栈（裸题）

Stack    Aizu - ALDS1_3_A

Write a program which reads an expression in the Reverse Polish notation and prints the computational result.

An expression in the Reverse Polish notation is calculated using a stack. To evaluate the expression, the program should read symbols in order. If the symbol is an operand, the corresponding value should be pushed into the stack. On the other hand, if the symbols is an operator, the program should pop two elements from the stack, perform the corresponding operations, then push the result in to the stack. The program should repeat this operations.

Input

An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character.

You can assume that +, - and * are given as the operator and an operand is a positive integer less than 10⁶

Output

Print the computational result in a line.

Constraints

2 ≤ the number of operands in the expression ≤ 100
1 ≤ the number of operators in the expression ≤ 99
-1 × 10⁹ ≤ values in the stack ≤ 10⁹

Sample Input 1

1 2 +

Sample Output 1

3

Sample Input 2

1 2 + 3 4 - *

Sample Output 2

-3

Notes

Template in C

本题就是一个裸的栈问题，计算后缀表答式，具体后追表达式是什么就不清楚的讲了，网上很多，只要知道从右往左一次遍历，遇到一个符号就取出他的前两个数来计算，因为还存在括号这种情况所以会使用到栈。而我的代码则是用数组模拟栈，老实说我觉得STL的栈真的很慢。

简单题不多说，直接上代码：

    #include <iostream>  
    #include <cstdio>  
    #include <cstdlib>  
    #include <cstring>  
    using namespace std;  
      
    const int maxn = 100 + 5;  
      
    int top = 0;  
    int S[1000];  
      
    void push(int key){  
      S[++top] = key;  
      return;  
    }  
      
    int pop(){  
      top--;  
      return S[top+1];  
    }  
      
    int main(){  
      int a,b;  
      top = 0;  
      char s[maxn];  
      while(scanf("%s",s) != EOF){  
        if(s[0] == '+'){  
          a = pop();  
          b = pop();  
          push(a+b);  
        }  
        else if(s[0] == '-'){  
          b = pop();  
          a = pop();  
          push(a-b);  
        }  
        else if(s[0] == '*'){  
          a = pop();  
          b = pop();  
          push(a*b);  
        }  
        else{  
          push(atoi(s));  
        }  
      }  
      printf("%d\n",pop());  
    }