HDU2212 DFS

DFS

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
......

——————————————————————————————————————————

名为DFS，实际上没有关系，直接暴力解决。

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int main()
{

int a[10],k,sum,s;

a[0]=1;
for(int i=1;i<10;i++)
{
a[i]=1;
for(int j=1;j<=i;j++)
{
a[i]*=j;
}
}

for(int i=1;i<100000;i++)
{
s=i;
sum=0;
while(s>0)
{
k=s%10;
sum+=a[k];
s/=10;
}
if(sum==i)
printf("%d\n",i);
}
return 0;
}