UVA Problem E - Camel trading（栈和队列）-优快云博客

本文介绍了一个基于驼队交易背景的数学问题，旨在通过编写程序确定不含括号的算术表达式的最大和最小可能值。问题源自公元800年左右的一段历史故事，涉及加法和乘法运算，探讨了不同运算优先级下表达式的多种解释。

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Description

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.

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：求该算术式所能得到的最大最小值（区分在于括号再哪）

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<queue>

using namespace std;

char a[1000100];

long long int min1;
long long int max1;

void minn()
{
    queue<long long int>q1;
    long long int num = 1;
    int p;
    long long int sum = 0;
    for(int i=0; a[i]!='\0'; i++)
    {
        if(a[i]>='0' && a[i]<='9')
        {
            long long int p = a[i] - '0';
            sum = sum * 10 + p;
        }
        else
        {
            if(a[i] == '+')
            {
                sum = sum * num;
                q1.push(sum);
                sum = 0;
                num = 1;
            }
            else if(a[i] == '*')
            {
                num = num * sum;
                sum = 0;
            }
        }
    }
    q1.push(sum*num);
    min1 = 0;
    while(!q1.empty())
    {
        num = q1.front();
        min1 = min1 + num;
        q1.pop();
    }
}

void maxx()
{
    queue<long long int>q2;
    long long int sum = 0;
    long long int num = 0;
    for(int i=0; a[i]!='\0'; i++)
    {
        if(a[i]>='0' && a[i]<='9')
        {
            long long int p = a[i] - '0';
            sum = sum *10 + p;
        }
        else
        {
            if(a[i] == '*')
            {
                sum = sum + num;
                q2.push(sum);
                sum = 0;
                num = 0;
            }
            else if(a[i] == '+')
            {
                num = num + sum;
                sum = 0;
            }
        }
    }
    q2.push(num+sum);
    max1 = 1;
    while(!q2.empty())
    {
        num = q2.front();
        max1 = max1 * num;
        q2.pop();
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(a,0,sizeof(a));
        scanf("%s",a);
        minn();
        maxx();
        printf("The maximum and minimum are %lld and %lld.\n",max1,min1);
    }
    return 0;
}