ACwing语法基础课第一节课例题与习题及个人总结

第一次课例题

若涉及到浮点数的计算，float一般是6到7位有效数字，double一般是15到16位有效数字，但是为了方便起见，建议直接设为double，因为若涉及浮点数的乘除运算，使用float类型很容易出现精度问题，从而造成答案偏差；

同时，一定要注意占位符与变量的对应关系，int，float，double在计算机中存储的方式是不同的，若学过计组的话应该很好理解，若是输出时，int型变量用%f去占位自然会出现错误，一定要用对应的%d占位。

//基础的运算
//AcWing 1. A + B  https://www.acwing.com/activity/content/problem/content/1822/
#include <iostream>
using namespace std;

int main(){
    int A,B;
    cin >> A >> B;
    cout << A+B;
    return 0;
}

//AcWing 608. 差  https://www.acwing.com/activity/content/problem/content/1823/
#include <iostream>
using namespace std;

int main(){
    int A,B,C,D;
    cin >> A >> B >> C >> D;
    cout << "DIFERENCA = " << (A*B-C*D);
    return 0;
}

//AcWing 604. 圆的面积  https://www.acwing.com/activity/content/problem/content/1824/
#include <iostream>
#include <math.h>
#define pi 3.14159
using namespace std;

int main(){
    double R, A;
    cin >> R;
    A = pi * pow(R, 2);
    printf("A=%.4f", A);  // 注意这里要用printf输出好一点，因为题目要求输出四位小数，而且printf比cout要快一点
    return 0;
}

//AcWing 606. 平均数1  https://www.acwing.com/activity/content/problem/content/1825/
#include <iostream>
#include <math.h>
using namespace std;

int main(){
    float A,B;
    cin >> A >>B;
    printf("MEDIA = %.5f", (3.5*A+7.5*B)/11); // 注意这里要用printf输出好一点，因为题目要求输出5位小数
    return 0;
}

//AcWing 609. 工资   https://www.acwing.com/activity/content/problem/content/1826/
#include <iostream>
using namespace std;

int main(){
    int tag, time;
    float wage;
    cin >> tag >> time >> wage;
    cout << "NUMBER = " << tag << endl;
    printf("SALARY = U$ %.2f", wage * time);
    return 0;

}

// AcWing 615. 油耗  https://www.acwing.com/activity/content/problem/content/1827/
#include <iostream>
using namespace std;

int main(){
    int km;
    float l;
    cin >> km >> l;
    printf("%.3f km/l", km/l);
    return 0;
}

//616. 两点间的距离  https://www.acwing.com/activity/content/problem/content/1828/
#include <iostream>
#include <math.h>
using namespace std;

int main(){
    double x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    printf("%.4f", sqrt(pow((x2 - x1), 2) + pow((y2 - y1), 2 )));
    return 0;
}

//AcWing 653. 钞票  https://www.acwing.com/activity/content/problem/content/1829/
#include <iostream>
using namespace std;

int main(){
    int dollar;
    cin >> dollar;
    cout << dollar << endl;
    cout << dollar/100 << " nota(s) de R$ 100,00" << endl , dollar %= 100;
    cout << dollar/50 << " nota(s) de R$ 50,00" << endl , dollar %= 50;
    cout << dollar/20 << " nota(s) de R$ 20,00" << endl , dollar %= 20;
    cout << dollar/10 << " nota(s) de R$ 10,00" << endl , dollar %= 10;
    cout << dollar/5 << " nota(s) de R$ 5,00" << endl , dollar %= 5;
    cout << dollar/2 << " nota(s) de R$ 2,00" << endl , dollar %= 2;
    cout << dollar/1 << " nota(s) de R$ 1,00" << endl ;
    return 0;
}

//AcWing 654. 时间转换  https://www.acwing.com/activity/content/problem/content/1830/
#include <iostream>
using namespace std;

int main(){
    int ++time;
    cin >> time;
    cout << time/3600 << ":" << (time%3600)/60 << ":" << (time%3600)%60;
    return 0;
}

第一次课练习

一些总结：float的有效表示范围是6到7位，double是15到16位，一般来说如果题目里面有小数，可以优先考虑用double，有时候用float会造成精度丢失。同时运算的时候，要注意变量类型的自动转换。

比如double l = time*s/12.0;：这里面若变量time与s是整数，而要求的结果是double类型，若除以/12则运算就是先用int计算完除法，再转换为double类型，输出的其实还是个整数，若数据大或者实际结果有小数则必然会出现精度丢失的问题，而若除以/12.0则两个int型变量会转换为浮点类型，从而保证精度。

同时，控制输出位有效位，或者说控制输出位的小数部分有效位十分关键，也许要求输出的是一个整数，但是若用int计算必然会造成精度丢失，那么可以考虑用float或者double计算，但在输出时使得其仅输出整数部分，即%.0f

// AcWing 605. 简单乘积  https://www.acwing.com/activity/content/problem/content/1831/
#include <iostream>

using namespace std;

int PROD;
int main(){
    int i, j;
    cin >> i >> j;
    cout << "PROD = " << i*j;
    return 0;
}
// AcWing 611. 简单计算  https://www.acwing.com/activity/content/problem/content/1832/
#include <iostream>

using namespace std;

int a ,b ,n ,m;
double a_value, b_value;  // 这里不能用float，而要用double，
int main()
{
    cin >> a >> n >> a_value ;
    cin >> b >> m >> b_value;
    printf("VALOR A PAGAR: R$ %.2f", n*a_value + m*b_value);
}
// AcWing 612. 球的体积  https://www.acwing.com/activity/content/problem/content/1833/
#include <iostream>

using namespace std;

#define pi 3.14159 

int main()
{
    double R;
    cin >> R;
    double v ;
    v = (4/3.0)*pi*(R*R*R);
    printf("VOLUME = %.3lf", v);
    return 0;
}
// AcWing 613. 面积  https://www.acwing.com/activity/content/problem/content/1834/
#include <iostream>
#define pi 3.14159
using namespace std;
double a, b, c;
int main(){
    cin >> a >> b >> c;
    //三角形
    double s1 = a*c/2;
    //圆
    double s2 = c*c*pi;
    //梯形
    double s3 = (a + b)*c/2;
    //正方形
    double s4 = b*b;
    //长方形
    double s5 = a*b;

    printf("TRIANGULO: %.3lf\n", s1);
    printf("CIRCULO: %.3lf\n", s2);
    printf("TRAPEZIO: %.3lf\n", s3);
    printf("QUADRADO: %.3lf\n", s4);
    printf("RETANGULO: %.3lf\n", s5);

    return 0;

}
//AcWing 607. 平均数2  https://www.acwing.com/activity/content/problem/content/1835/
#include <iostream>
using namespace std;
float A, B, C;

int main()
{
    cin >> A >> B >> C;
    float media = (A*2+B*3+C*5)/(2+3+5);
    printf("MEDIA = %.1f", media);
    return 0;
}
//AcWing 610. 工资和奖金  https://www.acwing.com/activity/content/problem/content/1836/
#include <iostream>

using namespace std;

char s[20];
float wage, month;
int main()
{
    cin >> s;
    cin >> wage >> month;
    float salary = wage + 0.15*month;
    printf("TOTAL = R$ %.2f", salary);
    return 0;
}
//AcWing 614. 最大值  https://www.acwing.com/activity/content/problem/content/1837/
#include <iostream>
#include <math.h>
using namespace std;
int a, b, c;
int main(){
    cin >> a >> b >> c;
    int max_ab = (a + b + abs(a - b)) / 2;  //求ab中最大值
    int max_abc = (max_ab + c + abs (max_ab - c)) / 2;  // 求abc最大值
    cout << max_abc << " eh o maior";
    return 0;
}
//AcWing 617. 距离  https://www.acwing.com/activity/content/problem/content/1838/
#include <iostream>

using namespace std;
int L;
int main(){
    cin >> L;

    double v = 30 / 60.0;

    double time = L / v;
    printf("%.lf minutos", time);
    return 0;
}
//AcWing 618. 燃料消耗  https://www.acwing.com/activity/content/problem/content/1839/
#include <iostream>

using namespace std;

double t, s;  // 这里t与s的取值范围在1到10^7，而在后续计算中t*s取极值时，会有14位有效数字，因此用double防止精度丢失

int main()
{
    cin >> t >> s;
    double l = t*s/12.0;  // 注意，这里一定要除以12.0而不是12，否则计算出来的是个整数，因为t与s没有转换为浮点数，而是直接以整数计算的
    printf("%.3lf", l);
    return 0;

}
//AcWing 656. 钞票和硬币  https://www.acwing.com/activity/content/problem/content/1839/
// 除了如下代码的这种方法，还可以将金额的单位从元转换为分，进而将浮点数转换为整数，就可以方便计算了，也不会出现丢失精度的情况
#include <iostream>

using namespace std;

int main()
{
    double N;
    cin >> N;
    // 分解为钞票

    int nota_100 = N / 100.00;
    N -= 100.00*nota_100;
    int nota_50 = N / 50.00;
    N -= 50.00*nota_50;
    int nota_20 = N / 20.00;
    N -= 20.00*nota_20;
    int nota_10 = N / 10.00;
    N -= 10.00*nota_10;
    int nota_5 = N / 5.00;
    N -= 5.00*nota_5;
    int nota_2 = N / 2.00;
    N -= 2.00*nota_2;
    // 分解为硬币

    int moeda_1 = N / 1.00;
    N -= moeda_1*1.00;
    int moeda_5 = N / 0.50;
    N -= moeda_5*0.5;
    int moeda_25 = N / 0.25;
    N -= moeda_25*0.25;
    int moeda_10 = N / 0.10;
    N -= moeda_10*0.10;
    int moeda_05 = N / 0.05;
    N -= moeda_05*0.05;
    // int moeda_01 = N / 0.01;  // 到达这一步，若N是0.01那么输出的moeda理论上应该为1，但是实际上为0，因为精度问题，0.01在double中存储的是0.99999999999999，故最后输出是0而不是1

    float moeda_01 = N / 0.01;  // 用float定义moeda_01变量，避免了int类型丢失0.01的情况
    cout << "NOTAS:" << endl;
    cout << nota_100 << " nota(s) de R$ 100.00" << endl;
    cout << nota_50 << " nota(s) de R$ 50.00" << endl;
    cout << nota_20 << " nota(s) de R$ 20.00" << endl;
    cout << nota_10 << " nota(s) de R$ 10.00" << endl;
    cout << nota_5 << " nota(s) de R$ 5.00" << endl;
    cout << nota_2 << " nota(s) de R$ 2.00" << endl;
    cout << "MOEDAS:" << endl;
    cout << moeda_1 << " moeda(s) de R$ 1.00" << endl;
    cout << moeda_05 << " moeda(s) de R$ 0.50" << endl;
    cout << moeda_25 << " moeda(s) de R$ 0.25" << endl;
    cout << moeda_10 << " moeda(s) de R$ 0.10" << endl;
    cout << moeda_05 << " moeda(s) de R$ 0.05" << endl;
    // cout << moeda_01 << " moeda(s) de R$ 0.01" << endl; // double可能会出现吧0.1丢失问题
    printf("%.0f moeda(s) de R$ 0.01", moeda_01);  // 这里输出因为moeda_01是float类型，然而要求的输出是整数个硬币，因此控制小数有效位为0,即可

    return 0;
}
//AcWing 655. 天数转换  https://www.acwing.com/activity/content/problem/content/1841/
#include <iostream>

using namespace std;

int day, ano, mes, dia;
int main(){
    cin >> day;
    ano = day / 365;
    mes = (day % 365) / 30;
    dia = ((day % 365) % 30) ;
    cout << ano << " ano(s)" << endl;
    cout << mes << " mes(es)" << endl;
    cout << dia << " dia(s)" << endl;
    return 0;
}