Smith Numbers
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 14521 | Accepted: 4906 |
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*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
0Sample Output
4937775
题意
定义一个Smith Numbers：对于一个数n，如果n中的各位数相加的和与n的每个质因数上的各位数字相加的和相等；
如：4937775= 3*5*5*65837 ； 4+9+3+7+7+7+5==3+5+5+6+5+8+3+7=42，即4937775是Smith Numbers
给出一个数n，求出大于n的最小的Smith Numbers
思路
由定义可推出，Smith Numbers一定不是素数，所以可以先判断一个数是否是素数，然后写个对数n的各位数求和的函数，分解质因数，暴力求解即可
AC代码
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const double E=exp(1);
const int maxn=1e6+10;
using namespace std;
int a[maxn];
int is(ll x)//判断素数
{
for(int i=2;i*i<=x;i++)
{
if(x%i==0)
return 0;
}
return 1;
}
ll sum(ll n)
{
ll ans=0;
while(n)
{
ans+=n%10;
n/=10;
}
return ans;
}
int main(int argc, char const *argv[])
{
ll n;
while(~scanf("%lld",&n))
{
if(n==0)
break;
ll _;
for(ll x=n+1;;x++)
{
ll ans;
ll flag=0;
ll y=x;
//如果不是素数
if(is(y)==0)
{
ans=sum(y);
for(int i=2;i*i<=y;i++)
{
if(y%i==0)
{
while(y%i==0)
{
flag+=sum(i);
y/=i;
}
}
}
if(y>1)
flag+=sum(y);
if(flag==ans)
{
_=x;
break;
}
}
}
printf("%lld
",_);
}
return 0;
}