hdu--DFS

DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4422    Accepted Submission(s): 2728

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

......

 

Author

zjt

水体.....暴搜....

代码：

//hdu 2212 
//@Gxjun coder 
#include<stdio.h>
int main()
{
    int val[10]={1,1},i;
    //求出1~9的阶乘
    for(i=2;i<=9;i++)
         val[i]=val[i-1]*i;
     int ans,tem;
    for( i=1 ; i<= 3628800 ; i++)
    {
        tem=i;
        ans=0;
        while(tem)
        {
            ans+=val[tem%10];
            tem/=10;
        }
        if(ans==i)
            printf("%d
",i);
    }
    return 0;
}