DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4923    Accepted Submission(s): 3029

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
......

这是一道很好玩的题，我根据题意写了一个程序，运行一下，发现居然输出只有4个数。然后提交，超时，我很无语。然后我就做了一件更无语的事，修改代码。

贴出代码：

#include <stdio.h>

int main()
{
    printf("1\n");
    printf("2\n");
    printf("145\n");
    printf("40585\n");

    return 0;
}

hdu 2212 DFS（水题）