ＨＤＵ 2212 DFS

DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3946    Accepted Submission(s): 2417

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

 

Recommend

lcy

这个题目看到[1, 2147483647] ，给我第一个感觉……果断会超时，我也去试了一下……我后来用自己的程序写出来有5个……满足条件的，于是我就用上了枚举，但是ＷＡ了，……后来到网上找只有四个满足条件的……。我顿时崩溃了……我代码肯定错了…………纠结……

#include<iostream>using namespace std;int t[10]={1,1,2,6,24,120,720,5040,40320,362880};//5个是我自己打表打错了……我纠结啊…………int main(){    long long n=1,s,sum;    long long t2[4]={1,2,145,40585};    for(int i=0;i<4;i++)    cout<<t2[i]<<endl;    return 0;}