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;
}