HDU-2212 DFS

DFS

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

Total Submission(s): 8889    Accepted Submission(s): 5439


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

 

水题

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <cmath>
#include <string>
using namespace std;

int f[10];
int t[10];
int sum;
int tmp;

void init(){
t[0]=1;
for(int i=1;i<=9;i++){
t[i]=t[i-1]*i;
}
}

void fun(int n){
sum=0;
tmp=n;
while(tmp){
sum+=t[tmp%10];
tmp/=10;
}
if(sum==n){
cout<<n<<endl;
}
}

int main(){
init();
for(int i=1;i<=10*t[9];i++){
fun(i);
}
return 0;
}


结果如下:

1
2
145
40585