poj 1684 Lazy Math Instructor(字符串)

题目链接：http://poj.org/problem?id=1686

思路分析：该问题为表达式求值问题，对于字母使用浮点数替换即可，因为输入中的数字只能是单个digit。

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
using namespace std;

const int MAX_N = ;
const double factor = 1.1;
double number[MAX_N];
char str[MAX_N];
int len, pos;

double Expression( );
double Term( );
double Factor( );

void InputExpression( )
{
  int i = ;
  pos = ;
  string str_in;
  getline(cin, str_in);

  for (i = , len = ; i < str_in.length( ); ++i, ++len)
    str[len] = str_in[i];
  str[len] = '\0';
}

char GetCurrentToken( )
{
  while (str[pos] == ' ' || str[pos] == '\t')
    pos++;
  return str[pos];
}

char GetNextToken( )
{
  ++pos;
  while (str[pos] == ' ' || str[pos] == '\t')
    pos++;
  if (pos == len)
    return '\n';
  else
    return str[pos];
}

double Expression( )
{
  char ch;
  double term1 = Term( );

  while ((ch = GetCurrentToken( )) == '+' || (ch == '-')) {
    GetNextToken( );
    double term2 = Term( );

    if (ch == '+')
      term1 += term2;
    else
      term1 -= term2;
  }

  return term1;
}

double Term( )
{
  char ch;
  double factor_1 = Factor( );

  while ((ch = GetCurrentToken( )) == '*') {
    GetNextToken( );
    double factor_2 = Factor( );

    factor_1 *= factor_2;
  }
  return factor_1;
}

double Factor( )
{
  double value = 0.0;
  char ch = GetCurrentToken( );

  if (ch == '(') {
    GetNextToken( );
    value = Expression( );
    GetNextToken( );
  } else if (isdigit(ch)) {
    value = ch - '';
    GetNextToken( );
  }
  else if (ch == '\n')
    value = -;
  else {
    if (number[ch - 'A'] == )
      value = number[ch - 'A'] = (ch - 'A') * 1.1;
    else
      value = number[ch - 'A'];
    GetNextToken( );
  }
  return value;
}

int main( )
{
  int case_time;

  scanf("%d\n", &case_time);
  while (case_time--) {
    InputExpression( );
    double ans_1 = Expression( );

    InputExpression( );
    double ans_2 = Expression( );

    if (abs(ans_1 - ans_2) < 1e-)
      printf("YES\n");
    else
      printf("NO\n");
  }

  return ;
}