Var
sum, k, n, i: integer;
begin
sum := 0;
write('n: ');
readln(n);
for i := 1 to n do
begin
readln(k);
sum := sum + k;
end;
writeln('sum = ', sum);
end.
Var
a:array [1..100] of integer;
n,l,i,j:integer;
begin
<span>for i:=1 to 20 do</span>
read (a[i]);
for j:=1 to N-1 do
for i:=1 to N-j do
<span>if a[i] > a[i+1] </span>
<span>then </span>
<span>begin </span>
l:=a[i];
a[i]:=a[i+1];
a[i+1]:=l;
end;
for i:=1 to 20 do
<span>if ((a[i] mod 2)=1) and ((a[i] mod 5)=0) </span>
<span>then </span>
begin
writeln (a[i]);
break;
else writeln('таких чисел нет');
end;
<span>end.</span>
Задачу удобно решать с помощью кругов Эйлера (см. рис.).
K1+K2+K3+K4+K5+K6 = 1000
K2+K4 = 250
K4+K5+K6 = 200
K3+K5 = 500
K4 = 20
K5 = 10
K2+K3+K4+K5+K6 - ?
K1 - ?
K2+K3+K4+K5+K6 = (K2+K4)+(K4+K5+K6)+(K3+K5)-K4-K5 = 250+200+500-20-10 = 920
K1 = (K1+K2+K3+K4+K5+K6)-(K2+K3+K4+K5+K6) = 1000-920 = 80