Reverse Polish notation

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

一开始出错 [Error] void value not ignored as it ought to be 原因是a = s.pop() 使用的一个函数的返回值类型是void，而对它进行了赋值处理;

使用到的C库函数atoi, 注意atoi()里传入的参数(atoi函数原型见下)

标准C库函数
#include <stdlib.h>

原型 : int atoi( const char *str );

功能：将字符串str转换成一个整数并返回结果。参数str 以数字开头，当函数从str
中读到非数字字符则结束转换并将结果返回。
例如：int num = atoi(“1314.012”);
int值为1314

解题思路: 边输入边处理, 输入的字符若是数字, 则入栈, 若是(+, -, *)其中一个, 便从栈中弹出两个元素进行相应的运算(注意运算的顺序, 次栈顶元素  运算符  栈顶元素), 再将运算结果入栈, 依次循环

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <stack>
using namespace std;

int main()
{
	stack<int> s;
	int a, b;
	char ch[110];

	while(scanf("%s", ch) != EOF)
	{
		if(ch[0] == '+')
		{
			a = s.top();
			s.pop();
			b = s.top();
			s.pop();
			s.push(a + b);
		}
		else if(ch[0] == '-')
		{
			a = s.top();
			s.pop();
			b = s.top();
			s.pop();
			s.push(b - a);
		}
		else if(ch[0] == '*')
		{
			a = s.top();
			s.pop();
			b = s.top();
			s.pop();
			s.push(a * b);
		}
		else
		{
			s.push(atoi(ch));
		}
	}

	printf("%d\n", s.top());

	while(!s.empty())
	{
		s.pop();
	}

	return 0;
}

　　

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int top, S[1000];

void push(int x)
{
	S[++top] = x;	// top 加1之后将元素插入top所在位置
}

int pop()
{
	top --;
	return S[top + 1];	// 返回top所指的元素
}

int main()
{
	int a, b;
	top = 0;
	char s[100];
	while(scanf("%s", s) != EOF)
	{
		if(s[0] == '+')
		{
			a = pop();
			b = pop();
			push(a + b);
		}
		else if(s[0] == '-')
		{
			a = pop();
			b = pop();
			push(b - a);
		}
		else if(s[0] == '*')
		{
			a = pop();
			b = pop();
			push(a * b);
		}
		else
		{
			push(atoi(s));
		}
	}

	printf("%d\n", pop());

	return 0;
}