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
題意：找一個大於當前給出數的數（例如4937774），其每位數和（ 4+9+3+7+7+7+5= 42）與其拆解成素數積（3*5*5*65837）後，每個素數每一位的和（3+5+5+6+5+8+3+7=42）相等。
思路：這道題的話，首先我們要找到這個數，這個數可以分解，所以它不能是素數，所以我們先判斷一下是否爲素數，然後我們先求所給數的每一位的和，再將所給數拆解爲素數積並求出每位和。最後判斷兩個是否相等，相等就是我們要找的數，不相等就繼續找。
AC代碼：
#include <stdio.h>
#include <string>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
typedef long long ll;
const int maxx=100010;
const int inf=0x3f3f3f3f;
const double eps=1e-5;
using namespace std;
int isprime(int n)//判斷n是否爲素數
{
for(int i=2; i*i<=n; i++)
if(n%i==0)
return 0;
return 1;
}
int sum1(int n)//求n每一位的和
{
int ans=0;
while(n)
{
ans+=n%10;
n/=10;
}
return ans;
}
int sum2(int n)//將n拆解爲素數積並求出每位和
{
if(isprime(n))
return sum1(n);
for(int i=2; i*i<=n; i++)
{
if(n%i==0)
return sum2(i)+sum2(n/i);
}
}
int main()
{
int n;
while(~scanf("%d",&n),n)
{
int m=n;
while(++m)
{
if(isprime(m))
continue;
int a=sum1(m);
int b=sum2(m);
if(a==b)
{
printf("%d\n",m);
break;
}
}
}
return 0;
}