#define is unsafe
 
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
 Total Submission(s): 255    Accepted Submission(s): 153 
 

 Problem 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
  
 
/*
hdoj 3350 栈的模拟题

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

读入整个字符串：
遇到MAX跳过，
1.遇到'（'，'+' 入符号栈 
2.遇到数字 入数字栈 
3，遇到','表示max的一半已经进入
    判断符号栈顶是否为'+', 是，就要弹出2个数字操作+
    
4,遇到')'，表示max结束，就要操作比较大小
    但是左边仍然有可能有+的操作 ，
    所以，先要判断符号栈顶是否为'+', 是，就要弹出2个数字操作+ 
然后开始比较大小： 
    注意这里默认的','前后的+操作为0，
    若有一方为1，且是比较大的数，则操作是需要乘以2的
    MAX(1+2,2)==>max(3,1)==>3 操作了2次+，同下： 
    ( 1 + 2 ) > 2 ? ( 1 + 2 ) : 2  
最后结束 仍然要判断符号栈顶是否有+操作，就是max中间的+ 
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stack>
using namespace std;
struct node{
	int x,count;
};

stack<node> ss;
stack<char> oper;

char str[1001];
int main()
{
	int t,len,i,sum;
	node tmp,tmp1,tmp2;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s",str);
		len=strlen(str);
		for(i=0;i<len;i++)
		{
			if(str[i]>='0'&&str[i]<='9')
			{
				sum=0;
				while(str[i]>='0'&&str[i]<='9')
				{
					sum=sum*10+str[i]-'0';
					i++;
				}
				i--;
				tmp.x=sum;
				tmp.count=0;
				ss.push(tmp);
			}
			else if(str[i]=='('||str[i]=='+')
			{
				oper.push(str[i]);
			}
			else if(str[i]==',')
			{
				while(!oper.empty()&&oper.top()=='+')
				{
					tmp1=ss.top();
					ss.pop();
					tmp2=ss.top();
					ss.pop();
					tmp1.x+=tmp2.x;
					tmp1.count+=tmp2.count+1;
					ss.push(tmp1);
					oper.pop();
				}
			}
			else if(str[i]==')')
			{
				while(!oper.empty()&&oper.top()=='+')
				{
					tmp1=ss.top();
					ss.pop();
					tmp2=ss.top();
					ss.pop();
					tmp1.x+=tmp2.x;
					tmp1.count+=tmp2.count+1;
					ss.push(tmp1);
					oper.pop();
				}
				oper.pop();//弹出 )
				tmp2=ss.top();
				ss.pop();
				tmp1=ss.top();
				ss.pop();
				if(tmp1.x>tmp2.x)//tmp1 保留最大的 
                    tmp1.count=tmp1.count*2+tmp2.count;
                else 
                {
                    tmp1.x=tmp2.x;
                    tmp1.count=tmp1.count+tmp2.count*2;
                }
                ss.push(tmp1);            				
			}
		}
		while(!oper.empty()&&oper.top()=='+')//完之后，还可能存在操作符没运算，
                                              //但只可能是‘+’
        {
                tmp1=ss.top();
				ss.pop();
				tmp2=ss.top();
				ss.pop();
				tmp1.x+=tmp2.x;
				tmp1.count+=tmp2.count+1;
				ss.push(tmp1);
				oper.pop();   
        }
        printf("%d %d\n",ss.top().x,ss.top().count);
       	ss.pop(); 
	}
	return 0;
}