栈 Stack Aizu - ALDS1_3_A

atoi()函数将字符串形式的数字转换为整形数值

string类函数c_str():返回当前字符串的首字符地址

题目：

Reverse Polish notation is a notation where every operator follows all of its operands. For example, an expression (1+2)*(5+4) in the conventional Polish notation can be represented as 1 2 + 5 4 + * in the Reverse Polish notation. One of advantages of the Reverse Polish notation is that it is parenthesis-free.

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

题意：已知逆波兰表达式，求值

#include<bits/stdc++.h>

typedef long long ll;

using namespace std;

int main(){
	ios::sync_with_stdio(false);
	stack<int> s;
	
	string str;
	while(cin>>str){
		if(str[0] == '+'){
			int t = s.top(); s.pop();
			t += s.top(); s.pop();
			s.push(t);
		}
		else if(str[0] == '-'){
			int t = s.top()*(-1) ; s.pop();
			t += s.top(); s.pop();
			s.push(t);
		}
		else if(str[0] == '*'){
			int t = s.top(); s.pop();
			t *= s.top(); s.pop();
			s.push(t);
		}
		else if(str[0] == '/'){
			int a = s.top(); s.pop();
			int b = s.top(); s.pop();
			s.push(b/a);
		}
		else{
			s.push(atoi(str.c_str()));  //atoi()函数将字符串形式的数字转换为整形数值 
                                                    //string类函数c_str():返回当前字符串的首字符地址
		}
	}
	cout<<s.top()<<endl;
	return 0;
}