HDU2212 DFS 2016-07-24 13:52 56人阅读 评论(0) 收藏

DFS

Problem Description

A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer.




For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number.



Now you should find out all the DFS numbers in the range of int( [1, 2147483647] ).



There is no input for this problem. Output all the DFS numbers in increasing order. The first 2 lines of the output are shown below.

Input

no input

Output

Output all the DFS number in increasing order.

Sample Output

1
2
......

——————————————————————————————————————————

名为DFS，实际上没有关系，直接暴力解决。

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int main()
{

    int a[10],k,sum,s;

    a[0]=1;
    for(int i=1;i<10;i++)
    {
        a[i]=1;
        for(int j=1;j<=i;j++)
        {
            a[i]*=j;
        }
    }

    for(int i=1;i<100000;i++)
    {
        s=i;
        sum=0;
        while(s>0)
        {
            k=s%10;
            sum+=a[k];
            s/=10;
        }
        if(sum==i)
            printf("%d\n",i);
    }
    return 0;
}