them from thedefinition.
Wilansky published an article about Smith numbers in the Two YearCollege Mathematics Journal and was able to present a wholecollection of different Smith numbers: For example, 9985 is a Smithnumber and so is 6036. However,Wilansky was not able to find aSmith number that was larger than the telephone number of hisbrother-in-law. It is your task to find Smith numbers that arelarger than 4937775!

Input

The input fileconsists of a sequence of positive integers, one integer per line.Each integer will have at most 8 digits. The input is terminated bya line containing the number 0.

Output

For every number n> 0 in the input, you are to compute the smallest Smith numberwhich is larger than n,and print it on a line by itself. You canassume that such a number exists.

Sample Input

49377740

Sample Output

Source

Mid-Central European Regional Contest 2000

题意：给定一个数n，若n的各位数之和与n的所有质因数各位之和相等，则n为simth数。现在对于输入的一个数n，找出一个大于n的最小的一个simth数，并输出。

思路：一个合数一定能被分解为几个比它的平方根小的素数和一个比它平方根大的一个素数的乘积。利用就能很快的求得它的所有质因子了。

#include
#include
#include
#include
using namespace std;
bool prime (int num)//快速判断素数
{
if (num == 2 || num == 3 || num == 5)
returntrue;
if (num % 2 == 0 || num % 3 == 0 || num % 5 == 0|| num == 1)
returnfalse;

unsigned long c = 7;
int maxc = int (sqrt((double)num));
while (c <= maxc)
    {
     if (num % c == 0)
       return false;
     c += 4;
     if (num % c == 0)
       return false;
     c += 2;
     if (num % c == 0)
       return false;
     c += 4;
     if (num % c == 0)
       return false;
     c += 2;
     if (num % c == 0)
       return false;
     c += 4;
     if (num % c == 0)
       return false;
     c += 6;
     if (num % c == 0)
       return false;
     c += 2;
     if (num % c == 0)
       return false;
     c += 6;
    }
return true;
}

int getsum(int x)
{
    intsum=0;
   while(x/10!=0)
    {
       sum+=x;
       x/=10;
    }
   sum+=x;
    returnsum;
}
bool is_Smith(int n)//判断smith数
{
    intk=n;
    intsum1=getsum(k);//sum1先获得原始个个数的和
    intsum2=0;
    for(inti=2;i<=(int)sqrt((double)n);)//注意这里的平方根(n)也要在不停的更新
    {
       if(n%i==0)//找到质因子
       {
           sum2+=getsum(i);
           n/=i;
           i=2;//重新更新i
       }
       else//找不到，则往后找
       {
           i++;
       }
    }
   sum2+=getsum(n);//最后还有一个质因子
   if(sum1==sum2)
    {
       return true;
    }
    returnfalse;
}
int main()
{
    int n;
   while(scanf("%d",&n)!=EOF)
    {
       if(n==0)
       {
           break;
       }
       while(++n)
       {
           if(!prime(n))
           {
               if(is_Smith(n))
               {
                   printf("%d\n",n);
                   break;
               }
           }
       }
    }
    return0;
}