UVA 10042 Smith Numbers（數論）

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Smith Numbers
BackgroundWhile 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.

ProblemWilansky 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.

InputThe 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.

OutputFor every input value n, you are to compute the smallest Smith number which is larger than nand print each number on a single line. You can assume that such a number exists.

Sample Input

14937774

Sample Output

4937775

題意：如果一個合數的各個數字之和等於該數所有素因子的各個數字之和，則稱這個數是Smith數。給出一個n，求大於n的最小的Smith數是多少。

分析：對要判斷的數進行素因子分解即可。因為所求數小於 10^9，若一個數是合數，則其素因子至少有一個小於或等於sqrt(10^9)，則可先把2 - sqrt(10^9) 之間的素數儲存起來。 

#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;}