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 ……
分析：9的阶乘为362880， 9！*10 而且由0~9的阶乘组成的最大数就是3628800。 而且0的阶乘是1，而不是0. 因为根据阶乘定义 n！=n*（n-1）！； 1！=1*0！=1； 所以人为规定了0！=1；
代码语言：javascript复制#include <stdio.h>
#include <stdlib.h>
#include <iostream>
int k[10]= {1,1};
void ff()
{
int i;
for(i = 2; i < 10; i ){
k[i] = k[i-1]*i;
// printf("%dn",k[i]);
}
}
int main()
{
ff();
long i;
long a,sum;
for(i=1; i<=3628800; i )
{
a=i;
sum=0;
while(a!=0)//a>0
{
sum =k[a];
a=a/10;
//printf("%dn",a);
}
if(sum==i)
printf("%ldn",i);
}
return 0;
}
