hdu 2212 dfs 简单的思维问题

DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2898    Accepted Submission(s): 1753

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

 

简单的思维问题 各个位上阶乘的和 要和这个数相匹配 这样才能得到正确的解。

各个位上阶乘的和 是最大是9*9！这样来求解。999999999   9个9 最大的各个位上的阶乘的和为3265920=9*9！

只需要遍历到上面这个数就可以了，不用遍历到21亿，否则会超时

只有四个数满足条件

post code

#include<stdio.h>
int a[11];
int fun(int n)  //1-9的阶乘存起来
{
   int i;
   int sum=1;
   for( i=1; i<=n; i++ )
       sum*=i;
   return sum;    
   
}

int main()
{
      a[0]=1;
     int i,num,ji,sum;
     
     for( i=1; i<=9; i++)
          {
               a[i]=fun(i);
          }
     for( i=1; i<=300500; i++ )  //直接进行遍历
     {
          num=i;
          sum=0;
          while(num)
          {
             ji = num%10;
             num/=10;
             sum+=a[ji];  
          } 
          if(i==sum){printf("%d\n",i); }   
     }
}