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;
}