HDU2212-DFS

DFS

                                                                             Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                        Total Submission(s): 7711    Accepted Submission(s): 4725

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

12......

 

Author

zjt

预处理，找出符合条件的数，然后再打出（避免超时）

#include <iostream>#include <stdio.h>using namespace std;int main(){    printf("1\n2\n145\n40585\n");    return 0;}