Определите количество полей записей в таблице​

Как-то так, наверное

Объяснение:

#include <iostream>

#include <cmath>

using namespace std;

int main()

{

   double a, x, b, c, D;

   

   cout<<"Введите a: "; cin >> a;

   cout<<"Введите b: "; cin >> b;

   cout<<"Введите c: "; cin >> c;

   

   D = pow(b,2)-4*a*c;

   if(D>0){

   cout << "x1 = " << ((-1)*b+sqrt(D))/(2*a)<<endl;

   cout << "x2 = " <<((-1)*b-sqrt(D))/(2*a)<<endl;

   }

   else if(D==0){

       cout << "x = " << ((-1)*b)/(2*a)<<endl;

   }

   else{

    D*=(-1);

    cout<< "x1 = " << ((-1)*b)/(2*a) <<"+"<< (sqrt(D))/(2*a) << "*i" <<endl;

    cout<< "x2 = " << ((-1)*b)/(2*a) <<"-"<< (sqrt(D))/(2*a) << "*i" <<endl;

   }

   return 0;

}

1)
from random import random
N = 10
a = []
for i in range(N):
    b = int(random() * 50)
    a.append(b)
print(a)

min = 101
for i in range(N-1):
    s = a[i]+a[i+1]
    if (s < min) and (s % 2 == 1):
        min=s
print(min)

Пример:[41, 35, 16, 7, 29, 9, 16, 28, 10, 6]
23

2)
from random import random
N = 10
a = []
for i in range(N):
    b = int(random() * 50)-25
    a.append(b)
print(a)

k = 0
for i in range(N-1):
    p = a[i]*a[i+1]
    s = a[i]+a[i+1]
    if (p % 2 != 0) and (s > 0):
        k = k+1
print(k)

Пример:[11, 23, 12, -16, 21, 15, -11, -10, 10, 17]
3