[UPC] Postfix Evaluation 后缀表达式求值 | 栈的简单应用

题目描述

In a postfix expression, operators follow their operands. For example, [ 5 2 * ] is interpreted as 5 * 2. If there are multiple operators, each operator appears after its last operand. Here are more examples, showing how postfix compares to parenthesized expressions:

6 3+5∗2− 等同 ( ( 6 + 3 ) ∗ 5 ) – 2 )

4 3 2∗+ 等同 4 + ( 3 ∗ 2 )

These examples show that the operand affected by an operator can be the result of a previous calculation.

Here is what you need to do: Given an integer postfix expression, you must calculate and print its value.

输入

Each input will consist of a single test case. Your program will be run multiple times on different inputs.

A single input line will contain a postfix expression containing single digit integers and operators from the following set: { +, –, /, * } (add, subtract, divide, multiply). There will be a single space between each digit and and operator. The line will contain no more than 64 operators and numbers, total.

输出

Output will be a single integer on a line by itself.

样例输入 Copy

【样例1】
6 3 + 5 * 2 –
【样例2】
6 3 + 5 2 * *
【样例3】
4 3 2 * + 

样例输出 Copy

【样例1】
43
【样例2】
90
【样例3】
10

stack<ll> st;
int main() {
  char c;
  while(cin >> c) {
    if(isdigit(c)){
      st.push(c - '0');
    }else{
      ll a = st.top();
      st.pop();
      ll b = st.top();
      st.pop();
      if(c == '+') st.push(a + b);
      else if(c == '-') st.push(b - a);
      else if(c == '*') st.push(a * b);
      else st.push(b / a);
    }
  }
  cout << st.top() << endl;
  return 0;
}
/**
**/