HDU 2212 DFS

DFS

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

Total Submission(s): 5671    Accepted Submission(s): 3503


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
no input
 

[align=left]Output[/align]
Output all the DFS number in increasing order.
 

[align=left]Sample Output[/align]

1
2
......

 

[align=left]Author[/align]
zjt
 
看这个题目真把我吓尿了，以为果断要超时的。没想到其实就4个数满足条件。上个代码、
 

#include <stdio.h>
int jieceng(int a)
{
int s,i;
s=1;
for(i=2;i<=a;i++)
s*=i;
return s;
}
int weishu(int a)
{
int sum=0;
int b;
b=a;
while(a)
{
sum+=jieceng(a%10);
a/=10;
}
if(sum==b)
return 1;
return 0;  //上面都是预处理。
}
int main()
{
int n,i;
for(i=1;i<=40585;i++)  //可以先用for(i=1;;i++)这个枚举来确定所有满足的数，然后控制区间长度。一共就四个数1,2,145,40585.
{
if(weishu(i))
printf("%d\n",i);
}
}
//      其实一行代码就能AC 了。  printf("1\n2\n145\n40585\n");