DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7486    Accepted Submission(s): 4578

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
......

所说最后觉得挺水的，但一来看见 [1, 2147483647] 就在想什么比较好的方法可以处理，结果想了半天也没想出来。一看题解，9！=3628801，也就是说10个9也就是36288010，1e7的复杂度，加上中间对每个位的处理，也就是不到1e8的复杂度，而题目给了2000ms，所以可以解决。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int mul[10];

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

bool cmp(int x)
{
    int a,tem=0;
    a=x;
    do
    {
        int mm=a%10;
        tem+=mul[mm];
        a/=10;
    }while(a>0);
    if(x==tem)
        return 1;
    else
        return 0;
}

int main()
{
    long long num1=0,num2=0;
    calcu();
    //cout<<mul[9];
    for(int i=1;i<=mul[9]*10;i++)
    {
        if(cmp(i))
            printf("%d\n",i);
    }
    return 0;
}