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题目描述

Smith Numbers

While skimming his phone directory in 1982, mathematician Albert Wilansky 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 named this type of number after his brother-in-law: the Smith numbers.
As this property is true for every prime number, Wilansky excluded them from the definition. Other Smith numbers include 6,036 and 9,985.

Wilansky was not able to find a Smith number which was larger than the telephone number of his brother-in-law. Can you help him out?

输入

The 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.

输出

For every input value n, compute the smallest Smith number which is larger than n and print it on a single line. You can assume that such a number exists.

样例输入

1
4937774

样例输出

4937775

提示

代码如下

#include <iostream>
#include <set>
#include <cmath>
using namespace std;
int work(int n)
{
    int c=0,t;
    for(int i=2; i*i<=n; ++i)
    {
        while(n%i==0)
        {
            n/=i;
            t=i;
            while(t)
            {
                c+=t%10;
                t/=10;
            }
        }
    }
    if(n>1&&c)
    {
        while(n)
        {
            c+=n%10;
            n/=10;
        }
    }
    return c;
}
int main()
{
    int T;
    int n;
    cin>>T;
    while(T--)
    {
        cin>>n;
        for(int i=n+1;; ++i)
        {
            int t=i;
            int c=0;
            while(t)
            {
                c+=t%10;
                t/=10;
            }
            t=work(i);
            if(t==c)
            {
                cout<<i<<endl;
                break;
            }
        }
    }
    return 0;
}

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