Hdoj2212DFS [water]

DFSTimeLimit: 50002000 MS (JavaOthers) & #160; & #160; & #160; & #160; MemoryLimit: 6553632768 K (JavaOthers) TotalSubmission (s): 5298 & #160; & #160; & #160; & #160; Accepted

DFSTime Limit: 5000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission (s): 5298 Accepted Submission (s): 3252

Problem DescriptionA 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.
Inputno input
OutputOutput all the DFS number in increasing order.
Sample Output

12......

This question is just a joke...

Analysis: the factorial of 9 is 362880, and it ranges from 0 ~ The maximum number of factorial 9 is 3628800, so you can just search for it, and you will be surprised...

Only 1, 2,145,405 85...

Code:

#include 
 
  int f[10] = {1, 1};int a[10];void ff(){    int i;    for(i = 2; i < 10; i ++) f[i] = f[i-1]*i;    //printf("%d", f[9]);}int judge(int n){    int sum1 = n, sum2;    sum2 = 0;    while(n){        int temp = n%10;        sum2 += f[temp];        n/= 10;    }    if(sum1 == sum2) return 1;    return 0;}int main(){    ff();    for(int i = 1; i <= 3628800; i ++)        if(judge(i)) printf("%d\n", i);    return 0;}
 

You can also directly output ....