Postfix expressions(Data Structure)

最新推荐文章于 2023-06-02 16:37:34 发布

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本文介绍了一种使用栈来解决后缀表达式求值问题的方法。通过遍历输入的后缀表达式，遇到字母则将其转换为对应的数值并压入栈中；遇到运算符，则从栈顶弹出两个元素进行相应的计算，并将结果压回栈中。最终输出结果保留两位小数。

Input

Integer is the first line,the number of the test caseis t.

Following t lines, each line is a string whose length does not exceed 100, representing a Postfix expression. Expression containing only +,-, *,/four operators and the 26 lowercase letters of the alphabet, does not contain other characters. Each letter represents a positive integer:
a = 1,b = 2,c = 3…y = 25,z = 26.

Output

Each input sample, single line output: Postfix expression value, a positive real number s, to two decimal places.

Case

Sample Input

2
ab+c*
int**py++

Sample Output

9.00
2561.00

Using the stack to solve the problem,if the traverse run into the letters, then push into the stack,otherwise, let the top element in the stackmake the operations to the next letter,then push the result into thestack.
Finally,output the number with two decimal places, using fixedand setprecision(2).

code (Hope have a help to you)

#include<iostream>
#include<string>
#include<stack>
#include <iomanip>
using namespace std;

int main() {
    int n;
    cin >> n;
    while (n--) {
        stack<float> temp;
        string str;
        cin >> str;
        int size = str.length();
        for (int i = 0; i < size; i++) {
            if (str[i] == '+' || str[i] == '-' || str[i] == '*' || str[i] == '/') {
                float a = temp.top();
                temp.pop();
                float b = temp.top();
                temp.pop();
                if (str[i] == '+') {
                    b += a;
                    temp.push(b);
                } else if (str[i] == '-') {
                    b -= a;
                    temp.push(b);
                } else if (str[i] == '*') {
                    b *= a;
                    temp.push(b);
                } else if (str[i] == '/') {
                    b /= a;
                    temp.push(b);
                }
            } else {
                temp.push(str[i]-'a'+1);
            }
        }
        cout<<fixed<<setprecision(2)<<temp.top()<<endl;
    }
}