Problem E - 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 12 numbers, 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.

题意：给你一个没有括号的表达式，只有数字和加乘号，怎么样组合使它最大最小？

思路：首先计算+号再计算乘号得到的结果最大，反之结果最小，可以用贪心方法处理，使用栈模拟，处理的时候遇到优先级大的先计算再压入栈中，最后处理优先级小的。

代码如下：

#include<iostream>
#include<stack>

using namespace std;

stack<long long>maxst;
stack<long long>minst;

int main()
{
    int num;
    cin>>num;
    while(num--)
    {
        while(!maxst.empty())
            maxst.pop();
        while(!minst.empty())
            minst.pop();
       long long val,x,y,z;
        char ch;
        cin>>val;
        maxst.push(val),minst.push(val);
        while(cin.get(ch)&&ch!=‘\n‘)
        {
            cin>>val;
            if(ch==‘+‘)
            {
                x=maxst.top()+val;
                maxst.pop();
                maxst.push(x);
                minst.push(val);
            }
            else if(ch==‘*‘)
            {
                x=minst.top()*val;
                minst.pop();
                minst.push(x);
                maxst.push(val);
            }
        }
        long long maxv=1,minv=0;
        while(!maxst.empty())
        {
            maxv=maxst.top()*maxv;
            maxst.pop();
        }
        while(!minst.empty())
        {
            minv=minst.top()+minv;
            minst.pop();
        }
        cout<<"The maximum and minimum are "<<maxv<<" and "<<minv<<"."<<endl;
    }
    return 0;
}

[贪心&&栈模拟]uva10700 Camel trading,布布扣,bubuko.com

[贪心&&栈模拟]uva10700 Camel trading

原文：http://blog.csdn.net/zju_ziqin/article/details/21073697