Program kek;
Var a, c:integer;
Begin
c:=0;
a:=1;
While a<>0 do begin
Writeln('Вводите целые числа. Когда закончите - введите "0"');
Readln(a);
If (a>0) and (a mod 10 = 1) then do begin
c:=c+a;
end;
End;
Writeln(c);
End.
Вроде так. Нет возможности проверить.
256 в 2-ую -100000<span>000;
256 в 8-ую - </span><span>400;
256 в 16-ую -100;
256 в 4-ую - 10000;
256 в 7-ую - 514;
</span>
1) 221(8)=145(10)
109(16)=265(10)
т.к. 145<265, то 221(8)<109(16)
2) А23(16)= 2595(10)
1223(8)=659(10)
т. к. 2595>659. то А23(16)> 1223(8)
Var
a, b, c, min : integer;
begin
read (a, b, c);
if (a <= b) and (a <= c) then min := a
else if (b <= a) and (b <= c) then min := b
else min := c;
write (min);
end.
<em>1)</em>
#include <stdio.h>
#include <stdlib.h>
int sover (int n)
{<span>
int sum = 0;</span><span>
for (int i = 1; i<=n/2; i++)</span><span>
if (n%i==0)</span><span> sum+=i;</span><span>
if (sum==n)</span><span>
return 1;</span><span>
else return 0;
</span>}
int main()
{<span>
int n;</span><span>
printf ("N = ");</span><span>
scanf ("%d",&n);</span><span>
if (sover(n)==1)</span><span>
printf ("Sovershennoe");</span><span>
else printf ("Ne sovershennoe");</span><span>
printf("\n");</span><span>
system ("PAUSE");</span><span>
return 0;
</span><span>}
<em>2)</em>
#include <stdio.h>
#include <stdlib.h>
int main()
{
for (int i = 1; i<=5; i++)
{
for (int j = 1; j<=5-i+1; j++)
printf ("%d ",5+i-1);
printf ("\n");
}
system ("PAUSE");
return 0;
}</span>