HDU 5323 DFS
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.
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
#include <iostream> #include<cstdio> using namespace std; int i,n; int a[11]; int main() { a[0]=1; a[1]=1; for(i=2;i<=9;i++) a[i]=a[i-1]*i; n=a[9]*10; //printf("%d\n",n); for(i=1;i<=n;i++) { int t=i; int sum=0; while(t>0) { sum+=a[t%10]; t=t/10; } if (sum==i) printf("%d\n",i); } return 0; }