NYOJ - 35 表达式求值 分类： NYOJ 2015-03-18 10:33 31人阅读 评论(0) 收藏

#include<iostream>
#include<string>
#include<stack>
#include<cstdio>
using namespace std;
stack<double>sn;
stack<char>sc;
double num, des, dou;
bool prt, flag;
int t, len;
string str;
bool isNum(char c)
{
if(c >= '0' && c <= '9')
return true;
if(c== '.')
{
flag = true;
return true;
}
return false;
}

bool isOp(char c)
{
if(c == '+' || c == '-' || c == '*' || c == '/')
return true;
return false;
}

void init()
{
while(sn.size())	sn.pop();
while(sc.size())	sc.pop();
str.clear();
len = num = des = 0;
dou = 1;
prt = false;
flag = false;
}

void calm()
{
if(sc.size() && isOp(sc.top()))
{
double ans;
double y = sn.top();
sn.pop();
double x = sn.top();
sn.pop();
char op = sc.top();
sc.pop();
switch(op)
{
case '+':	ans = x + y;	break;
case '-':	ans = x - y;	break;
case '*':	ans = x * y;	break;
case '/':	ans = x / y;	break;
}
//		cout<<op<<" "<<ans<<endl;
sn.push(ans);
}

}

int main()
{
//	freopen("in.in","r",stdin);
cin>>t;
while(t--)
{
init();
cin>>str;
int len = str.length();

for(int i = 0; i < len; i++)
{
if(isNum(str[i]))
{
if(str[i] == '.')	continue;
if(!flag)	num = num*10 + str[i] - '0';
else{
des = des*10 + str[i] - '0';
dou *= 10;
}
}
else
{
num = num + des/dou;
if(str[i] == '+' || str[i] == '-')
{
if(!prt)	sn.push(num);
while(sc.size() && sc.top() != '(')calm();
sc.push(str[i]);
prt = false;
}
else if(str[i] == '*' || str[i] == '/')
{
if(!prt)	sn.push(num);
if(sc.size())
if(sc.top()=='*' || sc.top()=='/')
calm();
sc.push(str[i]);
prt = false;
}
else if(str[i] == '(')
{
sc.push(str[i]);
}
else if(str[i] == ')')
{
if(!prt)	sn.push(num);
while(sc.top() != '(')	calm();

sc.pop();
prt = true;
}
else
{
if(!prt)	sn.push(num);
while(sn.size() > 1)	calm();
printf("%0.2lf\n",sn.top());
sn.pop();
break;
}
num = 0;
flag = false;
dou = 1;
des = 0;
}
}
}

return 0;
}