## hdu2212 DFS

2018年02月22日 ⁄ 综合 ⁄ 共 927字 ⁄ 字号 评论关闭
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
......#include<stdio.h>
int f[15],an;
int pass(unsigned int sum)//判断每一位的介层和是不是等于本身，反回是几位数
{
unsigned int su=0,a=sum;
int k,t=0;
while(a>0)
{
k=a%10;
su+=f[k];
a/=10;
t++;
}
return (su==sum)?t:0;
}
void DFS(unsigned int sum,int k,int t)//k是sum要达到多少位数，t是当前的数是几位数
{
int e;
if(pass(sum)==k)//只有当反回的位数等要求的位数才输出，不然会重复
printf("%d\n",sum);

if(t<k)//当前的位数达不到要求的位数才继续加
for(e=0;e<10;e++)
if((e!=0||t!=0)&&sum*10+e<=2147483647)//(e!=0||t!=0)是控制最高位不为0
{
DFS(sum*10+e,k,t+1);
}
}
int main()
{
int i;
f[0]=1;
for(i=1;i<10;i++)
f[i]=f[i-1]*i;

for(i=1;i<6;i++)
DFS(0,i,0);
}

```