UVA 10042 Smith Numbers（数论）

时间：2014-07-18 13:28:06 阅读：234 评论：0 收藏：0 [点我收藏+]

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:

The sum of all digits of the telephone number is4+9+3+7+7+7+5=42, and the sum of the digits of its prime factors is equally3+5+5+6+5+8+3+7=42. Wilansky was so amazed by his discovery that he named this type 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 and he excluded them from the definition.

Problem

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 give a Smith number which was larger than the telephone number of his brother-in-law. It is your task to find Smith numbers which are larger than 4937775.

1
4937774

Sample Output

题意：如果一个合数的各个数字之和等于该数所有素因子的各个数字之和，则称这个数是Smith数。给出一个n，求大于n的最小的Smith数是多少。

分析：对要判断的数进行素因子分解即可。因为所求数小于 10^9，若一个数是合数，则其素因子至少有一个小于或等于sqrt(10^9)，则可先把2 - sqrt(10^9) 之间的素数保存起来。

#include<stdio.h>
#include<string.h>
const int MAXN = 100005;
int vis[MAXN], prime[10000], num;

void get_prime()
{
    num = 0;
    memset(vis, 0, sizeof(vis));
    vis[0] = vis[1] = 1;
    for(int i = 2; i < MAXN; i++)
    {
        if(!vis[i])
        {
            prime[num++] = i;
            for(int j = i + i; j < MAXN; j += i)
                vis[j] = 1;
        }
    }
}

bool is_prime(int x)
{
    if(x == 0 || x == 1) return false;
    if(x == 2) return true;
    if(x % 2 == 0) return false;
    for(int i = 3; i * i <= x; i += 2)
        if(x % i == 0)
            return false;
    return true;
}

int sum(int x)
{
    int res = 0;
    while(x)
    {
        res += x % 10;
        x /= 10;
    }
    return res;
}

int main()
{
    get_prime();
    int n, t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i = n + 1; ; i++)
        {
            if(is_prime(i))
                continue;
            int s = 0, tmp = i, tmpsum = sum(i);
            for(int j = 0; j < num; j++)
            {
                if(tmp % prime[j] == 0)
                {
                    while(tmp % prime[j] == 0)
                    {
                        s += sum(prime[j]);
                        tmp /= prime[j];
                    }
                    if(is_prime(tmp))
                    {
                        s += sum(tmp);
                        break;
                    }
                }
            }
            if(tmpsum == s)
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}

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UVA 10042 Smith Numbers（数论）

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原文地址：http://blog.csdn.net/lyhvoyage/article/details/37913579

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