ПОМОГИТЕ ПОЖАЛУЙСТА С ИНФОРМАТИКОЙ, Надобно НАПИСАТЬ ВСЁ НА ЯЗЫКЕ СИ

Решения задач. (компилируется на MinGW GCC 8.1)
z1:
include lt;stdio.hgt; // Ввод-вывод
include lt;math.hgt; // Математические функции
double абс_double(double x) return x lt; 0 ? -x : x;
int main()
double K, L, x, T, Q;
 scanf("%lf %lf %lf", amp;K, amp;L, amp;x);
T = pow(cos(x)*(pow(K, 2) - pow(L, 2), 2) / (K * L * x);
 Q = sqrt(pow(T, 2) * абс_double(K - L) / 0.25);
printf("T = %f, Q = %f\n", T, Q);
return 0;

z2:
include lt;stdio.hgt; // Ввод-вывод
include lt;math.hgt; // Математические функции
double max(double a, double b) return a gt; b ? a : b;
int main() double a, n, l, x1, x2, y, z;
scanf("%lf %lf %lf", amp;a, amp;n, amp;l);
x1 = (a + n) / pow(a - l, 2);
x2 = a / (n * l);
if (x1 == x2)
 y = x1 * (a - n) / x2;

else if (x1 lt; x2)
 y = x1 + x2;
else
 y = a * x1 + n * x2;

z = max(x1, x2) / (x1 + x2);
printf("x1 = %f, x2 = %f, y = %f, z = %f\n", x1, x2, y, z);
return 0;


z3:
include lt;stdio.hgt;
include lt;math.hgt;
int main()
double x;
 double y = 0;
 scanf("%lf", amp;x);
for (int i = 1; i lt;= 11; i += 2)
 y += pow((x - 1) / (x + 1), i) / i;

printf("Y = %f", y);


z4:

include lt;stdio.hgt;
int main()
int n = 0, p;
int sum = 0;
do
p = n;
scanf("%d", amp;n);
sum += n;
while (n != 0);
printf("sum = %d, prev = %d\n", sum, p);
return 0;


z5:
include lt;math.hgt;
include lt;stdio.hgt;
int main()
const double PI = 3.141592;
double a; int b;
scanf("%lf %d", amp;a, amp;b);
double x, d, S, s;
if (b == 1)
 x = a;
 d = x * sqrt(2);
 S = pow(d / 2, 2) * PI;
 s = pow(x / 2, 2) * PI;
else if (b == 2)
 d = a;
 x = d / sqrt(2);
 S = pow(d / 2, 2) * PI;
 s = pow(x / 2, 2) * PI;
else if (b == 3)
 S = a;
 d = 2 * sqrt(S / PI);
 x = d / sqrt(2);
 s = pow(x / 2, 2) * PI;
else if (b == 4)
 s = a;
  x = 2 * sqrt(s / PI);
 d = x * sqrt(2);
 S = pow(d / 2, 2) * PI;
  else
 printf("error");
 return 0;

printf("x = %f, d = %f, s = %f, S = %f", x, d, s, S);
return 0;