\(\displaystyle{ x_{1} = x_{2} = x_{3} = 1}\)
\(\displaystyle{ x_{i} = (i+3)(x_{i-1}-1)+(i+4)x_{i-3}}\)
zacząłem tak:
Kod: Zaznacz cały
int a = 1;
for(int i=4; i <= n; i=i+1)
a = (i+3)(a-1)+(i+4)x;
Kod: Zaznacz cały
int a = 1;
for(int i=4; i <= n; i=i+1)
a = (i+3)(a-1)+(i+4)x;
Kod: Zaznacz cały
#include <iostream>
using namespace std;
int oblicz(int n)
{
if(1 == n || 2 == n || 3 == n)
{
return 1;
}
else
{
int xn3 = 1; //x_(n-3)
int xn2 = 1; //x_(n-2)
int xn1 = 1; //x_(n-1)
int xn; //obliczany wyraz ciagu
for(int i=4; i<=n; i++) {
xn3=xn2;
xn2=xn1;
xn1=xn;
xn=(i+3)*(xn1-1)+(i+4)*xn3;
}
return xn;;
}
}
int main() {
int n = 0;
cin >> n;
cout << oblicz(n);
return 0;
}
Kod: Zaznacz cały
#include <iostream>
using namespace std;
int oblicz(int n)
{
if(1 == n || 2 == n || 3 == n)
{
return 1;
}
else
{
return (n+3)*(oblicz(n-1)-1) + (n+4)*oblicz(n-3);
}
}
int main() {
int n = 0;
cin >> n;
cout << oblicz(n);
return 0;
}