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hdu 2212 DFS(水题)
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.
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......
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;}
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