poj 1142 Smith Numbers 【容斥原理】-优快云博客

本文链接：https://blog.youkuaiyun.com/chenzhenyu123456/article/details/45441439

本文介绍了一种特殊的数——Smith数，这种数的定义与其电话号码有关，即其各位数字之和等于其质因数各位数字之和。文章提供了一个判断Smith数的算法实现，并通过实例解释了如何找出大于特定数的最小Smith数。

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Smith Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12750		Accepted: 4356

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

漏了个重要条件，不能是素数。。。  思路：容斥原理的模板改变一下就ok了。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
using namespace std;
int sumf(int n)
{
	int sum=0;
	while(n)
	{
		sum+=n%10;
		n/=10;
	}
	return sum;
}
int getp(int n)
{
	int prime=n;
	int i,j;
	int sum=0;
	for(i=2;i*i<=n;i++)
	{
		while(n%i==0)
		{
			sum+=sumf(i);
			n/=i;
		}
	}
	if(prime==n)//为素数 
	return -1;
	if(n>1) sum+=sumf(n);
	return sum;
}
int main()
{
	int n;
	int i;
	while(scanf("%d",&n)&&(n!=0))
	{	
		for(i=n+1;;i++)
		{
			if(getp(i)!=-1&&sumf(i)==getp(i))//不是素数 
			{
				printf("%d\n",i);
				break;
			}
		}
	}
	return 0;
}