hdu 2212 DFS

最新推荐文章于 2025-08-22 12:09:46 发布

转载最新推荐文章于 2025-08-22 12:09:46 发布 · 43 阅读

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原文链接：http://www.cnblogs.com/forever4444/archive/2009/05/13/1456263.html

文章标签：

#c/c++ #java

本文介绍了一种特殊类型的整数——数字因子和数（DFS数），并提供了使用C++编程语言实现的两种方法来找出指定范围内的所有DFS数。一种是直接输出已知的DFS数，另一种是通过编写打表程序来找出这些数。

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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 ;
}

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/13/1456263.html