poj1142 Smith Numbers

Poj1142 Smith Numbers

Smith Numbers
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 13854 Accepted: 4716
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

题意 大于n满足你的各个位数之和等于质因子各位数之和。
题解 暴力过。

#include<cstdio>
int fun(long long a)
{
    int sum=0;
    while(a>0)
    {
        sum+=a%10;
        a/=10;
     }
     return sum; 
}

bool prime(long long a)
{
    int flag=1;
    if (a==1) return false;
    if(a==2) return true;
    for(int i=2;i*i<a+1;i++)
    if(a%i==0)
    {
        flag=0;
        break;
    }
    if(flag)
    return true;
    else
    return false;
}

int cnt(long long a)
{
    if(prime(a))
    return fun(a);
    else
    {
        for(int i=2;i*i<a+1;i++)
        {
            if(a%i==0)
            return cnt(i)+cnt(a/i);
        }
    }
}

int main()
{
    long long a;
    while(scanf("%lld",&a)!=EOF&&a)
    {
        while(a++)
        {
            int sum=fun(a);

            if(!prime(a)&&fun(a)==cnt(a))
            break;

        }
    printf("%lld\n",a);


    }
    return 0;
 } 

这时间倒也不是很多，79ms