HDOJ 2212 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
……

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；

#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("%d\n",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%10]; a=a/10; //printf("%d\n",a); } if(sum==i) printf("%ld\n",i); } return 0;
}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/49532795