#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
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 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;
}
