Beecrowd 1012-Area solution C, CPP and Python
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Beecrowd 1012-Area solution C, CPP and Python
beecrowd | 1012
Area
Make a program that reads three floating point values: A, B and C. Then, calculate and show:
a) the area of the rectangled triangle that has base A and height C.
b) the area of the radius’s circle C. (pi = 3.14159)
c) the area of the trapezium which has A and B by base, and C by height.
d) the area of the square that has side B.
e) the area of the rectangle that has sides A and B.
Input
The input file contains three double values with one digit after the decimal point.
Output
The output file must contain 5 lines of data. Each line corresponds to one of the areas described above, always with a corresponding message (in Portuguese) and one space between the two points and the value. The value calculated must be presented with 3 digits after the decimal point.
Input Samples | Output Samples |
3.0 4.0 5.2 | TRIANGULO: 7.800 CIRCULO: 84.949 TRAPEZIO: 18.200 QUADRADO: 16.000 RETANGULO: 12.000 |
12.7 10.4 15.2 | TRIANGULO: 96.520 CIRCULO: 725.833 TRAPEZIO: 175.560 QUADRADO: 108.160 RETANGULO: 132.080 |
Beecrowd 1012-Area solution in C
Solution :
#include<stdio.h> int main() { double A,B,C,t,pi,c,tra,s,rec; scanf("%lf%lf%lf",&A,&B,&C); pi=3.14159; t=.5AC; c=piCC; tra=.5(A+B)C; s=BB; rec=AB; printf("TRIANGULO: %.3lf\nCIRCULO: %.3lf\nTRAPEZIO: %.3lf\nQUADRADO: %.3lf\nRETANGULO: %.3lf\n",t,c,tra,s,rec); return 0; }
Beecrowd 1012-Area solution in Python
import math A, B, C = map(float, input().split()) pi = 3.14159 t = 0.5 * A * C c = pi * (C ** 2) tra = 0.5 * (A + B) * C s = B ** 2 rec = A * B print(f"TRIANGULO: {t:.3f}") print(f"CIRCULO: {c:.3f}") print(f"TRAPEZIO: {tra:.3f}") print(f"QUADRADO: {s:.3f}") print(f"RETANGULO: {rec:.3f}")
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