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:
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 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
转化一下题目意思就是说给定一个数n，找出比n大的最小的一个数m，m的所有位数之后等于m的所有质因子的位数之后，且m不是素数！此题涉及到整数的素数分解和一个整数的位数分离问题！
下面是我的C++代码，仅供参考！Run time：20MS， Run Memory：184KB
代码
using namespace std;
int CalDigitsSum(int num)
{
int sum=0;
while(num)
{
sum+=num%10;
num/=10;
}
return sum;
}
int PrimaryCal(int num)
{
int total=0;
int tempNum=num;
for(int i=2;i*i<=num;i++)
{
int temp;
if(num%i==0)//num可以整除i
{
temp=CalDigitsSum(i);
}
while(num%i==0)
{
total+=temp;
num/=i;
}
}
if(tempNum==num)//说明num是一个素数，不符合题目要求
{
return -1;
}
if(num!=1)//最后剩余的一个数也是素数
{
total+=CalDigitsSum(num);
}
return total;
}
int main(void)
{
int n,i;
while(cin>>n&&n)
{
for(i=n+1;;i++)
{
if(CalDigitsSum(i)==PrimaryCal(i))
{
cout<<i<<endl;
break;
}
}
}
return -1;
}
