#include <iostream>
using std::cout;
using std::cin;
using std::endl;
int main()
{
int n, m, counter = 0;
cout << "Enter n: ";
cin >> n;
cout << "Enter m: ";
cin >> m;
if(n >= m)
{
cout << "Invalid value" << endl;
}
else
{
while(n <= m)
{
if(n % 4 == 0)
{
counter++;
}
n++;
}
}
cout << counter << " value" << (counter != 1 ? "s" : "") << endl;
cin.get();
return 0;
}
Вывод:
Enter n: 2000
Enter m: 2012
4 values
На Паскале:
program test;
var n, m, counter: integer;
begin
write('Enter n: ');
readln(n);
write('Enter m: ');
readln(m);
counter := 0;
while n <= m do
begin
if n mod 4=0 then
counter := counter + 1;
n := n + 1;
end;
writeln(counter, ' values');
readln();
end.
#include <iostream>
using std::cout;
using std::cin;
using std::endl;
#include <cmath>
using std::pow;
using std::sqrt;
double vpisannayaOkryzhnost(double);
double opisannayaOkryzhnost(double);
int main()
{
double perimeter;
cout << "Vvedite perimeter: ";
cin >> perimeter;
cout << "r vpisannoi okruzhnosti = " << vpisannayaOkryzhnost(perimeter) << endl;
cout << "R opisannoi okruzhnosti = " << opisannayaOkryzhnost(perimeter) << endl;
cin.get();
return 0;
}
double vpisannayaOkryzhnost(double perimeter)
{
double a = perimeter / 4;
return (a / 2);
}
double opisannayaOkryzhnost(double perimeter)
{
double a = perimeter / 4;
return (sqrt(2) / 2 * a);
}
На Гэ-паскале можно написать так:
program math;
var perimeter, a: real;
begin
write('Vvedite perimeter: ');
read(perimeter);
a := perimeter / 4;
writeln('r vpisannoi okruzhnosti = ', a / 2:0:2);
writeln('R opisannoi okruzhnosti = ', sqrt(2) / 2 * a:0:2);
readln();
readln();
end.
