Depressed C small plus (ii)

Depressed C small plus (ii) time limit: +Ms | Memory Limit:65535KB Difficulty:4

Describe

Smart you help C small add to solve the infix expression to the suffix expression conversion (for details please refer to "Depressed C small plus (a)"), C small plus very happy. But C-Plus is a thoughtful person who wants to calculate the value of an expression in this way. That is, the expression is converted to a suffix expression, and then evaluated. It is also considered that the operand is a decimal and multi-digit situation.

Input

The first line enters an integer t, which is a total of T-Group test data (T<10).
Each set of test data has only one row, which is a string of not more than 1000 in length, representing the formula, each of which ends with a "=". This expression contains only +-*/and parentheses. Where parentheses can be used in a nested set. The data guarantees that no negative numbers are present in the operands of the input and less than 1000000.
The data guarantee divisor will not be 0.

Output

For each set of test data output consists of two lines, first output the converted suffix expression, and then output the results of the calculation, the result is reserved two decimal places. Two sets of test data are separated by a blank line.

Sample input

21+2= (19+21) *3-4/5=

Sample output

12+=3.001921+3*45/-=119.20

Code:

#include <stdio.h>
Double Fun (double A, double b, Char ch)
{
if (ch== ' + ')
return b+a;
if (ch== '-')
return b-a;
if (ch== ' * ')
return b*a;
if (ch== '/')
return b/a;
}
int main (void)
{
int i,n,top1,top2;
Double x,t,a,b,num[1000];
Char str[1000],ch[1000];
scanf ("%d", &n);
while (n--)
{
Top1=-1;
Top2=-1;
scanf ("%s", str);
for (i=0;str[i]!= ' n '; i++)
{
x=0;
if (str[i]>= ' 0 ' &&str[i]<= ' 9 ')
{
while (str[i]>= ' 0 ' &&str[i]<= ' 9 ')
{
printf ("%c", Str[i]);
x=x*10+str[i]-' 0 ';
i++;
}
if (str[i]== '. ')
{
printf ("%c", Str[i]);
i++;
t=0.1;
while (str[i]>= ' 0 ' &&str[i]<= ' 9 ')
{
printf ("%c", Str[i]);
x=x+ (str[i]-' 0 ') *t;
t=t*0.1;
i++;
}
}
Num[++top1]=x;
}
if (str[i]== ')
Break
if (str[i]== ' (')
{
Ch[++top2]=str[i];
}
else if (str[i]== ') ')
{
while (top2>=0&&ch[top2]!= ' (')
{
printf ("%c", Ch[top2]);
A=NUM[TOP1];
top1--;
B=NUM[TOP1];
Num[top1]=fun (A,b,ch[top2]);
top2--;
}
top2--;
}
else if (str[i]== ' * ' | | str[i]== '/')
{
while (ch[top2]== ' * ' | | ch[top2]== '/')
{
printf ("%c", Ch[top2]);
A=NUM[TOP1];
top1--;
B=NUM[TOP1];
Num[top1]=fun (A,b,ch[top2]);
top2--;
}
Ch[++top2]=str[i];
}
Else
{
while (top2>=0&&ch[top2]!= ' (')
{
printf ("%c", Ch[top2]);
A=NUM[TOP1];
top1--;
B=NUM[TOP1];
Num[top1]=fun (A,b,ch[top2]);
top2--;
}
Ch[++top2]=str[i];
}
}
while (top2>=0)
{
printf ("%c", Ch[top2]);
A=NUM[TOP1];
top1--;
B=NUM[TOP1];
Num[top1]=fun (A,b,ch[top2]);
top2--;
}
printf ("\n%.2f\n", num[0]);
}
return 0;
}

Depressed C small plus (ii)