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题目解答

题目:
Var
al,a2,a3,x:integer;
procedure swap (var a,b:integer);
Var
t:integer;
begin
var
t:integer;
begin
t:=a;
a:=b;
b:=t;
end;
begin
readln(al,a2,a3);
if al >a2 then
swap(a 1,a2);
if a2>a3 then
swap (a2,a3);
if al>a2 then
swap(a1,a2);
readlnl(x);
if x<a2 then
if x<al then
writeln(x, ‘ ‘,a1,’ ‘,a2,’ ‘,a3)
e1se
writeln(a1, ’ ‘,x,’ ‘,a2,’ ‘,a3)
else
if x<a3 then
writeln(al,’ ‘,a2,’ ‘,x,’ ‘,a3)
else
writeln(al,’ ‘, a2, ‘ ‘, a3, ‘ ‘.x);
end.

输入:
91 2 20
77
输出:2 20 77 91
考点:
分析:
解答:
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