Вспоминаем формулу Шеннона
![\displaystyle I=-\sum_{i=1}^n(p_i\times\log_2 p_i)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I%3D-%5Csum_%7Bi%3D1%7D%5En%28p_i%5Ctimes%5Clog_2%20p_i%29)
Общее количество карандашей равно 3+7 = 10.
p₁= 3/10 = 0.3; p₂ = 7/10 = 0.7
![\displaystyle I=-(0.3\log_2 0.3+0.7\log_20.7)\approx -(-0.521-0.360)=0.881 ](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I%3D-%280.3%5Clog_2%200.3%2B0.7%5Clog_20.7%29%5Capprox%20-%28-0.521-0.360%29%3D0.881%0A%0A)
1)
#include <iostream>
using namespace std;
int main()
{
int n = 0;
cin >> n;
for (int i = 0; i < n; ++i) {
cout << 2 * i << endl;
}
return 0;
}
2)
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int a = 0, b = 0;
cin >> a >> b;
for (int i = a; i <= b; ++i) {
cout << pow(i, 2) << endl;
}
return 0;
}
3)
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int a = 0, b = 0, sum = 0;
cin >> a >> b;
for (int i = a; i <= b; ++i) {
sum += pow(i, 2);
}
cout << sum << endl;
return 0;
}
128 - 1000 0000
256 - 1 0000 0000
512 - 10 0000 0000
1024 - 100 0000 0000