HDU 1237 简单计算器

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简单计算器

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15736 Accepted Submission(s): 5370

Problem Description

读入一个只包含 +, -, *, / 的非负整数计算表达式，计算该表达式的值。

Input

测试输入包含若干测试用例，每个测试用例占一行，每行不超过200个字符，整数和运算符之间用一个空格分隔。没有非法表达式。当一行中只有0时输入结束，相应的结果不要输出。

Output

对每个测试用例输出1行，即该表达式的值，精确到小数点后2位。

Sample Input

1 + 24 + 2 * 5 - 7 / 110

Sample Output

3.0013.36

Source

浙大计算机研究生复试上机考试-2006年

思路：要把乘除法在栈外处理，使得最后栈只用来处理加减，其实就是先算乘除。代码太乱，其实就是getRes返回l连续的乘除的结果，getNum就是返回这个数，最后进栈出栈顺序要搞一下，使得它们是从左到右计算的。

被这道题搞到眼瞎，只能承认自己代码功底无限低，还得再练练

#include <stdio.h>#include <stack>#include <queue>#include <string.h>using namespace std;#define maxn 250char s[maxn];double num[maxn];stack<double> n;stack<double> N;stack<char> f;stack<char> F;double getRes(int l, int r){    int i = l, In = 0;    long long int u = 10;    double sum = 0, res = 0.0;    char v = 'A';    while(s[i]==' ') i++;    while(i <= r&&s[i]>='0'&&s[i]<='9'){        res = res*u + (s[i] - '0');        i++;    }    //printf("%lf\n", res);    sum = res;    res = 0.0;    for(;i <= r;i++)    {        while(i <= r&&s[i]>='0'&&s[i]<='9'){            res = res*u + (s[i] - '0');            i++;            In = 1;        }        if(s[i]=='*'||s[i]=='/') v = s[i];        if(In)        {            if(v == '*') sum *= res;            else sum /= res;            u = 10, res = 0.0;            In = 0;        }    }    return sum;}double getNum(int tp){    long long int u = 1;    double res = 0.0;    for(int i = tp;i >= 0&&s[i] != ' ';i--)    {        if(s[i]>='0'&&s[i]<='9'){            res = res + (s[i] - '0') * u;            u *= 10;        }    }    return res;}int main(){    double res;    int len, l, r, j, i;    char last;    while(gets(s)&&strcmp(s, "0") != 0)    {        while(!n.empty()) n.pop();        while(!N.empty()) N.pop();        while(!f.empty()) f.pop();        while(!F.empty()) F.pop();        len = strlen(s);        last = 'A';        res = 0;        for(i = 0;i < len;i++)        {            if(s[i] == '+'||s[i] == '-')            {                if(last != '*'&&last != '/')                {                    n.push(getNum(i - 2));                    //printf("--%.2lf\n", getNum(i - 2));                }                f.push(s[i]);                last = s[i];            }            if(s[i] == '*'||s[i] == '/')            {                int l, r;                l = r = i;                last = s[i];                while(l >= 0&&s[l] != '-'&&s[l] != '+') l--;                while(r < len&&s[r] != '-'&&s[r] != '+') r++;                //printf("--%lf\n", getRes(l+1, r-1));                n.push(getRes(l+1, r-1));                last = s[i];                i = r - 1;            }        }        if(last == '+'||last == '-'){            n.push(getNum(len - 1));            //printf("--%.2lf\n", getNum(len - 1));        }        if(f.empty()&&n.empty())            res = getNum(len - 1);        else        {            while(!n.empty())            {                N.push(n.top());                n.pop();            }            while(!f.empty())            {                F.push(f.top());                f.pop();            }            while(!N.empty())            {                if(N.size()==1)                {                    res = N.top();                    N.pop();                    break;                }                double a, b;                char v;                if(!F.empty()){                    v = F.top();                    F.pop();                }                a = N.top();                N.pop();                b = N.top();                N.pop();                //printf("%lf %lf\n", a, b);                if(v == '+') N.push(a+b);                else if(v == '-') N.push(a-b);            }        }        printf("%.2lf\n", res);    }}

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