Напишите блок-схему к двум программам: var a: array [1..12] of integer; i: byte; min, indexmin: integer; begin i : = 1; indexmin : = 1; while (i < 13) do begin write('a[',i,'] = '); readln(a[i]); if i = 1 then min : = a[1] else if (a[i] < min) then begin min : = a[i]; indexmin : = i; end; i : = i + 1; end; i : = 0; repeat i : = i + 1; if (i = indexmin) then a[i] : = 1; write(a[i]: 4); until (i = 12); end. var a: array [1..5,1..4] of integer; i, j, min, imin, jmin: integer; begin for i: =1 to 5 do for j: =1 to 4 do a [i,j]: =random(21)+1; for i: =1 to 5 do begin for j: =1 to 4 do write (a[i,j]: 4); writeln; end; min: =a[1,1]; imin: =1; jmin: =1; for i: =1 to 5 do for j: =1 to 4 do if a[i,j] writeln; for j: =1 to 4 do write (a[imin,j],' '); writeln; for i: =1 to 5 do write (a[i,jmin],' '); end.
Ответы на вопрос:
c++ gnu. простое заполнение массива, а потом в цикле проверки.
в конце выводим результаты.
мой вариант решения:#include
using namespace std;
int main()
{
const int size = 20;
int arr[size];
for (int i = 0; i < size; ++i)
{
cout < < "enter " < < i < < " element -> ";
cin > > arr[i];
}
int allsum = 0;
int evsum = 0;
int oddsum = 0;
int evcounter = 0;
int oddcounter = 0;
int smallthenzero = 0;
for (int i = 0; i < size; ++i)
{
allsum += arr[i];
if (arr[i] < 0) smallthenzero++;
if (arr[i] % 2 == 0)
{
evcounter++;
evsum += arr[i];
}
else
{
oddcounter++;
oddsum += arr[i];
}
}
cout < < "even numbers counter - " < < evcounter < < endl;
cout < < "odd numbers counter - " < < oddcounter < < endl;
cout < < "even numbers sum - " < < evsum < < endl;
cout < < "odd numbers sum - " < < oddsum < < endl;
cout < < "all numbers sum - " < < allsum < < endl;
cout < < "small then zero numbers - " < < smallthenzero < < endl;
cin.get();
cin.get();
}
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