POJ 1472 Instant Complexity

模拟题：递归求解，有点像语法分析器

代码

   
     

    
      
    
       1 
    
      #include 
    
      <
    
      iostream
    
      >
    
      

    
       2 
    
       
    
      using
    
       
    
      namespace
    
       std;

    
       3 
    
       
    
      char
    
       ch[
    
      10
    
      ];

    
       4 
    
       
    
      int
    
       solve(
    
      int
    
      *
    
       exps)

    
       5 
    
      {

    
       6 
    
       scanf(
    
      "
    
      %s
    
      "
    
      , ch);

    
       7 
    
       
    
      if
    
      (ch[
    
      0
    
      ]
    
      ==
    
      '
    
      E
    
      '
    
      )
    
      return
    
       
    
      0
    
      ;

    
       8 
    
       
    
      if
    
      (ch[
    
      0
    
      ]
    
      ==
    
      '
    
      B
    
      '
    
      )

    
       9 
    
       
    
      while
    
      (solve(exps));

    
      10 
    
       
    
      else
    
       
    
      if
    
      (ch[
    
      0
    
      ]
    
      ==
    
      '
    
      L
    
      '
    
      )

    
      11 
    
       {

    
      12 
    
       scanf(
    
      "
    
      %s
    
      "
    
      , ch);

    
      13 
    
       
    
      int
    
       i,s,t 
    
      =
    
       
    
      -
    
      1
    
      ,texps[
    
      11
    
      ]
    
      =
    
      {
    
      0
    
      };

    
      14 
    
       
    
      if
    
      (ch[
    
      0
    
      ]
    
      !=
    
      '
    
      n
    
      '
    
      )t
    
      =
    
      atoi(ch);

    
      15 
    
       
    
      while
    
      (solve(texps));

    
      16 
    
       
    
      if
    
      (t
    
      ==-
    
      1
    
      ){

    
      17 
    
       
    
      for
    
      (i
    
      =
    
      10
    
      ;i
    
      >
    
      0
    
      ;i
    
      --
    
      )texps[i]
    
      =
    
      texps[i
    
      -
    
      1
    
      ];

    
      18 
    
       texps[
    
      0
    
      ]
    
      =
    
      0
    
      ;

    
      19 
    
       }

    
      20 
    
       
    
      else
    
      

    
      21 
    
       
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      11
    
      ;i
    
      ++
    
      )texps[i]
    
      *=
    
      t;

    
      22 
    
       
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      11
    
      ;i
    
      ++
    
      )exps[i]
    
      +=
    
      texps[i];

    
      23 
    
       }

    
      24 
    
       
    
      else
    
      

    
      25 
    
       {

    
      26 
    
       scanf(
    
      "
    
      %s
    
      "
    
      , ch);

    
      27 
    
       exps[
    
      0
    
      ]
    
      +=
    
      atoi(ch);

    
      28 
    
       
    
      return
    
       solve(exps);

    
      29 
    
       }

    
      30 
    
       
    
      return
    
       
    
      1
    
      ;

    
      31 
    
      }

    
      32 
    
      
    
      int
    
       main()

    
      33 
    
      {

    
      34 
    
       
    
      int
    
       n,i,j,t, exps[
    
      11
    
      ];

    
      35 
    
       scanf(
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      n);

    
      36 
    
       
    
      for
    
      (i
    
      =
    
      1
    
      ;i
    
      <=
    
      n;i
    
      ++
    
      )

    
      37 
    
       {

    
      38 
    
       memset(exps, 
    
      0
    
      , 
    
      sizeof
    
      (exps));

    
      39 
    
       solve(exps);

    
      40 
    
       printf(
    
      "
    
      Program #%d\nRuntime = 
    
      "
    
      , i);

    
      41 
    
       
    
      for
    
      (t
    
      =
    
      0
    
      ,j
    
      =
    
      10
    
      ;j
    
      >=
    
      0
    
      ;j
    
      --
    
      )

    
      42 
    
       
    
      if
    
      (exps[j]
    
      !=
    
      0
    
      )

    
      43 
    
       {

    
      44 
    
       t
    
      ++
    
      ;

    
      45 
    
       
    
      if
    
      (t
    
      !=
    
      1
    
      )printf(
    
      "
    
      +
    
      "
    
      );

    
      46 
    
       
    
      if
    
      (exps[j]
    
      !=
    
      1
    
      ||
    
      j
    
      ==
    
      0
    
      )printf(
    
      "
    
      %d
    
      "
    
      , exps[j]);

    
      47 
    
       
    
      if
    
      (exps[j]
    
      !=
    
      1
    
      &&
    
      j
    
      >
    
      0
    
      )printf(
    
      "
    
      *
    
      "
    
      );

    
      48 
    
       
    
      if
    
      (j
    
      >
    
      1
    
      )printf(
    
      "
    
      n^%d
    
      "
    
      , j);

    
      49 
    
       
    
      if
    
      (j
    
      ==
    
      1
    
      )printf(
    
      "
    
      n
    
      "
    
      );

    
      50 
    
       }

    
      51 
    
       
    
      if
    
      (
    
      !
    
      t)printf(
    
      "
    
      0
    
      "
    
      );

    
      52 
    
       printf(
    
      "
    
      \n\n
    
      "
    
      );

    
      53 
    
       }

    
      54 
    
       
    
      return
    
       
    
      0
    
      ;

    
      55 
    
      }