Ultraviolet A 10700-Camel trading (the maximum and minimum values of stack expressions)

Ultraviolet A 10700-Camel trading (the maximum and minimum values of stack expressions)

Camel tradingTime Limit: 1 second

Background

Aroud 800. D ., el Mamum, Calif of Baghdad was presented the formula 1 + 2*3*4 + 5, which had its origin in the financial accounts of a camel transaction. the formula lacked parenthesis and was ambiguous. so, he decided to ask savants to provide him with a method to find which interpretation is the most advantageous for him, depending on whether is buying or selling the camels.

The Problem

You are commissioned by El Mamum to write a program that determines the maximum and minimum possible interpretation of a parenthesis-less expression.

Input

The input consists of an integerN, FollowedNLines, each containing an expression. Each expression is composed of at most12Numbers, each ranging1And20, And separated by the sum and product operators+And*.

Output

For each given expression, the output will echo a line with the corresponding maximal and minimal interpretations, following the format given in the sample output.

Sample input

31+2*3*4+54*18+14+7*103+11+4*1*13*12*8+3*3+8

Sample output

The maximum and minimum are 81 and 30.The maximum and minimum are 1560 and 156.The maximum and minimum are 339768 and 5023.

Question: use parentheses to find the maximum and minimum values of an expression,

Idea: There is no clue at first, But if you calculate it carefully, the expression is calculated according to the priority order, and the result is the minimum value, if you calculate the addition first and then calculate the multiplication, it is the maximum value, so the rest can be simulated using the stack.

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include 
     
      #include #include 
      
       #include 
       
        #include 
        
         #include 
         using namespace std;int main(){ int T; char ch; long long a,t; long long min,max; scanf("%d",&T); while(T--) { stack
          
           minn; stack
           
            maxx; scanf("%lld",&a); minn.push(a); maxx.push(a); while((ch=getchar())!='\n') { scanf("%lld",&a); if(ch=='+') { minn.push(a); t=maxx.top(); maxx.pop(); t+=a; maxx.push(t); } else if(ch=='*') { maxx.push(a); t=minn.top(); minn.pop(); t*=a; minn.push(t); } } min=0; max=1; while(!minn.empty()) { min+=minn.top(); minn.pop(); } while(!maxx.empty()) { max*=maxx.top(); maxx.pop(); } printf("The maximum and minimum are %lld and %lld.\n",max,min); } return 0;}