杭电2212DFS

DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6381    Accepted Submission(s): 3930

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
......

 

我看见题目以为是深搜，吓死宝宝了，解释一下，9的阶乘=362880||*********||362880*7=一个7位数，就是3000000左右吧，362880*8=7位数还是3000000左右，所以3000000以后的就不需要考虑了，肯定没有，一个循环就解决了.

附代码：

#include<stdio.h>
#include<string.h>
int ac(int n)
{
	int ans=1;
	for(int i=1;i<=n;i++)
	ans*=i;
	return ans;
}
long i;
int main()
{
	
	for(i=1;i<=40585;i++)
	{
		long m=0,n=i;
		while(n)
		{
			int ans=n%10;
			m+=ac(ans);
			n/=10;
		}
		if(m==i)
		printf("%d\n",i);
	}
	return 0;
}