# Первый цикл поднимет a на 3, второй и третий на 0, 4-й на 2.
a = 1
for i in range(3):
a += 1
'''print(f'[for i in range(3) > RANGE][a = {a}]')'''
print(f'[for i in range(3) > TOTAL][a = {a}]')
a = 1
for i in range(3, 1):
a += 1
'''print(f'[for i in range(3, 1) > RANGE][a = {a}]')'''
print(f'[for i in range(3, 1) > TOTAL][a = {a}]')
a = 1
for i in range(1, 3, -1):
a += 1
'''print(f'[for i in range(1, 3, -1) > RANGE][a = {a}]')'''
print(f'[for i in range(1, 3, -1) > TOTAL][a = {a}]')
a = 1
for i in range(3, 1, -1):
a += 1
'''print(f'[for i in range(3, 1, -1) > RANGE][a = {a}]')'''
print(f'[for i in range(3, 1, -1) > TOTAL][a = {a}]')
nn = 100;
var
n, l1, l2, r1, r2, i: integer;
a, b: array[1..nn] of integer;
F: Text;
begin
Assign(F, 'input.txt');
Reset(F);
Readln(F, n);
for i := 1 to n do Read(F, a[i]);
Readln(F, l1, r1);
Readln(F, l2, r2);
Close(F);
for i := 1 to l1 - 1 do b[i] := a[i];
for i := l1 to r1 do b[i] := a[l1 + r1 - i];
if l2 <= r1 then
begin
for i := l1 to r1 do a[i] := b[i];
for i := l2 to r2 do b[i] := a[l2 + r2 - i]
end
else
begin
for i := r1 + 1 to l2 - 1 do b[i] := a[i];
for i := l2 to r2 do b[i] := a[l2 + r2 - i];
end;
for i := r2 + 1 to n do b[i] := a[i];
for i := 1 to n do Write(b[i],' ');
end.
Результат решения:
5 8 3 6 0 14 -6 -2 4 1