HDU 2646 栈的应用 STL

Expression

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 202    Accepted Submission(s): 61

Problem Description

  As a shopkeeper of a restaurant,everyday ,dandelion's mother needs to calculate the incomes.Sometimes,she may make some mistakes.So dandelion wants you to write a program to help her calculate the value of some expressions.   You may consider every expression is made of :left parenthesis '(' ,right parenthesis ')' , plus sign '+' , subtraction sign '-' , multiplication sign '*' ,positive sign '+' ,and negative sign '-'.There is no precursor 0.The length of expression will not exceed 100.The value produced during the calculating will not exceed 10^9.Every expression is legal.Ie. ((10+(-100))) ，-（-1），-3*(+5+7*2)-(0) ，-0 ，(-3)*(-5+7) are legal,while 1+-7 ，--3+8 ，-3+() are illegal.

 

Input

There are several cases,every line contains a expression.

 

Output

For every case ,print the answer of the expression.

 

Sample Input

-3*(+5-7*2)-(0)

 

Sample Output

27

 

Author

dandelion

 

Source

曾是惊鸿照影来

 

 

栈的运用 考虑的比较多 使用　　STL <stack>

详细的看注释

#include<bits/stdc++.h>
using namespace std;
char a[105];
stack<char> m;//运算符栈
stack<int> n;//操作室栈
map<char,int> mp;
//40  (
//41  )
//42  *
//43  +
//45  -
int main()
{
    mp['-']=1;
    mp['+']=1;
    mp['*']=2;
    mp['(']=-1;
    mp[')']=-1;
    memset(a,0,sizeof(a));
    while(gets(a))
    {

        while(!m.empty())
            m.pop();
        while(!n.empty())
            n.pop();
        n.push(0);//考虑初始有符号 放在栈底
        int len=strlen(a);
        int exm=0;
        int xx,yy;
        char what;
        for(int i=0; i<len; i++)
        {
            if(a[i]>=48&&a[i]<=57)
                exm=exm*10+a[i]-'0';
            else
            {
                if(a[i]=='(')// 前括号处理 添0 处理紧邻的符号
                {
                    m.push(a[i]);
                    n.push(0);
                    continue;
                }
                if(m.empty())//若运算符栈为空
                {
                    m.push(a[i]);
                    continue;
                }
                else
                {
                    if(mp[a[i]]>mp[m.top()]) //优先级大于栈顶运算符
                        m.push(a[i]);
                    else
                    {//直到优先级大于栈顶 或 栈空 或栈顶为后括号（这个没有验证）
                   while(!m.empty()&&mp[a[i]]<=mp[m.top()]&&m.top()!='(')// 这里理解
                        {
                            xx=n.top();
                            n.pop();
                            yy=n.top();
                            n.pop();
                            what=m.top();
                            m.pop();
                            if(mp[what]==1)
                            {
                                if(what=='+')
                                    n.push(yy+xx);
                                if(what=='-')
                                    n.push(yy-xx);
                            }
                            if(mp[what]==2)
                                n.push(yy*xx);
                        }
                        if(!m.empty()&&m.top()=='('&&a[i]==')') //当前后括号相遇pop
                            m.pop();
                        else          //否则插入
                            m.push(a[i]);

                    }
                    continue;
                }
            }
            if(i>=1)//处理前括号后若无符号
            {
                if(a[i-1]=='(')
                    n.pop();
            }
            if(mp[a[i+1]]!=0)//判断exm 积累结束
            {
                n.push(exm);
                exm=0;
            }
            if(i==len-1&&a[i]!=')')//考虑最后一个操作数
                n.push(exm);

        }
        while(!m.empty())//直到 运算符栈空
        {
            xx=n.top();
            n.pop();
            yy=n.top();
            n.pop();
            what=m.top();
            m.pop();
            if(mp[what]==1)
            {
                if(what=='+')
                    n.push(yy+xx);
                if(what=='-')
                    n.push(yy-xx);
            }
            if(mp[what]==2)
                n.push(yy*xx);
        }
        printf("%d
",n.top());//输出栈顶值
        memset(a,0,sizeof(a));
    }
    return 0;
}