Question

我在这个任务上的智慧结束了。我需要用户能够做一些事情。

输入他们想要藏羊的形状（按1或2）。
输入他们想要多少个畜栏; 3是最大值
输出每个畜栏的绵羊面积和数量。

圆形公式有效，但输入了多个符号后矩形会暂停。矩形部分的输出根本不显示。我尝试过使用while，switch，case等的组合，我无法让它工作。

如何显示公式部分，以及如何使用“while”循环矩形公式3次？有帮助吗？

// ConsoleApplication16.cpp : Defines the entry point for the console    application.
//

#include "stdafx.h"
#include <iostream>
#include <cstdio>
#include <cmath>
#include <iomanip>

using namespace std; 

int main()
{
const double pie = 3.141592;
double r;
double area = 0;
double length;
double width;
int shape;
double sheep = 2.54;
int corral;

cout << " Which shape do you want? " << endl;
cout << " 1. Circle.   2. Rectangle. ";
cin >> shape;
cout << " How many corrals do you want? " << endl;
cin >> corral;

while (corral <= 3)

    if (shape == 1)
    {
        cout << "Input radius. " << endl;
        cin >> r;
        cout << fixed << setprecision(2);
        area = pie * pow(r, r);
        cout << endl;

        cout << " Here's the area. << " endl;
        cout << area << endl;
        cout << fixed << setprecision(1) << "Number of sheep " << area / sheep << endl;
    }


else (shape = 2);
        {   
        cout << "Enter lenght and width. " << endl;
        cin >> length;
        cin >> width;
        area = length * width;

        cout << "Here's the area. " << endl;
        cout << area << endl;
        cout << "Number of sheep" << endl;
        cout << fixed << setprecision(2);
        cout << area / sheep;
        }




return 0;

}

Answer 1

我建议你先写一个计算区域的方法。然后使用循环来调用方法1,2或3的次数。

<强>逻辑

1）询问用户是否输入半径

2）调用计算区域方法或仅计算区域

顺便说一句，你不能使用＆＃34; if else＆＃34;像这样（shape = 2）;

if (condition) 
{}
else
{}

if you need more condition checks

if (condition){}
else if (condition) {}
else {}

Answer 2

else (shape = 2);

我觉得想写

else if (shape == 2)

循环矩形区域的公式

2 个答案: