UVA 10700 Camel trading

Camel trading

Time Limit: 1 second

Background

Aroud 800 A.D., El Mamum, Calif of Baghdad was presented the formula 1+2*3*4+5, which had its origin in the financial accounts of a camel transaction. The formula lacked parenthesis and was ambiguous. So, he decided to ask savants to provide him with a method
to find which interpretation is the most advantageous for him, depending on whether is is buying or selling the camels.

The Problem

You are commissioned by El Mamum to write a program that determines the maximum and minimum possible interpretation of a parenthesis-less expression.

Input

The input consists of an integer N, followed by N lines, each containing an expression. Each expression is composed of at most 12numbers, each ranging between 1 and 20, and
separated by the sum and product operators + and *.

Output

For each given expression, the output will echo a line with the corresponding maximal and minimal interpretations, following the format given in the sample output.

Sample input

3
1+2*3*4+5
4*18+14+7*10
3+11+4*1*13*12*8+3*3+8

Sample output

The maximum and minimum are 81 and 30.
The maximum and minimum are 1560 and 156.
The maximum and minimum are 339768 and 5023.

可以加括号，求表达式的最大最小值。

因为只有乘法和加法，所以最小值就是什么都不做，最大值是先算加法。

到了使用栈的时候了，c++真是方便啊。

#include<iostream>  
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<vector>
#include<functional>
#include<stack>
using namespace std;
long long t, sum1, sum2, c, d;
char ch;
stack<long long> a, b;

int main(){
	cin >> t;
	while (t--)
	{
		scanf("%lld", &c);	
		a.push(c);	b.push(c);
		while (scanf("%c", &ch) && (ch == '+' || ch == '*'))
		{
			scanf("%lld", &c);
			if (ch == '+'){ d = a.top() + c; a.pop(); a.push(d); b.push(c); }
			else{ d = b.top() * c; b.pop(); b.push(d); a.push(c); }
		}
		sum1 = 1; while (!a.empty()) sum1 *= a.top(), a.pop();
		sum2 = 0; while (!b.empty()) sum2 += b.top(), b.pop();
		printf("The maximum and minimum are %lld and %lld.\n", sum1, sum2);
	}
	return 0;
}