DFS
Time Limit : 5000/2000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 61 Accepted Submission(s) : 32
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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
......
Author
zjt
#include <iostream>
#include<cstdio>
using namespace std; int i,n;
int a[]; int main()
{
a[]=;
a[]=;
for(i=;i<=;i++) a[i]=a[i-]*i; n=a[]*;
//printf("%d\n",n);
for(i=;i<=n;i++)
{
int t=i;
int sum=;
while(t>)
{
sum+=a[t%];
t=t/;
}
if (sum==i) printf("%d\n",i);
}
return ;
}