Var i, j, k: integer; a, b: array [1..5] of array [1..5] of integer; begin for i : = 1 to 5 do for j : = 1 to 5 do read(a[i][j]); k : = 1; for i : = 1 to 5 do if (odd(i)) then for j : = 5 downto 1 do begin b[i][j] : = a[((k-1) div 5)+1][((k-1) mod 5)+1]; inc(k); end else for j : = 1 to 5 do begin b[i][j] : = a[((k-1) div 5)+1][((k-1) mod 5)+1]; inc(k); end; for i : = 1 to 5 do begin for j : = 1 to 5 do write(b[i][j]: 2, ' '); writeln; end; end.
viktorovna_Imamaevich679
15.08.2022
#include < iostream> #include < sstream> #include < vector> #include < string> using namespace std; int main() { cout < < "string: "; string s; getline(cin, s); istringstream ss(s); vector< string> v; while (ss > > s) v.push_back(s); int j=v.; for (int i=j; i> =0; cout< < v[i]< < " "; cout< < endl; system("pause"); return 0; }ввод - вывод string: one two three four fife fife four three two one
D-posilochka
15.08.2022
'basic консольное dim x as double, y as double, z as double, a as double, b as double a = 1.1 x = 0.2 b = 0.004 z = sin((x * x) + a) - sqr(x / b) y = (x * x) / a + cos(x + b) ^ 3 print "z = "; z print "y = "; y ' visual basic 2015 с формами dim x as double, y as double, z as double, a as double, b as double a = 1.1 x = 0.2 b = 0.004 z = math.sin((x * x) + a) - math.sqrt(x / b) y = (x * x) / a + math.cos(x + b) ^ 3 label1.text = "z = " + z.tostring label2.text = "y = " + y.tostring
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Программа в паскале. надо двумерный массив ввывести змейкой по горизонтали справа на лево(5*5)