Evaluating Simple C Expressions
The task in this problem is to evaluate a sequence of simple C expressions, buy you need not know C to solve the problem! Each of the expressions will appear on a line by itself and will contain no more than 110
characters. The expressions to be evaluated will contain only simple integer variables and a limited set of operators; there will be no constants in the expressions. There are 26 variables which may appear in our simple expressions, namely those with the names a through z (lower-case
letters only). At the beginning of evaluation of each expression, these 26 variables will have the integer values 1 through 26, respectively (that is, a = 1, b = 2, ..., n = 14, o = 15, ..., z = 26). Each variable
will appear at most once in an expression, and many variables may not be used at all.
The operators that may appear in expressions include the binary (two-operand) + and -, with the usual interpretation. Thus the expression a + c - d + b has the value 2 (computed as 1
+ 3 - 4 + 2). The only other operators that may appear in expressions are ++ and --. These are unary (one-operand) operators, and may appear before or after any variable. When the ++ operator appears before a variable, that
variable's value is incremented (by one) before the variable's value is used in determining the value of the entire expression. Thus the value of the expression ++c - b is 2, with c being incremented to 4 prior to evaluating the entire expression.
When the ++ operator appears after a variable, that variable is incremented (again, by one) after its value is used to determine the value of the entire expression. Thus the value of the expression c++ - b is 1, but c is incremented
after the complete expression is evaluated; its value will still be 4. The -- operator can also be used before or after a variable to decrement (by one) the variable; its placement before or after the variable has the same significance as for the ++ operator.
Thus the expression --c + b-- has the value 4, with variables c and b having the values 2 and 1 following the evaluation of the expression.
Here's another, more algorithmic, approach to explaining the ++ and -- operators. We'll consider only the ++operator, for brevity:
- Identify each variable that has a ++ operator before it. Write a simple assignment statement that increments the value of each such variable, and remove the ++ operator from before that variable in the expression.
- In a similar manner, identify each variable that has a ++ operator after it. Write a simple assignment statement that increments the value of each of these, and remove the ++ operator from after that variable in the expression.
- Now the expression has no ++ operators before or after any variables. Write the statement that evaluates the remaining expression after those statements written in step 1, and before those written in step 2.
- Execute the statements generated in step 1, then those generated in step 3, and finally the one generated in step 2, in that order.
Using this approach, evaluating the expression ++a + b++ is equivalent to computing
- a = a + 1 (from step 1 of the algorithm)
- expression = a + b (from step 3)
- b = b + 1 (from step 2)
where expression would receive the value of the complete expression.
Input and Output
Your program is to read expressions, one per line, until the end of the file is reached. Display each expression exactly as it was read, then display the value of the entire expression, and on separate lines, the
value of each variable after the expression was evaluated. Do not display the value of variables that were not used in the expression. The samples shown below illustrate the desired exact output format.
Blanks are to be ignored in evaluating expressions, and you are assured that ambiguous expressions like a+++b(ambiguous because it could be treated as a++ + b or a + ++b) will not appear
in the input. Likewise, ++ or --operators will never appear both before and after a single variable. Thus expressions like ++a++ will not be in the input data.
Sample Input
a + b b - z a+b--+c++ c+f--+--a f-- + c-- + d-++e
Sample Output
Expression: a + b
value = 3
a = 1
b = 2
Expression: b - z
value = -24
b = 2
z = 26
Expression: a+b--+c++
value = 6
a = 1
b = 1
c = 4
Expression: c+f--+--a
value = 9
a = 0
c = 3
f = 5
Expression: f-- + c-- + d-++e
value = 7
c = 2
d = 4
e = 6
f = 5
题意：就是给你一堆表达式，然后a~z一开始分别代表1~26。而这些表达式当中会有+，—，++，— —。其中自加和自减可以分别为前缀和后缀。前缀代表先使该值自加或自减，然后再用该值。后缀则是先用，后进行自加自减。最后要得到表达式的值，以及出现过的字母的最后值。
这道题我没有看出和树有什么很大的关联。所以就直接模拟做了。
先判断每个字母是否有前缀，然后用一个标记符来记录加减号。具体实现看代码吧
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<ctype.h>
using namespace std;
int main ()
{
char str[1000];
int len,i,j,t;
while(gets(str))
{
t=i=0;
int a[30],b[30]={0};
for (i=1; i<=26; i++)
a[i]=i;
len=strlen(str);
char right[120];
for (i=0; i<len; i++)
{
if (str[i]!=' ')
right[t++]=str[i];
}
right[t]='\0';
i=0;
int sum=0,k=1;
for (i=0; i<t; i++)
{
if (isalpha(right[i]))
{
b[right[i]-'a'+1]=1;
bool flag=false;
if (i>=2)
{
if(right[i-1]==right[i-2])
{
if (right[i-1]=='+') a[right[i]-'a'+1]+=1;
else a[right[i]-'a'+1]-=1;
flag=true;
sum+=k*a[right[i]-'a'+1];
if (right[i+1]=='+') k=1;
else k=-1;
}
}
if (i<=t-3)
{
if (right[i+1]==right[i+2])
{
sum+=k*a[right[i]-'a'+1];
if (right[i+1]=='-') a[right[i]-'a'+1]-=1;
else a[right[i]-'a'+1]+=1;
flag=true;
if (right[i+3]=='+') k=1;
else k=-1;
}
}
if (!flag)
{
sum+=k*a[right[i]-'a'+1];
if (right[i+1]=='+') k=1;
else k=-1;
}
}
}
printf("Expression: %s\n",str);
cout<<" value = "<<sum<<endl;
for (i=1; i<=26; i++)
if (b[i]) printf(" %c = %d\n",i+96,a[i]);
}
return 0;
}