Smith Numbers
时间： 1ms 内存：64M
描述：
Smith Numbers
While skimming his phone directory in 1982, mathematician Albert Wilansky 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 named this type of number after his brother-in-law: the Smith numbers.
As this property is true for every prime number, Wilansky excluded them from the definition. Other Smith numbers include 6,036 and 9,985.Wilansky was not able to find a Smith number which was larger than the telephone number of his brother-in-law. Can you help him out?
输入：
The input consists of several test cases, the number of which you are given in the first line of the input. Each test case consists of one line containing a single positive integer smaller than 109.
输出：
For every input value n, compute the smallest Smith number which is larger than n and print it on a single line. You can assume that such a number exists.
示例输入：
1
4937774
示例输出：
4937775
提示：
参考答案（内存最优[748]）：
#include"stdio.h"
#include"string.h"
#include"math.h"
int CalDigitsSum(int num)
{
int sum=0;
while(num)
{
sum+=num%10;
num/=10;
}
return sum;
}
int main()
{
int n,i,j,t,s,x;
int sum,sum2;
scanf("%d",&t);
while(t--)
{
sum=0;
scanf("%d",&n);
for(i=n+1;;i++)
{
x=i;
sum=CalDigitsSum(i);
sum2=0;
for(j=2; j*j<=x; j++)
{
if(x%j==0)
s=CalDigitsSum(j);
while(x%j==0)
{
sum2+=s;
x/=j;
}
}
if(x==i)
continue;
if(x!=1)
sum2+=CalDigitsSum(x);
if(sum==sum2)
{
printf("%d\n",i);
break;
}
}
}
return 0;
}
参考答案（时间最优[32]）：
#include <iostream>
#include <set>
#include <cmath>
using namespace std;
int work(int n)
{
int c=0,t;
for(int i=2; i*i<=n; ++i)
{
while(n%i==0)
{
n/=i;
t=i;
while(t)
{
c+=t%10;
t/=10;
}
}
}
if(n>1&&c)
{
while(n)
{
c+=n%10;
n/=10;
}
}
return c;
}
int main()
{
int T;
int n;
cin>>T;
while(T--)
{
cin>>n;
for(int i=n+1;; ++i)
{
int t=i;
int c=0;
while(t)
{
c+=t%10;
t/=10;
}
t=work(i);
if(t==c)
{
cout<<i<<endl;
break;
}
}
}
return 0;
}
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