hdu 2212 DFS

DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 622    Accepted Submission(s): 380

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

 

Recommend

lcy

 

Statistic | Submit | Back

//1364495 2009-05-13 20:42:44 Time Limit Exceeded 2212 2000MS 232K 426 B C++ Wpl
//1364655 2009-05-13 21:03:44 Accepted 2212 0MS 204K 299 B C++ Wpl  
/*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] ).*/
#include <iostream>
#define MAX 5
using namespace std;
int data[MAX];
int main()
{
    data[0]=1;
    data[1]=2;
    data[2]=145;
    data[3]=40585;
    int i;
    for(i=0;i<=3;i++)
        printf("%d\n",data[i]);
    return 0;
}

附上打表的程序

 

//1364495 2009-05-13 20:42:44 Time Limit Exceeded 2212 2000MS 232K 426 B C++ Wpl 
#include <iostream>
#include <fstream>
#define MAX 10000
using namespace std;
int f[11],data[MAX];
bool DFS(int n)
{
    int sum=0,x=n;
    while(x!=0)
    {
        sum+=f[x%10];
        x=x/10;
        if(sum>n)
            return false;
    }
    if(sum==n)
        return true;
    else
        return false;
}
int main()
{
    int i,j;
    f[0]=1;
    for(i=1;i<=10;i++)
        f[i]=i*f[i-1];
    ofstream outfile("ans.txt");
    j=0;
    for(i=1;i<2147483647;i++)
    {
        if(DFS(i))
        {
            outfile<<"data["<<j++<<"]="<<i<<endl;
        }
    }
    return 0;
}