#define is unsafe

时间：2023-03-09 05:19:45
#define is unsafe

#define is unsafe
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 66 Accepted Submission(s): 52
 
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
Author
madfrog
Source
HDU2010省赛集训队选拔赛（校内赛）
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lcy
 
/*

题意：模拟题,给你一行只有MAX函数，和加法运算的表达式，输出表达式的值，并且统计在本质MAX运算中+运算的次数

初步思路：一行表达式中有用的有 (:表示一个运算开始了 , +:表示有一个加法运算, ):表示一个运算结束了

*/

#include<bits/stdc++.h>

using namespace std;

struct node{

    int a;//用来存储数字

    int res;//用来存储加法运算的次数

    node(){};

    node(int b,int c){

        a=b;

        res=c;

    }

};

stack<node>Q;//用来存储数字

stack<char>oper;//用来存储运算符

node x1,x2;

int n;

char str[];

int main(){

    // freopen("in.txt","r",stdin);

    scanf("%d",&n);

    getchar();

    while(n--){

        gets(str);

        //cout<<str<<endl;

        for(int i=;i<strlen(str);i++){

            if(str[i]>=''&&str[i]<=''){//如果当前位置的是数字的话，直接将他压进栈中

                int cur=str[i]-'';

                i++;

                while(str[i]>=''&&str[i]<=''){

                    cur=cur*+str[i]-'';

                    i++;

                }

                i--;//这位不是数字了但是不能继续往前走，要让for循环来承担这个任务

                Q.push(node(cur,));//将这个数字装进栈中，并且初始化经过乘法运算的次数为零

            }else if(str[i]=='+'||str[i]=='('){

                oper.push(str[i]);

            }else if(str[i]==','){//如果遇见','就要将','左边的加法运算进行计算之后重新压到栈中

                while(!oper.empty()&&oper.top()=='+'){//提取到一个加号就要运算一次

                    x1=Q.top();

                    Q.pop();

                    x2=Q.top();

                    Q.pop();

                    x1.a+=x2.a;

                    x1.res+=(x2.res+);//这地方将max函数展开看就知道为什么加法运算是累加了

                    Q.push(x1);

                    oper.pop();//将这个运算符出栈

                }

            }else if(str[i]==')'){//遇到一个外括号肯定是一个max运算结束了

                //将这之前的加法运算全部算完

                while(!oper.empty()&&oper.top()=='+'){//提取到一个加号就要运算一次

                    x1=Q.top();

                    Q.pop();

                    x2=Q.top();

                    Q.pop();

                    x1.a+=x2.a;

                    x1.res+=(x2.res+);//这地方将max函数展开看就知道为什么加法运算是累加了

                    Q.push(x1);

                    oper.pop();//将这个运算符出栈

                }

                oper.pop();

                x2=Q.top();

                Q.pop();

                x1=Q.top();//后取的才是第一个

                Q.pop();

                if(x1.a>x2.a){//说明x1的部分要运算两次

                    x1.res=x1.res*+x2.res;

                    Q.push(x1);

                }else{//否则的话就是x2的部分要运算两次

                    x1.res=x1.res+x2.res*;

                    x1.a=x2.a;

                    Q.push(x1);

                }

            }

        }

        //将所有的操作进行之后，栈中如果还有运算，只能是MAX函数之间的加法运算

        while(!oper.empty()){

            x1=Q.top();

            Q.pop();

            x2=Q.top();

            Q.pop();

            x1.a+=x2.a;

            x1.res+=(x2.res+);

            Q.push(x1);

            oper.pop();

        }

        printf("%d %d\n",Q.top().a,Q.top().res);

        Q.pop();//将最后一个元素出栈

    }

    return ;

}

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