DFS 注意点

1、打表问题

 1 //1364495 2009-05-13 20:42:44 Time Limit Exceeded 2212 2000MS 232K 426 B C++ Wpl
 2 
 3 //1364655 2009-05-13 21:03:44 Accepted 2212 0MS 204K 299 B C++ Wpl  
 4 
 5 /*For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number.
 6 
 7 Now you should find out all the DFS numbers in the range of int( [1, 2147483647] ).*/
 8 
 9 #include <iostream>
10 
11 #define MAX 5
12 
13 using namespace std;
14 
15 int data[MAX];
16 
17 int main()
18 
19 {
20 
21     data[0]=1;
22 
23     data[1]=2;
24 
25     data[2]=145;
26 
27     data[3]=40585;
28 
29     int i;
30 
31     for(i=0;i<=3;i++)
32 
33         printf("%d\n",data[i]);
34 
35     return 0;
36 
37 }
View Code

2、记忆数组（省去不必要的操作）

 1 //1364495 2009-05-13 20:42:44 Time Limit Exceeded 2212 2000MS 232K 426 B C++ Wpl 
 2 
 3 #include <iostream>
 4 
 5 #include <fstream>
 6 
 7 #define MAX 10000
 8 
 9 using namespace std;
10 
11 int f[11],data[MAX];
12 
13 bool DFS(int n)
14 
15 {
16 
17     int sum=0,x=n;
18 
19     while(x!=0)
20 
21     {
22 
23         sum+=f[x%10];
24 
25         x=x/10;
26 
27         if(sum>n)
28 
29             return false;
30 
31     }
32 
33     if(sum==n)
34 
35         return true;
36 
37     else
38 
39         return false;
40 
41 }
42 
43 int main()
44 
45 {
46 
47     int i,j;
48 
49     f[0]=1;
50 
51     for(i=1;i<=10;i++)
52 
53         f[i]=i*f[i-1];
54 
55     ofstream outfile("ans.txt");
56 
57     j=0;
58 
59     for(i=1;i<2147483647;i++)
60 
61     {
62 
63         if(DFS(i))
64 
65         {
66 
67             outfile<<"data["<<j++<<"]="<<i<<endl;
68 
69         }
70 
71     }
72 
73     return 0;
74 
75 }
View Code

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

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 bool DFS(int x);
 5 int jie(int n)
 6 {
 7     if(n == 1 || n == 0) return 1;
 8     return n * jie(n-1);
 9 }
10 int main()
11 {
12     int i;
13     for(i = 1; i <= 40585; i++)//2147483647
14         if(DFS(i))
15             cout << i << endl;
16     return 0;
17 }
18 bool DFS(int x)
19 {
20     int xx = x, sum = 0;
21     while(x != 0)
22     {
23         int tmp = x % 10;
24         sum += jie(tmp);
25         x /= 10;
26         if(sum > xx)
27             return false;
28     }
29     if(xx == sum)
30         return true;
31     return false;
32 }
View Code

 

posted @ 2015-04-09 16:28  PastLIFE  阅读(244)  评论(0)  编辑  收藏  举报