ответ:
#include
#include
using namespace std;
double fact(int n)
{
if (n == 0)
return 1;
else
return(n * fact(n - 1));
}
int main()
{
int n;
double s = 0;
cin > > n;
for (int i = 1; i < = n; i++) {
s += pow(-1 , 2 + i) / fact(i);
}
cout < < s < < endl;
return 0;
}
объяснение:
ответ:
n, m, t = map(int,
x = list(map(int,
def build(name):
if name in built:
pass
elif b[a.index(name)] == ["0"]:
built.append(name)
else:
for i in b[a.index(: ]:
if i not in built:
build(i)
built.append(name)
a = []
b = []
built = []
for i in range(0, n):
a.append(
b.append(
for i in range(0, b.:
built.append(a[b.
a.pop(b.
b.pop(b.
[build(str(o)) for o in x]
print(len(built))
объяснение:
Поделитесь своими знаниями, ответьте на вопрос: