HDU 3350 #define is unsafe 栈的模拟题 宏问题

#define is unsafe

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 252    Accepted Submission(s): 152

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

6MAX(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 02 13 14 25 228 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 number{    int x,count;};stack<number> num_stack;stack<char> oper;char str[1001];int main(){        int t,len,i,y;    number temp1,temp2,num;    //    freopen("test.txt","r",stdin);    scanf("%d",&t);    while(t--)    {        scanf("%s",str);        len=strlen(str);                for(i=0;i<len;i++)        {            if(str[i]>='0'&&str[i]<='9') //数字 提取完整入数字栈             {                y=str[i]-'0';                i++;                while(str[i]>='0'&&str[i]<='9')                {                    y=y*10+str[i]-'0';                    i++;                }                i--;                num.count=0;                num.x=y;                num_stack.push(num);            }            else if(str[i]=='('||str[i]=='+')//( ,+ 符号 直接入符号栈                     {                        oper.push(str[i]);                    }            else if(str[i]==',')//如果是‘，’，则把‘，’左边连续的‘+’进行运算，                                 //运算结果重新压入栈中                 {                    while(!oper.empty()&&oper.top()=='+')                    {                        temp1=num_stack.top();                        num_stack.pop();                        temp2=num_stack.top();                        num_stack.pop();                        temp1.x+=temp2.x;                        temp1.count+=temp2.count+1;                        num_stack.push(temp1);                        oper.pop();                        }                                }            else if(str[i]==')')//遇到‘）’，则说明有一个MAX 可以执行，                                //但要先把左边连续的‘+’运算完                {                    while(!oper.empty()&&oper.top()=='+')                    {                        temp1=num_stack.top();                        num_stack.pop();                        temp2=num_stack.top();                        num_stack.pop();                        temp1.x+=temp2.x;                        temp1.count+=temp2.count+1;                        num_stack.push(temp1);                        oper.pop();                        }                    oper.pop();                    temp2=num_stack.top();                    num_stack.pop();                    temp1=num_stack.top();                    num_stack.pop();                    if(temp1.x>temp2.x)                        temp1.count=temp1.count*2+temp2.count;                    else                     {                        temp1.x=temp2.x;                        temp1.count=temp1.count+temp2.count*2;                    }                    num_stack.push(temp1);                                }        }        while(!oper.empty()&&oper.top()=='+')//完之后，还可能存在操作符没运算，                                                //但只可能是‘+’        {                temp1=num_stack.top();                num_stack.pop();                temp2=num_stack.top();                num_stack.pop();                temp1.x+=temp2.x;                temp1.count+=temp2.count+1;                num_stack.push(temp1);                oper.pop();            }        printf("%d %d\n",num_stack.top().x,num_stack.top().count);        num_stack.pop();        }    return 0;}