Hdu 2212 DFS

DFS

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7020 Accepted Submission(s): 4316



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


思路：题目是DFS，但是内容不是，而且题目比较坑，直接打一个0-9的阶乘表就能解决，或者也可以直接输出四个数字，也就是答案

AC代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

long long a[10];
void init(){
    long long tmp=1;
    for(int i=1;i<=9;i++){
        tmp*=i;
        a[i]=tmp;
    }
    a[0]=1;
}

bool cal(long long  i){
    long long ttmp=i;
     long long sum=0;
    while(i){
        long long  tx=i%10;
        sum+=a[tx];
        i/=10;
    }
    if(sum==ttmp) return true;
    else return false;
}

int main(){
    init();
   for(long long  i=1;i<=300000;i++){
     if(cal(i))
        printf("%I64d\n",i);
   }
}