#define is unsafe hdu3350

#define is unsafe

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 545    Accepted Submission(s): 341

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

 

思路：

如果是数字就进栈，如果是加号或者是(代表运算的开始，也放入栈中，如果遇到了一个逗号，代表左边的已经输入完毕，要运算左边的值，如果遇到）代表max运算结束了，先把逗号右边的算出来，左边的值与右边的进行比较。进行完这一系列之后，如果运算符的栈里面还有值的话就是max之间的运算。

 

#include<stack>
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
string s;
struct number
{
	int a;int res;
	number(){
	}
	number(int x,int y){
		a=x;res=y;
	}
};
number x1,x2;
stack<number>q;
stack<char>oper;
int main()
{
	int ca;
	scanf("%d",&ca);
	while(ca--)
	{
		cin>>s;
		int cur=0;
		for(int i=0;i<s.length();i++)
		{
			if(s[i]>='0'&&s[i]<='9')
			{
				cur=s[i]-'0';
				i++;
				while(s[i]>='0'&&s[i]<='9')
				{
					cur=cur*10+s[i]-'0';
					i++;
				}
				i--;
				q.push(number(cur,0));
			}
			else if(s[i]=='+'||s[i]=='(')oper.push(s[i]);
			else if(s[i]==',')
			{
				while(!oper.empty()&&oper.top()=='+')
				{
					x1=q.top();q.pop();
					x2=q.top();q.pop();
					x1.a+=x2.a;
					x1.res+=(x2.res)+1;
					q.push(x1);
					oper.pop(); 
				}
			}
			else if(s[i]==')')//说明一个max运算结束 
			{
				while(!oper.empty()&&oper.top()=='+')
				{
					x1=q.top();q.pop();
					x2=q.top();q.pop();
					x1.a+=x2.a;
					x1.res+=(x2.res)+1;
					q.push(x1);
					oper.pop(); 
				}
				oper.pop();//弹出(
				x2=q.top();q.pop();
				x1=q.top();q.pop();
				if(x1.a>x2.a)//大的那个加法算了两次 
				{
					x1.res=x1.res*2+x2.res;
					q.push(x1);
				}
				else 
				{
					x1.res=x1.res+x2.res*2;
					x1.a=x2.a;
					q.push(x1);
				}
			}
		}
		while(!oper.empty()){//max之间的加法运算 
			x1=q.top();q.pop();
			x2=q.top();q.pop();
			x1.a+=x2.a;
			x1.res+=(x2.res+1);
			q.push(x1);
			oper.pop();
		}  
		printf("%d %d\n",q.top().a,q.top().res);
		q.pop();
	}
	return 0;
}