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:
1 //hdu 2212
2 //@Gxjun coder
3 #include<stdio.h>
4 int main()
5 {
6 int val[10]={1,1},i;
7 //求出1~9的阶乘
8 for(i=2;i<=9;i++)
9 val[i]=val[i-1]*i;
10 int ans,tem;
11 for( i=1 ; i<= 3628800 ; i++)
12 {
13 tem=i;
14 ans=0;
15 while(tem)
16 {
17 ans+=val[tem%10];
18 tem/=10;
19 }
20 if(ans==i)
21 printf("%d\n",i);
22 }
23 return 0;
24 }