HDOJ 1237 simple calculator (stack)

HDOJ 1237 simple calculator (stack)
Simple Calculator

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 14833 Accepted Submission (s): 5050

 

Problem Description reads a non-negative integer calculation expression that only contains +,-, *,/, and calculates the value of the expression.

The Input of the Input test contains several test cases. Each test case occupies one row. Each row contains no more than 200 characters. integers and operators are separated by a space. No invalid expression. When a row contains only 0, the input ends, and the corresponding results are not output.

Output outputs one line for each test case, that is, the value of this expression, accurate to 2 digits after the decimal point.

Sample Input

1 + 24 + 2 * 5 - 7 / 110

Sample Output

3.0013.36

 

Just use the stack. The c code is as follows:

 

#include
 
  #include
  
   double a[210];int main(){int i,sign,count,j;double sum;char c;while(1){sign=1;i=0;memset(a,0,sizeof(a));scanf(%lf,&a[0]);while(getchar()!=''){sign=0;scanf(%c %d,&c,&count);if(c=='+')  a[++i]=count;else if(c=='-')  a[++i]=-count;else if(c=='*')  a[i]=a[i]*count;else if(c=='/')  a[i]=a[i]/(count*(1.0));}if(sign)  break;sum=0;for(j=0;j<=i;j++)  sum+=a[j];printf(%0.2lf,sum);}return 0;}
  
 

 

C ++ STL -- stack solution:

 

 

#include
 
  #include
  
   using namespace std;int main(){int i,sign,j;double sum,num;char c;while(1){sign=1;stack
   
    count;scanf(%lf,&num);count.push(num);while(getchar()!=''){sign=0;scanf(%c %lf,&c,&num);if(c=='+')   count.push(num);else if(c=='-')   count.push(-num);else if(c=='*'){double x=count.top();x=x*num;count.pop();count.push(x);}else{double x=count.top();x=x/num;count.pop();count.push(x);}}if(sign)   break;sum=0;while(!count.empty()){sum+=count.top();count.pop();}printf(%.2lf,sum);}return 0;}