Автор: vombat

vombat — Mon, 08 May 2017 09:12:45 +0000

#include<stdio.h>
#include<math.h>

int main(void)
{	
	float x;
	int number;
	const float pi = 3.14;

	printf("x:");
	scanf("%f", &x);
	printf("number:");
	scanf("%i", &number);

	switch(number)
	{
		case 1: 
			x = x; 
			printf("result: D=%.2f L=%.2f S=%.2f\n", 2 * x, 2 * pi * x, pi * (x * x)); 
			break;
		case 2: 
			x = x / 2; 
			printf("result: R=%.2f L=%.2f S=%.2f\n", x, 2 * pi * x, pi * (x * x)); 
			break;
		case 3: 
			x = x / (2*pi); 
			printf("result: R=%.2f D=%.2f S=%.2f\n", x, 2 * x, pi * (x * x)); 
			break;
		case 4: 
			x = sqrt(x / pi); 
			printf("result: R=%.2f D=%.2f L=%.2f\n", x, 2 * x, 2 * pi * x); 
			break;
	}

	return 0;
}

#include

int main(void)

{

float x;

int number;

const float pi = 3.14;

printf("x:");

scanf("%f", &x);

printf("number:");

scanf("%i", &number);

switch(number)

{

case 1:

x = x;

printf("result: D=%.2f L=%.2f S=%.2f\n", 2 * x, 2 * pi * x, pi * (x * x));

break;

case 2:

x = x / 2;

printf("result: R=%.2f L=%.2f S=%.2f\n", x, 2 * pi * x, pi * (x * x));

break;

case 3:

x = x / (2*pi);

printf("result: R=%.2f D=%.2f S=%.2f\n", x, 2 * x, pi * (x * x));

break;

case 4:

x = sqrt(x / pi);

printf("result: R=%.2f D=%.2f L=%.2f\n", x, 2 * x, 2 * pi * x);

break;

}

return 0;

}

Комментарии: Решаем задачи Абрамян на C. Case12

Автор: vombat