HDOJ 2212 DFS

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

……

分析:9的阶乘为362880, 9!*10 而且由0~9的阶乘组成的最大数就是3628800。

而且0的阶乘是1,而不是0.

因为根据阶乘定义 n!=n*(n-1)!;

1!=1*0!=1;

所以人为规定了0!=1;

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
int k[10]= {1,1};
void ff()
{
int i;
for(i = 2; i < 10; i ++){
k[i] = k[i-1]*i;
// printf("%d\n",k[i]);
}
}
int main()
{
ff();
long i;
long a,sum;
for(i=1; i<=3628800; i++)
{
a=i;
sum=0;
while(a!=0)//a>0
{
sum+=k[a%10];
a=a/10;
//printf("%d\n",a);
}
if(sum==i)
printf("%ld\n",i);
}
return 0;
}
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