sincerit-杭电oj 1237 简单计算器

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1237 简单计算器

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26856 Accepted Submission(s): 9741

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 + 2
4 + 2 * 5 - 7 / 11
0

Sample Output
3.00
13.36

#include <iostream>
#include <cstring>
#include <iomanip>
#include <stdio.h> 
#include <stack>
using namespace std;
int main() {double x;char sign;while (scanf("%lf", &x)) {stack<double> num;num.push(x);char ch;ch = getchar(); if (x == 0 && ch == '\n') break; // 特殊情况 0 + 1所以要加上ch=='\n'的判断 scanf("%c", &sign);getchar();while (scanf("%lf", &x)) {if (sign == '+') {num.push(x);} else if (sign == '-') {num.push(-x);} else if (sign == '*') {double temp = num.top();num.pop();num.push(temp * x);} else if (sign == '/') {double t = num.top();num.pop();num.push(t*1.0 / x);}ch = getchar(); // 结束时数字后面有一个换行 if (ch == '\n') break;scanf("%c", &sign);getchar();}double sum = 0.0;while (num.size()) {sum += num.top();num.pop();}//cout << fixed << setprecision(2) << sum << "\n";printf("%.2f\n", sum);}return 0;
}

原先两个栈模拟的不知道哪里错了

#include <iostream>
#include <stdio.h>
#include <stack> 
#include <cstring>
#include <iomanip>
double cal(double x, double y, char ch) {if (ch == '+') return y + x;  else if (ch == '-') return y - x;else if (ch == '*') return y * x;   else return y / x;// 入栈再出来的数据刚好相反是y/x 不是x/y 
}
int compare(char a, char b) {if (a == '#') return 1;else if (b == '*' || b == '/') {if (a == '+' || a == '-') return 1;}return 0;
}
using namespace std;
int main() {string str;while (getline(cin, str)) {if (str == "0") break;stack<char> sign;sign.push('#');stack<double> num;int k = 0;for (int i = 0; i < str.size(); i++) {if (str[i] >= '0' && str[i] <= '9') {k = k * 10 + str[i] - '0';} else if (str[i] == ' ') {//cout << k << "\n";num.push(k*1.0);k = 0;} else {if (compare(sign.top(), str[i]) == 1) {sign.push(str[i]);} else {double x = num.top();num.pop();double y = num.top();num.pop();num.push(cal(x, y, sign.top()));sign.pop();sign.push(str[i]);  // 计算完了之后再把当前符号入栈 }i++;  // 符号后面的空格要跳过去 }}num.push(k*1.0);  // 最后一个数据还在k中 while (sign.size() > 1) {double a = num.top();num.pop();double b = num.top();num.pop();num.push(cal(a, b, sign.top()));sign.pop();}//cout << fixed << setprecision(2) << num.top() << "\n";printf("%.2f\n", num.top());}return 0;
}