http://acm.hdu.edu.cn/showproblem.php?pid=1237

表达式计算,方法是中缀转后缀,再计算。中间处理用栈操作

讲解看http://blog.csdn.net/antineutrino/article/details/6763722

这题是简易版本的,不用处理括号

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>

using namespace std;

int cmp(char a, char b) {
    if((a == '*' || a == '/') && (b == '+' || b == '-')) return 1;
    return 0;
}

double cal(double a, double b, char c) {
    if(c == '+') return a + b;
    if(c == '-') return a - b;
    if(c == '*') return a * b;
    if(c == '/') return a / b;
}

int main() {
    double a;
    while(~scanf("%lf", &a)) {
        char c;
        c = getchar();
        stack <char> s1;
        stack <double> s2;
        if(!a && c=='\n') break;
        s2.push(a);
        c = getchar(); 
        while(~scanf("%lf", &a)) {
            if(s1.empty()) {
                s1.push(c);
            }
            else {
                if(cmp(c, s1.top())) s1.push(c);
                else {
                    while(1) {
                        double t1 = s2.top();
                        s2.pop();
                        double t2 = s2.top();
                        s2.pop();
                        char t3 = s1.top();
                        s1.pop();
                        double t4 = cal(t2, t1, t3);
                        s2.push(t4);
                        if(s1.empty() || cmp(c, s1.top())) {
                            s1.push(c);
                            break;
                        }
                    }
                }
            }
            s2.push(a);
            if(getchar() == '\n') break;
            c = getchar();
        }
        while(!s1.empty()) {
            double t1 = s2.top();
            s2.pop();
            double t2 = s2.top();
            s2.pop();
            char t3 = s1.top();
            s1.pop();
            double t4 = cal(t2, t1, t3);
            s2.push(t4);
        }
        printf("%.2lf\n", s2.top());
    }
    return 0;
}
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