Задание 1
#include <iostream>
using namespace std;
int main() {
int a, b;
cin >> a >> b;
cout << a - b;
}
Задание 2
#include <math.h>
int a;
cin >> a;
cout << pow(a, 2) << ' ' << pow(a, 3)<< ' ' << pow(a, 5);
Задание 3
cout << "The next number for the number " << a << " is " << a + 1 <<"!\n" <<"The previous number for the number "<< a << " is " << a - 1 <<"!";
Задание 4
cout << (3 * pow(a, 3) + 18 * pow(a, 2)) * a + 12 * pow(a, 2) - 5;
Задание 5
int a, b, c;
cin >> a >> b >> c;
cout << a % 7 << "\n" << b % 7 << "\n" << c % 7;
Задание 6
int a, b, c, a1, b1, c1;
cin >> a >> a1 >> b >> b1 >> c >> c1;
cout << a1 % (8 - a) << "\n" << b1 % (8 - b) << "\n" << c1 % (8 - c);
Задание 1
#include <iostream>
using namespace std;
int main() {
int a, b;
cin >> a >> b;
cout << a - b;
}
Задание 2
#include <iostream>
#include <math.h>
using namespace std;
int main() {
int a;
cin >> a;
cout << pow(a, 2) << ' ' << pow(a, 3)<< ' ' << pow(a, 5);
}
Задание 3
#include <iostream>
#include <math.h>
using namespace std;
int main() {
int a;
cin >> a;
cout << "The next number for the number " << a << " is " << a + 1 <<"!\n" <<"The previous number for the number "<< a << " is " << a - 1 <<"!";
}
Задание 4
#include <iostream>
#include <math.h>
using namespace std;
int main() {
int a;
cin >> a;
cout << (3 * pow(a, 3) + 18 * pow(a, 2)) * a + 12 * pow(a, 2) - 5;
}
Задание 5
#include <iostream>
#include <math.h>
using namespace std;
int main() {
int a, b, c;
cin >> a >> b >> c;
cout << a % 7 << "\n" << b % 7 << "\n" << c % 7;
}
Задание 6
#include <iostream>
#include <math.h>
using namespace std;
int main() {
int a, b, c, a1, b1, c1;
cin >> a >> a1 >> b >> b1 >> c >> c1;
cout << a1 % (8 - a) << "\n" << b1 % (8 - b) << "\n" << c1 % (8 - c);
}
using namespace std;
int gcd(int a, int b);
int main()
{
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
int n;
cin >> n;
for (int i = 0; i < n; ++i)
{
int a, b, c, d;
scanf("%d/%d+%d/%d=", &a, &b, &c, &d);
int num = a * d + b * c;
int den = b * d;
int cur_gcd = gcd(num, den);
num /= cur_gcd;
den /= cur_gcd;
cout << num;
if(den != 1)
cout << '/' << den;
cout << endl;
}
fclose(stdin);
fclose(stdout);
return 0;
}
int gcd(int a, int b)
{
if(a == 0)
return b;
return gcd(b % a, a);
}