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 }