0
点赞
收藏
分享

微信扫一扫

HDOJ(HDU) 2212 DFS(阶乘相关、)

梯梯笔记 05-14 12:50 阅读 1


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;
}
}



举报

相关推荐

0 条评论