C++入门小练习

好记性不如烂笔头,分享下我做的练习:

1、设货物运费每吨单价 p（元）与运输,距离 s（公里）之间有如下关系：

输入要托运的货物重量为 w
吨，托运距离 s 公里，计算总运费 t ：

t = p * w * s

按照一般思路范围从小到大进行筛选,要写五个判断,而观察一下里程与单价的变化规律可以发现,从100公里开始每加100公里减少2.5元单价,据此可以只写三个判断,以规律囊括中间100到400的单价变化,实现代码:

float w, s, t, p;
    cout << "托运重量(/t): ";
    cin >> w;
    cout << "托运里程(/km): ";
    cin >> s;
    if (s / 100 <1) 
        p = 30;     
    else 
        if (s / 100>= 4) 
            p = 20;     
        else 
            p = 30 -(int (s / 100))*(2.5); 
            
    t = p*w*s;
    cout << "托运费用: " << t << endl;

2、输入三个整数，按从小到大顺序输出。（if else/三目）
这题要求两种写法，我们先把if else写法写出来：

int x, y, z, a=0,b=0, c=0;

    cout << "输入三个整数: ";

    cin >> x >>y >> z;

    if (x >= y) {

        if (x >=z) {
            a = x;
            if (y>= z) {
                b =y, c = z;
            }
            else {
                b =z, c = y;
            }
        }
        else {
            a = z, b= x, c = y;
        }   
    }
    else if(y>=z) {
        a = y;
        if (x >=z) {
            b = x, c= z;
        }
        else {
            b = z, c= x;
        }
    }
    else {
        a = z, b =y, c = x;
    }
cout << "a:" << a << " b: " << b << " c:" << c << endl;
cout << "x:" << x << " y: " << y << " z:" << z << endl;

但是三目的稍微麻烦一些，运算符优先级判断要熟练

int x, y, z, a=0,b=0, c=0;
    cout << "输入三个整数: ";
    cin >> x >>y >> z;

x >= y ? (x >=z ? (a = x, (y >= z ? (b = y, c = z) : (b = z, c = y))) : (a = z, b = x, c =y)) : (y >= z ? (a = y, (x >= z ? (b = x, c = z) : (b = z, c = x))) : (a= z, b = y, c = x)); 

    cout << "a:" << a << " b: " << b << " c:" << c << endl;

cout << "x:" << x << " y: " << y << " z:" << z << endl;

由于三目的优先级高于赋值符，经常容易漏括号导致赋值错误，所以还是要养成随手加括号的习惯啊