FZU 2108 Mod problem

参考这里：http://blog.csdn.net/q775968375/article/details/8828952
大神说的没错，这跟建树没有半毛线的关系，就是一DFS。。。
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 
 5 #define LL long long int
 6 
 7 const int MAXN = 1010;
 8 
 9 struct node
10 {
11     LL val, BitWide;
12     node(): val(0), BitWide(0) { }
13 };
14 
15 char str[MAXN];
16 int MOD;
17 
18 LL Qmod( LL a, LL k )   //快速幂模板
19 {
20     LL v = 1;
21     while ( k )
22     {
23         if ( k & 1 ) v = ( v * a ) % MOD;
24         a = ( a * a ) % MOD;
25         k >>= 1;
26     }
27     return v % MOD;
28 }
29 
30 node DFS( int l, int r )
31 {
32     node ans;   //本层括号的数值
33 
34     int left, right, cnt;
35     int top = 0;    //括号配对专用栈顶指针
36     for ( int i = l; i <= r; ++i )
37     {
38         if ( str[i] == '[' )
39         {
40             if ( !top ) left = i + 1;
41             ++top;
42         }
43         else if ( str[i] == ']' )
44         {
45             --top;
46             if ( !top )  //找到一对配对括号
47             {
48                 right = i - 1;
49                 ++i;
50                 cnt = str[i] - '0';   //本括号内的内容的重复次数
51                 node temp = DFS( left, right );   //求得本括号内的数的值，以及位数(10进制下)
52 
53                 LL base = Qmod( 10, temp.BitWide );   //括号内的数是10的多少次方
54 
55                 ans.BitWide += cnt * temp.BitWide;    //位数加上
56                 ans.val = ( ans.val * Qmod( base, cnt ) ) % MOD;  //此括号的重复次数是10的多少次方
57 
58                 LL st = 1, BB = base;
59 
60                 while ( --cnt )        //这里不好解释，我费解了半天，等等画个图贴上来
61                 {
62                     st += BB;
63                     BB = ( BB * base ) % MOD;
64                     st %= MOD;
65                 }
66 
67                 temp.val = (temp.val * st) % MOD;
68                 ans.val = ( ans.val + temp.val ) % MOD;
69             }
70         }
71         else if ( !top )
72         {
73             ans.val = ans.val * 10 + str[i] - '0';
74             ans.val %= MOD;
75             ++ans.BitWide;
76         }
77     }
78 
79     return ans;
80 }
81 
82 int main()
83 {
84     int T;
85     scanf( "%d", &T );
86     while ( T-- )
87     {
88         scanf( "%s%lld", str, &MOD );
89         node ans = DFS( 0, strlen(str) - 1 );
90         printf("%lld\n", ans.val );
91     }
92     return 0;
93 }