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
……
其实你输出后就会知道。。输出就只有4个数，你可以直接输出。
在这里，我是写了过程的。
public class Main{ static int fact[] = new int[10]; public static void main(String[] args) { dabiao(); //9!*7 7位数-比9999999小，后面的数字更不用说了，肯定小。 for(int i=1;i<=9999999;i++){ if(isTrue(i)){ System.out.println(i); } } } private static void dabiao() { //求阶乘的，注意：0的阶乘为1 fact[0]=1; for(int i=1;i<fact.length;i++){ fact[i]=1; for(int j=1;j<=i;j++){ fact[i]=fact[i]*j; } } } //判断是不是相等 private static boolean isTrue(int i) { if(i==1||i==2){ return true; } int sum=0; int n=i; while(n!=0){ int k=n%10; sum+=fact[k]; n=n/10; } if(sum==i){ return true; } return false; } }
