Пусть x это основание сс 1*x^3=1*x^2+2*x + 1*x^2+1*x x^3-x^2-2x-x^2-x=0 x^3-2x^2-3x=0 x(x^2-2x-3) = 0 x^2-2x-3=0 x=3
i7aster26
30.07.2021
// pascalabc.net 3.1, сборка 1174 от 22.02.2016 begin var z: =arrrandom(8,1,20); z.println; writeln(z.where(x-> x> z[4]).count); var d: =arrrandom(12,-20,20); d.println; writeln(d.where(x-> x< 0).sum); var r: =arrrandom(10,-9,9); r.println; var s: =r.sum; for var i: =0 to 9 do if r[i]=0 then r[i]: =s; r.println end. тестовое решение: 15 10 10 8 8 20 19 12 6 -17 -11 -3 -15 -11 -12 -16 19 1 1 11 -15 -100 2 -2 1 0 -8 1 -1 0 -5 9 2 -2 1 -3 -8 1 -1 -3 -5 9
Sharap
30.07.2021
Program u2; const n=6; var i,max: integer; a,b: array [1..n] of integer; begin a[1]: =2; a[2]: =5; a[3]: =7; a[4]: =-2; a[5]: =0; a[6]: =8; writeln('a: '); for i: =1 to n do write(a[i]: 3); writeln; writeln('b: '); for i: =1 to n do begin b[i]: = (a[i]-1); write(b[i]: 3); end; writeln; max: =a[1]; for i: =1 to n do begin if a[i]> max then max: = a[i]; b[i]: =max; end; writeln('new b: '); for i: =1 to n do write(b[i]: 3); writeln; end. результат: a: 2 5 7 -2 0 8 b: 1 4 6 -3 -1 7 new b: 2 5 7 7 7 8