DFS

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

Total Submission(s): 5298    Accepted Submission(s): 3252

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!&＃43;4!&＃43;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
......

这道题就是逗你玩呢。。。

分析：9的阶乘为362880， 而且由0~9的阶乘组成的最大数就是3628800，所以搜索一下就好了，结果你会很惊讶。。。

只有1， 2， 145， 40585符合。。。

代码：

#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;
}

也可以直接输出。。。。