HDU 1237 简单计算器(栈+stringstream)

提供几份代码,这题的输入可以用stringsteam处理,先处理乘除后处理加减,正常思路,但是后面统计加减法的时候,对栈的运用错了,我用的时候相当于给它多加了几个括号就错了。
正确的简单解法就是,加法,就让正数入栈,减法就让该数的相反数入栈,之后的操作也不会影响该数的正负,这样处理最简单。
单栈

#include <iostream>
#include <sstream>
#include <stack>
using namespace std;

stack<double> stn;

int main()
{
    string line;
    while (getline(cin,line)) {
        while (!stn.empty()) {
            stn.pop();
        }
        if (line=="0") {
            break;
        }
        
        stringstream ss(line);
        long long x;
        string oper;
        
        ss>>x;
        stn.push((double)x);
        while (ss>>oper>>x) {
            //cout<<oper<<endl<<x<<endl;
            if (oper=="*") {
                double num=stn.top();
                stn.pop();
                stn.push(num*(double)x);
            }
            else if (oper=="/") {
                double num=stn.top();
                stn.pop();
                stn.push(num/(double)x);
            }
            else if (oper=="-") {
                stn.push((double)-x);
            }
            else {
                stn.push((double)x);
            }

        }
        double sum=0;
        while (!stn.empty()) {
            sum+=stn.top();
            stn.pop();
        }
        printf("%.2lf
",sum);
    }
    return 0;
}

双栈
AC

#include <iostream>
#include <sstream>
#include <stack>
using namespace std;

stack<string> sts;
stack<double> stn;

int main()
{
    string line;
    while (getline(cin,line)) {
        while (!stn.empty()) {
            stn.pop();
        }
        while (!sts.empty()) {
            sts.pop();
        }
        if (line=="0") {
            break;
        }
        
        stringstream ss(line);
        long long x;
        string oper;
        
        ss>>x;
        double sum=0;
        stn.push(x);
        while (ss>>oper>>x) {
            //cout<<oper<<endl<<x<<endl;
            if (oper=="*") {
                double num=stn.top();
                stn.pop();
                stn.push(num*(double)x);
            }
            else if (oper=="/") {
                double num=stn.top();
                stn.pop();
                stn.push(num/(double)x);
            }
            else {
                stn.push((double)x);
                sts.push(oper);
            }

        }
        while (!sts.empty()) {
            string op=sts.top();
            sts.pop();
            if (op=="+") {
                sum+=stn.top();
            }
            else {
                sum-=stn.top();
            }
            stn.pop();
        }
        sum+=stn.top();
        printf("%.2lf
",sum);
    }
    return 0;
}

WA D了好久…

#include <iostream>
#include <sstream>
#include <stack>
using namespace std;

stack<string> sts;
stack<double> stn;

int main()
{
    string line;
    while (getline(cin,line)) {
        while (!stn.empty()) {
            stn.pop();
        }
        while (!sts.empty()) {
            sts.pop();
        }
        
        stringstream ss(line);
        long long x;
        string oper;
        
        ss>>x;
        if (x==0&&line.length()==1) {
            break;
        }
        stn.push((double)x);
        while (ss>>oper>>x) {
            //cout<<oper<<endl<<x<<endl;
            if (oper=="*") {
                double num=stn.top();
                stn.pop();
                stn.push(num*(double)x);
            }
            else if (oper=="/") {
                double num=stn.top();
                stn.pop();
                stn.push(num/(double)x);
            }
            else {
                stn.push((double)x);
                sts.push(oper);
            }

        }
        while (!sts.empty()) {
            string op=sts.top();
            sts.pop();
            double num2=stn.top();
            stn.pop();
            double num1=stn.top();
            stn.pop();
            if (op=="+") {
                stn.push(num1+num2);
            }
            else {
                stn.push(num1-num2);
            }
        }
        printf("%.2lf
",stn.top());
    }
    return 0;
}
原文地址:https://www.cnblogs.com/xyqxyq/p/12328867.html