PYTHON

1041 uri solution

//uri problem 1041 || coordinates of a point in python 3.9

inp = input()
x, y = inp.split()
x = float(x)
y = float(y)
if x == 0 and y ==0:
    print("Origem")
elif(x==0):
    print("Eixo Y")
elif(y==0):
    print("Eixo X")
elif x > 0 and y > 0:
    print("Q1")
elif x < 0 and y > 0:
    print("Q2")
elif x < 0 and y < 0:
    print("Q3")
elif x > 0 and y < 0:
    print("Q4")

1042 uri solution

#uri 1042 solution || simple sort
x, y, z = map(int,input().split())
arr = [x,y,z]
arr.sort()
print(f"{arr[0]}
{arr[1]}
{arr[2]}")
print(f"
{x}
{y}
{z}")

1043 uri solution

# 1043 triangle
a, b, c = map(float,input().split())
if a+b > c and b+c > a and a+c > b:
    peri = a+b+c
    print("Perimetro = {:.1f}".format(peri))
else:
    area = ((a+b)/2) *c
    print("Area = {:.1f}".format(area))

1044 uri solution

# 1044 Multiples
a, b = map(int,input().split())
if a%b ==0 or b%a == 0:
    print("Sao Multiplos")
else:
    print("Nao sao Multiplos")

1046 uri solution

a,b=list(map(int,input().split()))
if(a<b):
    time=b-a
else:
    time=b+24-a
print(f"O JOGO DUROU {time} HORA(S)")

1045 uri solution

data = list(map(float, input().split( )))

data.sort(reverse=True) 

a,b,c = data

if a>=b+c:
    print("NAO FORMA TRIANGULO")
elif a**2 == b**2 + c**2:
    print("TRIANGULO RETANGULO")
elif a**2 > b**2 + c**2:
    print("TRIANGULO OBTUSANGULO")
elif a**2 < b**2 + c**2:
    print("TRIANGULO ACUTANGULO")
    
if a==b and a==c and b==a and b==c and c==a and c==b:
    print("TRIANGULO EQUILATERO")
elif a==b or a==c or b==a or b==c or c==a or c==b:
    print("TRIANGULO ISOSCELES")

1047 uri solution

a,c,b,d=list(map(int,input().split()))

dif=((b*60)+d)-((a*60)+c)
if(dif<=0):
    dif+=24*60
    
time=dif//60
minute=dif%60
print(f"O JOGO DUROU {time} HORA(S) E {minute} MINUTO(S)")

1043 uri solution c

#include <stdio.h>
int main()
{
    float A, B, C;
    scanf("%f %f %f", &A, &B, &C);
    if (A < B+C && B < A+C && C < B+C){
        printf("Perimetro = %.1f
",A+B+C);
    }
    else
        printf("Area = %.1f
",((A+B)*C) / 2);
    return 0;
}

uri 1045 solution in c

#include <stdio.h>

int main(){
    double a,b,c;

    scanf("%lf %lf %lf", &a, &b, &c);

    if(a >= (b+c) || b >= (a+c) || c >= (a+b))
         printf("NAO FORMA TRIANGULO
");
    else if(a*a == (b*b + c*c) || b*b == (a*a + c*c) || c*c == (a*a + b*b))
           printf("TRIANGULO RETANGULO
");
    else if(a*a > (b*b + c*c) || b*b > (a*a + c*c) || c*c > (a*a + b*b))
           printf("TRIANGULO OBTUSANGULO
");
    else if(a*a < (b*b + c*c) || b*b < (a*a + c*c) || c*c < (a*a + b*b))
           printf("TRIANGULO ACUTANGULO
");
    if(a == b && a == c)
            printf("TRIANGULO EQUILATERO
");
    if((a == b && a != c)  || (a == c && a != b) || (b == c && b != a))
            printf("TRIANGULO ISOSCELES
");

 return 0;
    
}