Smith Numbers（算术基本定理）

Smith Numbers

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= 355*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
这道题求比输入的数大的第一个史密斯数；
史密斯数既满足上述三个条件
ac代码：

#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;

const int MAXN=100005;
int prime[MAXN+1];
int factor[1000][2];
int isprime(int n){//判断是否是素数的函数 
	for(int i=2;i*i<=n;i++){
		if(n%i==0) return 0;
	}
	return 1;
}
void getprime(){//素数打表 
	memset(prime,0,sizeof(prime));
	for(int i=2;i<=MAXN;i++){
		if(!prime[i]) prime[++prime[0]]==i;
		for(int j=2;j<=MAXN/i;j++){
			prime[i*j]=1;
		}
	}
} 
int getfactor(int n){
	int cnt=0;
	memset(factor,0,sizeof(factor));
	for(int i=1;prime[i]*prime[i]<=n;i++){
		factor[cnt][1]=0;
		if(n%prime[i]==0){
			factor[cnt][0]=prime[i];
			while(n%prime[i]==0){//每当整除一次，幂次方位就加一 
				n/=prime[i];
				factor[cnt][1]++;
			}
			cnt++;
		}
	}
	if(n!=1){
		factor[cnt][0]=n;
		factor[cnt][1]=1;
		cnt++;
	}
	return cnt;
} 
int f(int n){//该函数求一个数字的每位的和 
	int ans=0;
	while(n){
		ans+=n%10;
		n/=10;
	}
	return ans;
} 
int cnt(int n){//以递归的方法求至起因子为素数并求出其个数位之和
	if(isprime(n)) return f(n);//n的是素数，返回位数和 
	for(int i=(int)sqrt(n+0.5);i>1;i--){
		if(n%i==0){
			//cout<<i<<"  "<<n-i<<endl;
		return cnt(i)+cnt(n/i);//n不是素数 
	}
	} 
} 
int main(){
	ios::sync_with_stdio(0);
	cin.tie(),cout.tie();
	int n;
	getprime();
	while(cin>>n){
		int m=n;
		while(m++){
			if(isprime(m)) continue;
			int ans1=f(m);
			int ans2=cnt(m);
			if(ans1==ans2){
				cout<<m<<endl;
				break;
			}
		}
	}
	return 0;
}