acm pku 1142 Smith Numberd的模拟实现方法
Smith Numbers
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
0
Sample Output
4937775
Source
Mid-Central European Regional Contest 2000
分析： 这道题目的题意非常的明确，就是求某个数使得该数的各位数之和等于其素因式分解的各数的各位数之和。还有一个条件需要注意：所求之数为满足上述条件的，比给出数大的最小数，不能包括等于给出数的数（作者就是没有注意到这个条件，杯具了好久）。
具体实现：
#include "iostream"
#include "math.h"
using namespace std;
bool IsPrime(int n)
{
int i, sq;
sq =(int)sqrt((double)n);
if(n == 1) return true;
for(i = 2; i < sq+1 && n%i != 0; i++);
if(i == sq+1) return true;
else return false;
}
int DigitSum(int n)
{
int sum = 0;
while(n)
{
sum += n%10;
n /= 10;
}
return sum;
}
bool SmithNumber(int n, int tmp, int max)
{
int i, sum, itmp=tmp;
sum = 0;
if(IsPrime(itmp)) return false;
for(i = 2; i < (int)sqrt((double)itmp)+1 && itmp != 1; i++)
{
while(itmp%i == 0) { sum += DigitSum(i); itmp /= i;}
if(sum > max) return false;
}
if(itmp > 1) sum += DigitSum(itmp);
if(sum == max && tmp > n) return true;
else return false;
}
int main(void)
{
int n, tmp;
cin >> n;
while(n != 0)
{
tmp = n;
while(!SmithNumber(n, tmp, DigitSum(tmp))) tmp++ ;
cout << tmp << endl;
cin >> n;
}
return 0;
}
执行结果：
| Problem: 1142 |
|
User: uestcshe |
| Memory: 712K |
|
Time: 47MS |
| Language: G++ |
|
Result: Accepted |