HDOJ(HDU) 2212 DFS(階乘相關、)

標籤：

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
……

其實你輸出後就會知道。。輸出就只有4個數，你可以直接輸出。
在這裡，我是寫了過程的。

public class Main{    static int fact[] = new int[10];    public static void main(String[] args) {        dabiao();        //9!*7 7位元-比9999999小，後面的數字更不用說了，肯定小。        for(int i=1;i<=9999999;i++){            if(isTrue(i)){                System.out.println(i);            }        }    }    private static void dabiao() {        //求階乘的，注意：0的階乘為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;            }        }    }    //判斷是不是相等    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;    }}

HDOJ(HDU) 2212 DFS(階乘相關、)