#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 " , 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 " , 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

Author

madfrog

Source

HDU2010省赛集训队选拔赛（校内赛）

Recommend

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[1010];
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&n);
    getchar();
    while(n--){
        gets(str);
        //cout<<str<<endl;
        for(int i=0;i<strlen(str);i++){
            if(str[i]>='0'&&str[i]<='9'){//如果当前位置的是数字的话，直接将他压进栈中
                int cur=str[i]-'0';
                i++;
                while(str[i]>='0'&&str[i]<='9'){
                    cur=cur*10+str[i]-'0';
                    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,0));//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+1);//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+1);//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*2+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*2;
                    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+1);
            Q.push(x1);
            oper.pop();
        }
        printf("%d %d
",Q.top().a,Q.top().res);
        Q.pop();//将最后一个元素出栈
    }
    return 0;
}