【语法基础练习】1.变量、输入输出、表达式与顺序语句

文章简介：下面的题目是AcWing网站语法基础练习篇的第一小节，内容基础，难度：易。

主要都是简单的输入输出练习。仅以此记录自己的学习。

1.A+B

代码如下：

#includeusing namespace std;int main(){    int A,B;    cin>>A>>B;    cout    int A,B,C,D;    cinA;    cinB;    cin>>C;    cin>>D;    cout    double r,A;    cinr;    A=3.14159*r*r;    printf("A=%.4f",A);    return 0;}    double A,B,S;    cinA;    cinB;    S=(A*3.5+B*7.5)/11;    printf("MEDIA = %.5f",S);    return 0;}

5.工资

代码如下：

#include using namespace std;int main(){    int id;//编号    int hour; //时长    double salary;//时薪    cin>>id;    cin>>hour;    cin>>salary;    cout    int X;//行驶总路程    double Y;//油量    cinX;    cinY;    printf("%.3f km/l",X/Y);    return 0;}    double x1,x2,y1,y2,d;    cinx1>>y1;    cin>>x2>>y2;    d=sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));    printf("%.4f",d);    return 0;}

8.钞票

代码如下：

#include using namespace std;int main(){    int N;    cin>>N;    int R100,R50,R20,R10,R5,R2,R1;    int t;//临时变量    R100=N/100;    t=N-R100*100;    R50=t/50;    t=t-R50*50;    R20=t/20;    t=t-R20*20;    R10=t/10;    t=t-R10*10;    R5=t/5;    t=t-R5*5;    R2=t/2;    t=t-R2*2;    R1=t;    cout    int N;    int hour,min,sec;    int t;    cinN;    hour=N/3600;    t=N-hour*3600;    min=t/60;    t=t-min*60;    sec=t;    cout    int A,B;    cinA;    cinB;    cout    int num1,num2,amount1,amount2;    double price1,price2;    cinnum1amount1price1;    cinnum2amount2price2;    printf("VALOR A PAGAR: R$ %.2f",amount1*price1+amount2*price2);    return 0;}    double  R,V;    cinR;    V=(4/3.0)*3.14159*R*R*R;    printf("VOLUME = %.3f",V);    return 0;}    double A,B,C;    double S1,S2,S3,S4,S5;    cinABC;    S1=0.5*A*C;    S2=3.14159*C*C;    S3=0.5*(A+B)*C;    S4=B*B;    S5=A*B;    printf("TRIANGULO: %.3f\n",S1);    printf("CIRCULO: %.3f\n",S2);    printf("TRAPEZIO: %.3f\n",S3);    printf("QUADRADO: %.3f\n",S4);    printf("RETANGULO: %.3f\n",S5);    return 0;}    double A,B,C,S;    cinA;    cinB;    cinC;    S=(A*2+B*3+C*5)/10;    printf("MEDIA = %.1f",S);    return 0;}    string name;    double salary,sales,P;    cinnamesalarysales;    P=salary+sales*0.15;    printf("TOTAL = R$ %.2f",P);    return 0;}    int A,B,C;    cinABC;    int max;    max=0.5*(A+B+abs(A-B));    max=0.5*(max+C+abs(max-C));    cout    int L,X;    cinL;    X=L*2;    cout    int v;    double t;    cintv;    double total;    total=t*v/12.0;    printf("%.3f",total);    return 0;}    int N;    cinN;    int t;    int ano,mes,dia;    ano=N/365;    t=N-ano*365;    mes=t/30;    t=t-mes*30;    dia=t;    cout