DFS 注意点

1、打表问题

//1364495 2009-05-13 20:42:44 Time Limit Exceeded 2212 2000MS 232K 426 B C++ Wpl

//1364655 2009-05-13 21:03:44 Accepted 2212 0MS 204K 299 B C++ Wpl  

/*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] ).*/

#include <iostream>

#define MAX 5

using namespace std;

int data[MAX];

int main()

{

    data[0]=1;

    data[1]=2;

    data[2]=145;

    data[3]=40585;

    int i;

    for(i=0;i<=3;i++)

        printf("%d
",data[i]);

    return 0;

}

2、记忆数组（省去不必要的操作）

//1364495 2009-05-13 20:42:44 Time Limit Exceeded 2212 2000MS 232K 426 B C++ Wpl 

#include <iostream>

#include <fstream>

#define MAX 10000

using namespace std;

int f[11],data[MAX];

bool DFS(int n)

{

    int sum=0,x=n;

    while(x!=0)

    {

        sum+=f[x%10];

        x=x/10;

        if(sum>n)

            return false;

    }

    if(sum==n)

        return true;

    else

        return false;

}

int main()

{

    int i,j;

    f[0]=1;

    for(i=1;i<=10;i++)

        f[i]=i*f[i-1];

    ofstream outfile("ans.txt");

    j=0;

    for(i=1;i<2147483647;i++)

    {

        if(DFS(i))

        {

            outfile<<"data["<<j++<<"]="<<i<<endl;

        }

    }

    return 0;

}

3、如果超时，可以尝试后面的不用输出，将结果找出之后打表

#include <iostream>
#include <cstdio>
using namespace std;
bool DFS(int x);
int jie(int n)
{
    if(n == 1 || n == 0) return 1;
    return n * jie(n-1);
}
int main()
{
    int i;
    for(i = 1; i <= 40585; i++)//2147483647
        if(DFS(i))
            cout << i << endl;
    return 0;
}
bool DFS(int x)
{
    int xx = x, sum = 0;
    while(x != 0)
    {
        int tmp = x % 10;
        sum += jie(tmp);
        x /= 10;
        if(sum > xx)
            return false;
    }
    if(xx == sum)
        return true;
    return false;
}