#define is unsafe

 Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

Description

Have you used #define in C/C++ code like the code below?

#include <stdio.h>
#define MAX(a , b) ((a) > (b) ? (a) : (b))
int main()
{
printf(“%d\n” , MAX(2 + 3 , 4));
return 0;
}

Run the code and get an output: 5, right?
You may think it is equal to this code:

#include <stdio.h>
int max(a , b) {  return ((a) > (b) ? (a) : (b));  }
int main()
{
printf(“%d\n” , max(2 + 3 , 4));
return 0;
}

But they aren’t.Though they do produce the same anwser , they work in two different ways.
The first code, just replace the MAX(2 + 3 , 4) with ((2 + 3) > (4) ? (2 + 3) : 4), which calculates (2 + 3) twice.
While the second calculates (2 + 3) first, and send the value (5 , 4) to function max(a , b) , which calculates (2 + 3) only once.

What about MAX( MAX(1+2,2) , 3 ) ?
Remember “replace”.
First replace: MAX( (1 + 2) > 2 ? (1 + 2) : 2 , 3)
Second replace: ( ( ( 1 + 2 ) > 2 ? ( 1 + 2 ) : 2 ) > 3 ? ( ( 1 + 2 ) > 2 ? ( 1 + 2 ) : 2 ) : 3).
The code may calculate the same expression many times like ( 1 + 2 ) above.
So #define isn’t good.In this problem,I’ll give you some strings, tell me the result and how many additions(加法) are computed.

Input

The first line is an integer T(T<=40) indicating case number.
The next T lines each has a string(no longer than 1000), with MAX(a,b), digits, ‘+’ only(Yes, there’re no other characters).
In MAX(a,b), a and b may be a string with MAX(c,d), digits, ‘+’.See the sample and things will be clearer.

Output

For each case, output two integers in a line separated by a single space.Integers in output won’t exceed 1000000.

Sample Input

6

MAX(1,0)

1+MAX(1,0)

MAX(2+1,3)

MAX(4,2+2)

MAX(1+1,2)+MAX(2,3)

MAX(MAX(1+2,3),MAX(4+5+6,MAX(7+8,9)))+MAX(10,MAX(MAX(11,12),13))

Sample Output

1 0

2 1

3 1

4 2

5 2

28 14

Source

HDU2010省赛集训队选拔赛（校内赛）

AC代码：

#include <cstdio>
#include <cstring>
#include <stack>
using namespace std;
char buf[1010];
struct node
{
};
stack <node> num;
stack <char> sig;
void input()
{
scanf("%s", buf);
}
void solve()
{
while (!num.empty())
num.pop();
while (!sig.empty())
sig.pop();
int len = strlen(buf);
node tmp;
for (int i = 0; i < len; i++)
{
if (buf[i] == '(' || buf[i] == '+')
sig.push(buf[i]);
else if (buf[i] >= '0' && buf[i] <= '9')
{
while (buf[i] >= '0' && buf[i] <= '9')
tmp.val = tmp.val * 10 + buf[i++] - '0';
i--;
num.push(tmp);
}
else if (buf[i] == ',')
{
while (!sig.empty() && sig.top() == '+')
{
sig.pop();
tmp = num.top();
num.pop();
tmp.val += num.top().val;
num.pop();
num.push(tmp);
}
}
else if (buf[i] == ')')
{
while (!sig.empty() && sig.top() == '+')
{
sig.pop();
tmp = num.top();
num.pop();
tmp.val += num.top().val;
num.pop();
num.push(tmp);
}
sig.pop();
tmp = num.top();
num.pop();
if (tmp.val < num.top().val)
{
tmp.val = num.top().val;
}
else
num.pop();
num.push(tmp);
}
}
while (!sig.empty() && sig.top() == '+')
{
sig.pop();
tmp = num.top();
num.pop();
tmp.val += num.top().val;
num.pop();
num.push(tmp);
}
}
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
input();
solve();
}
return 0;
}