AcWing语法基础课学习笔记第二讲 printf语句与判断结构

桃扇骨 2024-03-17 09:50 14阅读 0赞

### AcWing语法基础课学习笔记 ###

#### 文章目录 ####

*   *  AcWing语法基础课学习笔记
     *   *   *  第二讲 printf语句与判断结构
             *   *  1. \`printf\`输出格式
                 *  2. if语句
                 *  3. 条件表达式

![在这里插入图片描述][629bd37a553d4d7ca39161a9d5ae18a4.png_pic_center]

##### 第二讲 printf语句与判断结构 #####

学习语言最好的方式就是实践，每当掌握一个新功能时，就要立即将这个功能应用到实践中。——闫学灿

###### 1. `printf`输出格式 ######

注意：使用`printf`时最好添加头文件 `#include <cstdio>`。

#include <iostream>
    #include <cstdio>
    
    using namespace std;
    
    int main()
    {
        printf("Hello World!");
    
        return 0;
    }

1.  `int`、`float`、`double`、`char`等类型的输出格式：
    
    (1) `int`：`%d`  
    (2) `float`: `%f`, 默认保留6位小数  
    (3) `double`: `%lf`， 默认保留6位小数  
    (4) `char`: `%c`, 回车也是一个字符，用`'\n'`表示
2.  所有输出的变量均可包含在一个字符串中【格式化输出】
3.  扩展功能
    
    (1) `float`, `double`等输出保留若干位小数时用：`%.4f`, `%.3lf`
    
    (2) 最小数字宽度
    
     *  `%8.3f`, 表示这个浮点数的最小宽度为8，保留3位小数，当宽度不足时在前面补空格。
     *  `%-8.3f`，表示最小宽度为8，保留3位小数，当宽度不足时在后面补上空格
     *  `%08.3f`, 表示最小宽度为8，保留3位小数，当宽度不足时在前面补上0

###### 2. if语句 ######

1.  基本if-else语句
    
    当条件成立时，执行某些语句；否则执行另一些语句。
    
        if (a > 5)
        {
            printf("%d is big!\n", a);
            printf("%d + 1 = %d\n", a, a + 1);
        }
        else
        {
            printf("%d is small!\n", a);
            printf("%d - 1 = %d\n", a, a - 1);
        }
    
    else 语句可以省略：
    
        if (a > 5)
        {
            printf("%d is big!\n", a);
            printf("%d + 1 = %d\n", a, a + 1);
        }
    
    当只有一条语句时，大括号可以省略：
    
        if (a > 5)
            printf("%d is big!\n", a);
        else
            printf("%d is small!\n", a);
    
    `if-else`语句内部也可以是`if-else`语句。
    
        #include <iostream>
        #include <cstdio>
        
        using namespace std;
        
        int main()
        {
            int a, b, c;
            cin >> a >> b >> c;
        
            if (a > b)
            {
                if (a > c) cout << a << endl;
                else cout << c << endl;
            }
            else
            {
                if (b > c) cout << b << endl;
                else cout << c << endl;
            }
        
            return 0;
        }
2.  常用比较运算符
    
    (1) 大于 `>`  
    (2) 小于 `<`  
    (3) 大于等于 `>=`  
    (4) 小于等于 `<=`  
    (5) 等于 `==`  
    (6) 不等于 `!=`
3.  `if-else`连写：
    
    【栗】简单计算器输入两个数，以及一个运算符`+`, `-`, `*`, `/`，输出这两个数运算后的结果。  
    当运算符是`/`，且除数是`0`时，输出`Divided by zero!`; 当输入的字符不是`+`, `-`, `*`, `/`时，输出`Invalid operator!`。
    
        #include <iostream>
        #include <cstdio>
        
        using namespace std;
        
        int main()
        {
            int a, b;
            char c;
            cin >> a >> b >> c;
        
            if (c == '+') cout << a + b << endl;
            else if (c == '-') cout << a - b << endl;
            else if (c == '*') cout << a * b << endl;
            else if (c == '/')
            {
                if (b != 0)
                {
                    cout << a / b << endl;
                }
                else
                {
                    cout << "Divided by zero!" << endl;
                }
            }
            else
            {
                cout << "Invalid operator!" << endl;
            }
        
            return 0;
        }

###### 3. 条件表达式 ######

(1) 与 `&&` 或者 `and`

(2) 或 `||` 或者 `or`

(3) 非 `!` 或者 `not`

优先级：`not` > `and` > `or`

闰年判断：【(四年一闰、百年不闰)、四百年又闰】

if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0)
        cout << "yes" << endl;
    else
        cout << "no" << endl;

> 判断 `!= 0` 其实不写

变量作用域问题：简单地说

*  外面的可以用里面的
 *  里面的不能用外面的【越往内，作用域越小】

`AcWing 665. 倍数`

#include <iostream>
    
    using namespace std;
    
    int main()
    {
        
        int a,b;
        
        cin >> a >> b;
        
        if(a % b == 0 || b % a == 0){
            cout << "Sao Multiplos" << endl;
        }else{
            cout << "Nao sao Multiplos" << endl;
        }
        
        return 0;
    }

`AcWing 660. 零食`

#include <cstdio>
    
    int main(){
        
        int x, y;
        
        scanf("%d%d", &x, &y);
        
        double price;
        
        if(x == 1) price = 4;
        else if(x == 2) price = 4.5;
        else if(x == 3) price = 5;
        else if(x == 4) price = 2;
        else if(x == 5) price = 1.5;
        
        printf("Total: R$ %.2lf", price * y);
        
        return 0;
    }

`AcWing 659. 区间`

#include <iostream>
    
    using namespace std;
    
    int main(){
        
        double x;
        
        cin >> x;
        
        if(x >= 0 && x <= 25) cout << "Intervalo [0,25]" << endl;
        else if(x > 25 && x <= 50) cout << "Intervalo (25,50]" << endl;
        else if(x > 50 && x <= 75) cout << "Intervalo (50,75]" << endl;
        else if(x > 75 && x <= 100) cout << "Intervalo (75,100]" << endl;
        else cout << "Fora de intervalo" << endl;
        
        return 0;
    }

`AcWing 664. 三角形`

#include <cstdio>
    #include <iostream>
    
    using namespace std;
    
    int main(){
        
        
        double a, b, c;
        
        cin >> a >> b >> c;
        
        if(a + b > c && a + c > b && b + c > a){
            printf("Perimetro = %.1lf", a + b + c);
        }else {
            printf("Area = %.1lf", (a + b) * c / 2);
        }
        
        return 0;
    }

`AcWing 667. 游戏时间`

#include <iostream>
    #include <cstdio>
    
    using namespace std;
    
    int main(){
        
        int a,b;
        
        cin >> a >> b;
        
        if(a == b) printf("O JOGO DUROU %d HORA(S)", 24);
        else if(a > b) printf("O JOGO DUROU %d HORA(S)", b - a + 24);
        else printf("O JOGO DUROU %d HORA(S)", b - a);
        
        return 0;
    }

`AcWing 669. 加薪`

#include <iostream>
    #include <cstdio>
    
    using namespace std;
    
    int main(){
        
        double salary;
        
        cin >> salary;
        
        double x;
        
        if(salary <= 400) x = 0.15;
        else if(salary <= 800) x = 0.12;
        else if(salary <= 1200) x = 0.10;
        else if(salary <= 2000) x = 0.07;
        else x = 0.04;
        
        printf("Novo salario: %.2lf\nReajuste ganho: %.2lf\nEm percentual: %.0lf %%\n", salary * (1 + x), salary * x , x * 100);
        
        return 0;
    }

> 这道题注意最后输出的涨薪幅度是保留0 位小数的浮点数。【不然又会有精度问题】
> 
> ![在这里插入图片描述][c394d44843b9464db95b10b9bdd62012.png_pic_center]
> 
> 然后就是百分号 `%` 的转义，可以写`%%`
> 
> ![在这里插入图片描述][53b07b4bf85545ab83d6b8d1d6f23171.png_pic_center]
> 
> 这里不转义还是过了…

`AcWing 670. 动物`

#include <iostream>
    
    using namespace std;
    
    int main(){
        
        string x, y, z;
        
        cin >> x >> y >> z;
        
        if(x == "vertebrado"){
            if(y == "ave"){
                if(z == "carnivoro"){
                    cout << "aguia" << endl;
                }else{
                    cout << "pomba" << endl;
                }
            }else{
                if(z == "onivoro"){
                    cout << "homem" << endl;
                }else{
                    cout << "vaca" << endl;
                }
            }
        }else{
            if(y == "inseto"){
                if(z == "hematofago"){
                    cout << "pulga" << endl;
                }else{
                    cout << "lagarta" << endl;
                }
            }else{
                if(z == "hematofago"){
                    cout << "sanguessuga" << endl;
                }else{
                    cout << "minhoca" << endl;
                }
            }
        }
        
        
        return 0;
    }

> 这个题就毫无技巧可言

`AcWing 657. 选择练习1`

#include <iostream>
    
    using namespace std;
    
    int main(){
        
        int a, b, c, d;
        
        cin >> a >> b >> c >> d;
        
        if(b > c && d > a && (c + d) > (a + b) && c > 0 && d > 0 && a % 2 == 0) cout << "Valores aceitos" << endl;
        else cout << "Valores nao aceitos" << endl;
        
        return 0;
    }

`AcWing 671. DDD`

#include <iostream>
    
    using namespace std;
    
    int main(){
        
        int x;
        
        cin >> x;
        
        if(x == 61) cout << "Brasilia" << endl;
        else if(x == 71) cout << "Salvador" << endl;
        else if(x == 11) cout << "Sao Paulo" << endl;
        else if(x == 21) cout << "Rio de Janeiro" << endl;
        else if(x == 32) cout << "Juiz de Fora" << endl;
        else if(x == 19) cout << "Campinas" << endl;
        else if(x == 27) cout << "Vitoria" << endl;
        else if(x == 31) cout << "Belo Horizonte" << endl;
        else cout << "DDD nao cadastrado" << endl;
        
        return 0;
    }

`AcWing 662. 点的坐标`

#include <iostream>
    
    using namespace std;
    
    int main(){
        
        double x, y;
        
        cin >> x >> y;
        
        if(x > 0 && y > 0) cout << "Q1" << endl;
        else if(x < 0 && y > 0) cout << "Q2" << endl;
        else if(x < 0 && y < 0) cout << "Q3" << endl;
        else if(x > 0 && y < 0) cout << "Q4" << endl;
        
        else{
            if(!x && !y) cout << "Origem" << endl;
            else if(!y) cout << "Eixo X" << endl;
            else cout << "Eixo Y" << endl;
        }
        
        return 0;
    }

> 等于0 可以直接写成`!x`，不等于0 可以直接写成`x`

`AcWing 666. 三角形类型`

#include <iostream>
    
    using namespace std;
    
    int main(){
        
        double a, b, c;
        cin >> a >> b >> c;
        
        // 降序排序
        if(b > a){
            double t = a;
            a = b;
            b = t;
        }
        if(c > a){
            double t = a;
            a = c;
            c = t;
        }
        if(c > b){
            double t = b;
            b = c;
            c = t;
        }
        
        
        if(a >= b + c) cout << "NAO FORMA TRIANGULO" << endl;
        else{
            if(a * a == b * b + c * c) cout << "TRIANGULO RETANGULO" << endl;
            if(a * a > b * b + c * c) cout << "TRIANGULO OBTUSANGULO" << endl;
            if(a * a < b * b + c * c) cout << "TRIANGULO ACUTANGULO" << endl;
            if(a == b && b == c) cout << "TRIANGULO EQUILATERO" << endl;
            else if(a == b || b == c || a == c) cout << "TRIANGULO ISOSCELES" << endl;
        }
        
        
        return 0;
    }

> 需要注意的就是一个样例可能会有多个输出，else 中的判断不能直接用 `else if` ，不然它就只走一个了。使用多个`if` 单独拎出来判断

`AcWing 668. 游戏时间2`

#include <cstdio>
    
    int main(){
        
        int a, b, c, d;
        
        scanf("%d%d%d%d", &a, &b, &c, &d);
        
        int start =  a * 60 + b;
        int end =  c * 60 + d;
        
        int spend_time = end - start;
        if(spend_time <= 0 ) spend_time += 1440;
        
        printf("O JOGO DUROU %d HORA(S) E %d MINUTO(S)", spend_time / 60, spend_time % 60);
        
        return 0;
    }

> 注意也有可能跨天

`AcWing 672. 税`

#include <iostream>
    #include <cstdio>
    
    using namespace std;
    
    int main(){
        
        double salary;
        
        cin >> salary;
        
        double sum = 0;
        
        if (salary > 2000){
            double y = 3000;
            if(salary < 3000) y = salary;
            sum += (y - 2000) * 0.08;
        }
        if (salary > 3000){
            double y = 4500;
            if(salary < 4500) y = salary;
            sum += (y - 3000) * 0.18;
        }
        if (salary > 4500) sum += (salary - 4500) * 0.28;
        
        if(sum == 0) cout << "Isento" << endl;
        else printf("R$ %.2lf", sum);
        
        return 0;
    }

> 这道题还是注意多级判断，不能直接上`else if`

`AcWing 663. 简单排序`

#include <iostream>
    
    using namespace std;
    
    int main(){
        
        int a, b, c;
        
        cin >> a >> b >> c;
        
        int x = a, y = b, z = c;
        
        if(b < a){
            int tmp = a;
            a = b;
            b = tmp;
        }
        if(c < a){
            int tmp = a;
            a = c;
            c = tmp;
        }
        if(c < b){
            int tmp = b;
            b = c;
            c = tmp;
        }
        
        cout << a << endl << b << endl << c << endl << endl;
        cout << x << endl << y << endl << z << endl;
        
        return 0;
    }

> 事先存一下原来的数，引入中间变量进行两数交换

`AcWing 658. 一元二次方程公式`

#include <iostream>
    #include <cstdio>
    #include <cmath>
    
    using namespace std;
    
    int main(){
        
        double a, b, c;
        
        cin >> a >> b >> c;
        
        double r1, r2;
        
        if(b * b - 4 * a * c < 0 || a == 0) cout << "Impossivel calcular" << endl;
        else{
            r1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
            r2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);
            
            printf("R1 = %.5lf\n", r1);
            printf("R2 = %.5lf", r2);
        }
        
        return 0;
    }

> 主要是那个结果的表达式，优先级，小括号的使用

`AcWing 661. 平均数3`

#include <iostream>
    #include <cstdio>
    
    using namespace std;
    
    int main(){
        
        double n1, n2, n3, n4;
        
        cin >> n1 >> n2 >> n3 >> n4;
        
        double x = (n1 * 2 + n2 * 3 + n3 * 4 + n4 * 1) / (2 + 3 + 4 + 1);
        
        printf("Media: %.1lf\n", x);
        
        if(x >= 7) cout << "Aluno aprovado." << endl;
        else if(x < 5) cout << "Aluno reprovado." << endl;
        
        else{
            
            cout << "Aluno em exame." << endl;
            double y, z;
            
            cin >> y;
            printf("Nota do exame: %.1lf\n", y);
            
            z = (x + y) / 2;
            if(z >= 5) cout << "Aluno aprovado." << endl;
            else cout << "Aluno reprovado." << endl;
            
            printf("Media final: %.1lf", z);
        }
        
        return 0;
    }

> 这道题也是主打一个翻译题目。

![在这里插入图片描述][a9bdfbef90e4453a9e3d87e96475bbbf.png_pic_center]

OK。

[629bd37a553d4d7ca39161a9d5ae18a4.png_pic_center]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/17/a928254567994a48b1918ea05d029da1.png
[c394d44843b9464db95b10b9bdd62012.png_pic_center]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/17/16307cc0390e426fb10195cccc01a81d.png
[53b07b4bf85545ab83d6b8d1d6f23171.png_pic_center]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/17/24c1f809f6144c2a9218bfd1a9509d07.png
[a9bdfbef90e4453a9e3d87e96475bbbf.png_pic_center]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/17/4fe89cab79bb46aaaab3e52b854c3da0.png