AcWing 第一讲 打卡&例题&习题题目

1. AcWing 1. A + B

#include<iostream>using namespace std;int main(){int a,b;cin >> a >> b;cout << a+b << endl;return 0;
}

2. AcWing 608. 差

#include<iostream>using namespace std;int main(){int A,B,C,D;cin>>A>>B>>C>>D;cout<<"DIFERENCA = "<<(A*B-C*D)<<endl;return 0;
}

3. AcWing 604.圆的面积

#include<cstdio>using namespace std;int main(){double pi,R;double ans;pi=3.14159;scanf("%lf",&R);ans=R*R*pi;printf("A=%.4f",ans);return 0;
}

4. AcWing 606.平均数1

#include<iostream> 
#include<cstdio>using namespace std;int main(){double A,B;double MEDIA;cin>>A>>B;MEDIA=(3.5*A+7.5*B)/(3.5+7.5);printf("MEDIA = %.5lf",MEDIA);return 0; 
}

5. AcWing 609. 工资

#include<iostream>
#include<cstdio>using namespace std;int main(){int Num,Total_Time;float Salary;cin>>Num>>Total_Time>>Salary;Salary=Total_Time*Salary;cout<<"NUMBER = "<< Num << endl;printf("SALARY = U$ %.2f",Salary);return 0;
}

6. AcWing 615. 油耗

#include<iostream>
#include<cstdio> using namespace std;int main(){int X;float Y;cin>>X>>Y;printf("%.3f km/l",X/Y);return 0;
}

7. AcWing 616. 两点间的距离

#include<cstdio>
#include<iostream>
#include<cmath>using namespace std;int main(){double x1,y1;double x2,y2;double ans;cin>>x1>>y1;cin>>x2>>y2;ans=sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));printf("%.4lf",ans);return 0;
}

8.AcWing 653. 钞票

#include<iostream> 
#include<cstdio>using namespace std;int main(){int N,n;int nota_100,nota_50,nota_20,nota_10,nota_5,nota_2,nota_1;cin>>N;n=N;nota_100=int(N/100);N=N-100*nota_100;nota_50=int(N/50);N=N-50*nota_50;nota_20=int(N/20);N=N-20*nota_20;nota_10=int(N/10);N=N-10*nota_10;nota_5=int(N/5);N=N-5*nota_5;nota_2=int(N/2);N=N-2*nota_2;nota_1=int(N/1);N=N-1*nota_1;printf("%d\\n",n);printf("%d nota(s) de R$ 100,00\\n",nota_100);printf("%d nota(s) de R$ 50,00\\n",nota_50);printf("%d nota(s) de R$ 20,00\\n",nota_20);printf("%d nota(s) de R$ 10,00\\n",nota_10);printf("%d nota(s) de R$ 5,00\\n",nota_5);printf("%d nota(s) de R$ 2,00\\n",nota_2);printf("%d nota(s) de R$ 1,00",nota_1);return 0;
}

9.AcWing 654. 时间转换

#include<iostream> 
#include<cstdio>using namespace std;int main(){int T,t;int H=0,M=0,S;cin>>T;t=T;H=int(t/3600);t=t-3600*H;M=int(t/60);t=t-60*M;S=t; printf("%d:%d:%d",H,M,S);return 0;
}

10. AcWing 605. 简单乘积

#include<iostream> 
#include<cstdio>using namespace std;int main(){int a,b;cin>>a>>b;printf("PROD = %d",a*b);return 0;
}

11.AcWing 611. 简单计算

#include<iostream>
#include<cstdio>using namespace std;int main(){int n1,q1,n2,q2;double p1,p2;cin>>n1>>q1>>p1;cin>>n2>>q2>>p2;printf("VALOR A PAGAR: R$ %.2f",q1*p1+q2*p2);return 0;
}

12.AcWing 612. 球的体积

#include<iostream>
#include<cstdio>using namespace std;int main(){int R;double pi;pi=3.14159;cin>>R;printf("VOLUME = %.3lf",(4/3.0)*pi*R*R*R);return 0;
}

13.AcWing 613. 面积

#include<iostream>
#include<cstdio>using namespace std;int main(){double A,B,C;cin>>A>>B>>C;printf("TRIANGULO: %.3lf\\n",0.5*A*C);printf("CIRCULO: %.3lf\\n",3.14159*C*C);printf("TRAPEZIO: %.3lf\\n",0.5*C*(A+B));printf("QUADRADO: %.3lf\\n",B*B);printf("RETANGULO: %.3lf",A*B);return 0;
}

14.AcWing 607. 平均数2

#include<iostream>
#include<cstdio>using namespace std;int main(){double A,B,C;cin>>A>>B>>C;printf("MEDIA = %.1lf",(A*2+B*3+C*5)/10);return 0;
}

15.AcWing 610. 工资和奖金

#include<iostream>
#include<cstdio>using namespace std;int main(){string name;double dixin,xiaoshoue;cin>>name;cin>>dixin>>xiaoshoue;printf("TOTAL = R$ %.2lf",dixin+0.15*xiaoshoue);return 0;
}

16.AcWing 614. 最大值

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;int main(){int A,B,C;int max1,max2;cin>>A>>B>>C;max1=(A+B+abs(A-B))/2;max2=(max1+C+abs(max1-C))/2;printf("%d eh o maior",max2);return 0;
}

17.AcWing 617. 距离

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;int main(){int L,t;double ans;cin>>L;ans=(L/30.0)*60;t=int(ans);printf("%d minutos",t);return 0;
}

18.AcWing 618. 燃料消耗

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;int main(){double S,T;double youhao;cin>>S>>T;youhao=(S*T)/12.0;printf("%.3lf",youhao);return 0;
}

19.AcWing 656. 钞票和硬币

#include <cstdio>
int main(){double n;scanf("%lf",&n);int money=n*100;printf("NOTAS:\\n");printf("%d nota(s) de R$ 100.00\\n",money/10000); money%=10000;printf("%d nota(s) de R$ 50.00\\n",money/5000); money%=5000;printf("%d nota(s) de R$ 20.00\\n",money/2000); money%=2000;printf("%d nota(s) de R$ 10.00\\n",money/1000); money%=1000;printf("%d nota(s) de R$ 5.00\\n",money/500); money%=500;printf("%d nota(s) de R$ 2.00\\n",money/200); money%=200;printf("MOEDAS:\\n");printf("%d moeda(s) de R$ 1.00\\n",money/100); money%=100;printf("%d moeda(s) de R$ 0.50\\n",money/50); money%=50;printf("%d moeda(s) de R$ 0.25\\n",money/25); money%=25;printf("%d moeda(s) de R$ 0.10\\n",money/10); money%=10;printf("%d moeda(s) de R$ 0.05\\n",money/5); money%=5;printf("%d moeda(s) de R$ 0.01\\n",money/1); money%=1;return 0;
}

利用放大法解决double数据类型精度的问题。

20.AcWing 655. 天数转换

#include<iostream>
#include<cstdio>using namespace std;int main(){int N;int a,m,d;cin>>N;a=N/365;m=(N-a*365)/30;d=N-a*365-m*30;cout<<a<<" "<<"ano(s)"<<endl;cout<<m<<" "<<"mes(es)"<<endl;cout<<d<<" "<<"dia(s)";return 0;
}