NYOJ 35表达式求值

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stack>
#include<sstream>
#include<algorithm>
#include<cstdlib>
using namespace std;

char pri[7][7] = {  // 判断符号优先级
{'>','>','<','<','<','>','>'},
{'>','>','<','<','<','>','>'},
{'>','>','>','>','<','>','>'},
{'>','>','>','>','<','>','>'},
{'<','<','<','<','<','=',' '},
{'>','>','>','>',' ','>','>'},
{'<','<','<','<','<',' ','='}};

int shine(char ch)  //映射符号到整型
{
switch(ch)
{
case '+':return 0;
case '-':return 1;
case '*':return 2;
case '/':return 3;
case '(':return 4;
case ')':return 5;
case '=':return 6;
default:return 7;
}
}
char judge(char a,char b) //判断两个字符的优先级
{
return pri[shine(a)][shine(b)];
}
char str1[3000];
int pos = 0,p = 0;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(str1,0,sizeof(str1));
stack<char> ope;//操作符
stack<double> ope_num;//操作数
string str;
cin>>str; getchar();
double res = 0; pos = 0;
int n; p = 0;
while(pos < str.size()-1)
{
double midnum = 0;
if(str[pos] <= '9' && str[pos] >= '0')
{
//for(int i = pos;shine(str[i]) == 7;++i) //输出数据
//printf("%c",str[i]);
sscanf(&str[pos],"%lf%n",&midnum,&n);  //读取数据
sprintf(&str1[p],"%lf",midnum);  //将数据输入到STR1数组中，STR1数组存的是后缀表达式
p += n;str1[p++] = '@'; // 每个数据后符号后加个特殊字符，方便以后计算后缀表达式
pos += n; ope_num.push(midnum);
}
else
{
char ch = str[pos];  pos++;
if(ope.empty()) ope.push(ch);
else
{
loop:if(!ope.empty())
switch(judge(ope.top(),ch))  //判断新读取的运算符与栈顶运算符的优先级
{
case '>':{                //如果优先级小于等于栈顶元素，弹出栈顶运算符,直至大于栈顶运算符的优先级
//printf("%c",ope.top());
sprintf(&str1[p++],"%c",ope.top());
str1[p++] = '@';
ope.pop();goto loop;
}
case '<':ope.push(ch);break;  // 将新加入的运算符压入操作符栈内
case '=':ope.pop();break;
default:printf("error\n");
}
}
if(ope.empty() && shine(ch) != 5) ope.push(ch); //将新加入的运算符压入操作符栈内
}
}
while(!ope.empty()) //弹出操作符栈内的操作符
{
//printf("%c",ope.top());
str1[p++] = ope.top();
str1[p++] = '@';
ope.pop();
}str1[--p] = '\0';
while(!ope_num.empty()) ope_num.pop();
double num1 = 0,num2 = 0;
char op;
for(int i = 0;i < p;)   //计算后缀表达式
{
if(str1[i] <= '9' && str1[i] >= '0')
{
sscanf(&str1[i],"%lf%n",&num1,&n);
ope_num.push(num1); i += n;
}
else if(str1[i] == '@') i++;
else
{
num2 = ope_num.top(); ope_num.pop();
num1 = ope_num.top(); ope_num.pop();
switch(str1[i])
{
case '+':num1 += num2;break;
case '-':num1 -= num2;break;
case '*':num1 *= num2;break;
case '/':num1 /= num2;break;
}
++i;
ope_num.push(num1);
}
}
//printf("=\n");
printf("%.2lf\n",ope_num.top());
}
return 0;
}