POJ2248 Addition Chains

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <cstring>
 5 #define min(a, b) ((a) < (b) ? (a) : (b))
 6 #define max(a, b) ((a) > (b) ? (a) : (b)) 
 7 
 8 inline void read(int &x)
 9 {
10     x = 0;char ch = getchar(), c = ch;
11     while(ch < ‘0‘|| ch > ‘9‘)c = ch, ch = getchar();
12     while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘, ch = getchar();
13     if(c == ‘-‘)x = -x; 
14 }
15 
16 const int MAXN = 100 + 10;
17 
18 int num[MAXN], n, b[MAXN], ID;
19 
20 int dfs(int step)
21 {
22     if(num[step - 1] == n) return 1;
23     if(step > ID)return 0;
24     register int tmp;
25     for(register int i = step - 1;i >= 1;-- i)
26         for(register int j = i;j >= 1;-- j)
27         {
28             tmp = num[i] + num[j];
29             if(tmp <= num[step - 1] || b[tmp] || tmp > n)continue;
30             num[step] = tmp;
31             b[tmp] = 1;
32             if(dfs(step + 1)) return 1;
33             b[tmp] = 0;
34         }
35     return 0;
36 }
37 
38 int main()
39 {
40     num[1] = 1;
41     while(scanf("%d", &n) != EOF && n)
42     {
43         for(register int i = 1;i <= n;++ i)
44         {
45             ID = i;
46             memset(b, 0, sizeof(b));
47             if(dfs(2)) 
48             {
49                 for(register int j = 1;j <= i;++ j)
50                     printf("%d ", num[j]);
51                 putchar(‘\n‘);
52                 break;
53             }
54         }
55     }
56     return 0;
57 }