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题目只是提示四则运算，是否存在负数，是否存在括号等情况并没详细说明，因此为保证程序的健壮性，编写代码时将这两种情况考虑进去，具体的详细注释在代码中体现。
补充测试样例：
13
(-1)*(-1)-1
(-1)*(-1)-(1)*(-1)
(-1*-3*(1*4))*1
(((1+9)))
-1+((-2*-3)*-1)
1+1
2.2/3
1+2*3
1+2*-300
-1+2+0
(-1*(-3+-4-(-1*-2)))
(-2)
2*(-300)/(-2)*2
#include<iostream>
#include<stack>
#include<algorithm>
#include<cstdlib>
#include<cmath>
using namespace std;
int a[300][300]={0};//用来判断该当前字符串是进行入栈操作，还是先出栈再入栈
int book[300]={0};//用来判断'-'是减号还是负号
void init(){//当前符号是左括号直接入栈 a['(']['('] = 1;a['+']['('] = 1;a['-']['('] = 1;a['*']['('] = 1;a['/']['('] = 1;//栈内没有操作符则将该操作符入栈 a['#']['('] = 1;a['#']['+'] = 1;a['#']['-'] = 1;a['#']['*'] = 1;a['#']['/'] = 1;//栈顶为左括号，四则运算符号入栈 a['(']['+'] = 1;a['(']['-'] = 1;a['(']['*'] = 1;a['(']['/'] = 1;//栈顶为加号，乘除入栈 a['+']['*'] = 1;a['+']['/'] = 1;//栈顶为减号，乘除入栈 a['-']['*'] = 1;a['-']['/'] = 1;//以下三种符号后面紧跟着的'-'为负号 book['('] = 1;book['+'] = 1;book['-'] = 1;book['*'] = 1;book['/'] = 1;
}
//运算操作
double acc(double a,double b,char c){if(c == '+'){return a + b;}if(c == '-'){return a - b;}if(c == '*'){return a * b;}if(c == '/'){return a / b;}
}
int main(){init();int n;cin>>n;for(int k = 0;k < n;k++){string s;cin>>s;cout<<s<<endl;stack<double>s1;//操作数栈 stack<char>s2;//操作符栈 s2.push('#');//放入一个非操作符 int cnt = 0;//用来统计当前操作数的长度 bool flag = true;//第一个字符就是'-'则该符号肯定为负号 for(int i = 0;i < s.length();i++){//以下三种情况表示该字符属于操作数的一部分//1、阿拉伯数字 //2、小数点//3、负号 if( (s[i] >= '0'&&s[i] <= '9') || (s[i] == '.') || (flag&&s[i] == '-')){cnt ++;//操作数长度加一 flag = false;//跟在操作数后面的'-'为减号 //字符串末尾是操作数时需要进行额外判断 if(i == s.length()-1&&cnt > 0){s1.push( atof( s.substr(i-cnt+1,cnt).c_str() ) );// cout<<s1.top()<<endl;}}else{//该字符前面出现了操作数，将字符串转化成浮点数入栈 if(flag==false&&cnt > 0){s1.push( atof( s.substr(i-cnt,cnt).c_str() ) );// cout<<s1.top()<<endl;cnt = 0;//操作数长度清零 }if(book[s[i]] == 1){//该字符后面紧跟着的'-'是负号 flag = true;}//cout<<"操作符"<<s[i]<<endl;if(s[i] == ')'){//两个括号之间只有一个操作数，没有操作符，将左括号出栈 if(s2.top()=='('){s2.pop();}else{//两个括号中间有操作符和两个操作数，应该先把这两个数进行四则运算，再将左括号出栈 //进行四则运算最少需要两个数，并且需要一个操作符，左括号已经排除了 if(s1.size() > 1&&book[s2.top()]){//先出栈的在操作符后面，后出栈的在操作符前面//栈的原则，先进后出 double c = s1.top();s1.pop();double b = s1.top();s1.pop();char top = s2.top();s2.pop();s2.pop();s1.push(acc(b,c,top));// cout<<b<<" "<<c<<" "<<top<<" "<<s[i]<<endl;}}}else{if(a[s2.top()][s[i]] == 1){s2.push(s[i]);//当前操作符的优先级比栈顶操作符优先级高 }else{//进行四则运算最少需要两个数，并且需要一个操作符，如果左括号不会进入该步 if(s1.size() > 1&&book[s2.top()]){double c = s1.top();s1.pop();double b = s1.top();s1.pop();char top = s2.top();s2.pop();s1.push(acc(b,c,top));// cout<<b<<" "<<c<<" "<<top<<" "<<s[i]<<endl;}s2.push(s[i]);//将当前操作符入栈 }} }}//完场剩余的四则运算 while(s1.size() > 1&&book[s2.top()]){double c = s1.top();s1.pop();double b = s1.top();s1.pop();char top = s2.top();s2.pop();s1.push(acc(b,c,top));// cout<<b<<" "<<c<<" "<<top<<endl;}printf("%.1f\n",s1.top());}return 0;
}