HDu2212

DFS

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

Total Submission(s): 6792 Accepted Submission(s): 4180

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

......

方法一；

其实大多数的都是通过打表来实现的，这里我也讲一种！其实对付这种题目，最好的方法也就是打表！

附代码

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

附打表代码

#include<stdio.h>
int main()
{

__int64 i,j,k,l,a[10],s;
a[0]=1;
for(i=1;i<=10;i++)
a[i]=a[i-1]*i;
for(i=1;i<=2147483647;i++)
{
k=i;s=0;
while(k)
{
j=k%10;
k=k/10;
s=s+a[j];
}
if(s==i)
printf("%I64d\n",i);
}
return 0;
}

方法二

其实方法二和方法一差不多，就是去求最大的一个数是多少！我从9999999开始的，然后就8888888…………往前面求，当我求到999999时我发现可以作为上界了；所以……

附代码

#include<stdio.h>
int main()
{
__int64 i,j,k,l,a[10],s;
a[0]=1;
for(i=1;i<=10;i++)
a[i]=a[i-1]*i;
for(i=1;i<=999999;i++)
{
k=i;s=0;
while(k)
{
j=k%10;
k=k/10;
s=s+a[j];
}
if(s==i)
printf("%I64d\n",i);
}

return 0;
}