hdu 2212 DFS

DFS

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6502 Accepted Submission(s): 3998



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
no input

[align=left]Output[/align]
Output all the DFS number in increasing order.

[align=left]Sample Output[/align]

1
2
......

[align=left]Author[/align]
zjt

#include<stdio.h>
#include<string.h>
int dfs(int n){
if(n==0 || n==1) return 1;
else return n*dfs(n-1);
}
int main(){
int i;
for(i=1;i<50000;i++){
int a=i,c,sum=0,ok=0;
while(a){
c=a%10;
sum+=dfs(c);
a/=10;
if(sum>i){
ok=1;
break;
}
}
if(!ok && sum==i)
printf("%d\n",i);
}
return 0;
}