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*65837The 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
思路：
本题的关键是质因数分解
首先有如下性质：
- 任意合数都可被分解为几个质因数的乘积
- 给定合数的质因数分解表达式唯一
根据上述性质，我们的质因数分解思路如下：
设被分解合数为N，则分解步骤如下：
- 初始状态，M = 2
- 用M试除N，若能整除，说明M为N的质因数，则更新N = N / M，M不变；若不能整除，则N不变，M++
上述方法蕴涵两个特性：
- 被当前M整除的N其所有质因数均大于等于M。譬如：N若能被5整除，则其所有质因数均大于等于5，即其不可能再被2或3整除
- 不需要判断当前M是否为素数，因为若为合数则必然不能整除。证明如下：假设M当前为合数，且M的一个质因数为P，则若N能被M整除则必然能被P整除，这与特性1矛盾-------摘抄的。
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
#define MAXN 10000
#define LL int
bool a[MAXN+1];
LL save[MAXN+1];
LL Smith(LL x)
{
LL sum=0;
while(x)
{
sum=sum+x%10;
x=x/10;
}
return sum;
}
void prime()
{
LL i,j;
//memset(a,1,MAXN+1);//曾今用这个判断素数，超时。
a[1]=1;
for(i=2,j=2;i*j<=MAXN;j++)
a[i*j]=1;
for(i=3;i<100;i+=2)
{
if(!a[i])
{
for(j=2;i*j<=MAXN;j++)
a[i*j]=1;
}
}
}
bool myth(LL x)
{
LL i;
if(x==2)return true;
if(x%2==0||x==1)return false;
for(i=3;i*i<=x;i+=2)
if(x%i==0)return false;
return true;
}
int main()
{
LL x,sum,j,i,temp;
prime();
j=0;
for(i=2;i<=MAXN;i++)
if(!a[i])
save[j++]=i;
while(scanf("%d",&x)&&x)
{
while(++x)
{
sum=0;
temp=x;
if(myth(temp))
continue;
i=0;
while(temp!=1)
{
while(temp%save[i]==0)
{
sum+=Smith(save[i]);
temp=temp/save[i];
}
i++;
if(myth(temp))//大于10000的素数最多出现一次
{
sum+=Smith(temp);
break;
}
}
if(sum==Smith(x))
{
printf("%d\n",x);
break;
}
}
}
return 0;
}