HDU--2212 DFS【水题】

DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5753    Accepted Submission(s): 3552

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
......

 

先求出都是那些数

#include<cstdio>
#include<cstring>

int change(int n){
    int i,ans=1;
    for(i=1;i<=n;++i)
        ans=ans*i;
    return ans;
}

int main(){
    int i,t,sum;
    for(i=1;i<=2147483647;++i){
        t=i;sum=0;
        while(t){
           sum+=change(t%10);
           t/=10;
        }
       if(sum==i) printf("%d\n",sum);
    }
   return 0;
}

再直接打表

#include <cstdio>
#include <cstring>
int main (){
    printf("1\n2\n145\n40585\n");
    return 0;
}