我是java的新手,我正在构建一个计算器,它需要一个等式并对其进行评估. 我正在使用Scanner方法获取输入,但这意味着我的输入是扫描仪类型.我应该怎么做这个输入,以便我可以评估它？一旦我可以评估它,我怎样才能优先考虑括号？ 例如,对于等式(5 *(4 3))* 2,我想首先评估(4 3)的程序,然后
我是JavaScript的新手,所以我甚至不知道这是否是尝试这个的正确语言,但我想我会尝试.我已经阅读了其他几篇文章,我找不到真正让我知道如何做到这一点的东西,所以我在这里问.我读过的所有例子都涉及用户输入的数字和/或选择.但是,我要采用静态的计算方法.我在HTML中输入信息到表中,J
software: CFD POST ANSYS menu bar, select Tools > Macro Calculator.
Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . Example 1: Input: "1 + 1"Output: 2
In the distant space, there is a technologically advanced planet. One day they provided the Earth with a code that could achieve the ultimate meaning of the universe. People were very happy, but found that this code can only run on computers with a word l
可以利用Magma calculator在线计算椭圆曲线点乘。使用help文档，可以计算256位GF(p)点乘如下：a:=-3;b:= 41058363725152142129326129780047268409114441015993725554835256314039467401291;E:= EllipticCurve([GF(115792089210356248762697446949407573530086143415290314195533631308
本文介绍类和简单使用，类是需要class这个关键字来声明的，一般如下面的语法： class className： def fun1(): pass def fun2(): pass 看下面demo.py # 练习类和使用'''
page { height:100%;} .calculator { width: 100%; height: 100vh; border:solid 1px; background: rgb(238, 5, 5); position: relative; box-shadow: 0px 0px 20px 0px rgb(211, 41, 41); display: flex; flex-direction: column; box-sizing: border-box;} .calc
On a broken calculator that has a number showing on its display, we can perform two operations: Double: Multiply the number on the display by 2, or; Decrement: Subtract 1 from the number on the display. Initially, the calculator is displaying the number
On a broken calculator that has a number showing on its display, we can perform two operations: Double: Multiply the number on the display by 2, or; Decrement: Subtract 1 from the number on the display. Initially, the calculator is displaying the number
1 class Solution 2 { 3 public: 4 int brokenCalc(int X, int Y) 5 { 6 int n = X, m = Y; 7 if (n>=m) 8 return n-m; 9 else10 {11 int count = 0;12
