# Including
& ltiostream & gt
use
Namespace
std
Internationalorganizations (same as international organizations)
Partition (int
*s，
Internationalorganizations (same as international organizations)
Low,
Internationalorganizations (same as international organizations)
High)
{
Internationalorganizations (same as international organizations)
I
j，
t；
Internationalorganizations (same as international organizations)
pivot
=
S[ low];
j
=
Low;
for
(I
=
low
+
1;
I
& lt=
High;
i++)
{
if
(s[i]
& lt
Pivot)
{
j++；
t
=
s[I]；
s[i]
=
s[j]；
s[j]
=
t；
}
}
t
=
s[j]；
//Swap pivot and the last number of the sequence less than pivot.
s[j]
=
Pivot;
S[ low]
=
t；
return
j；
}
empty
Quick sort (integer
*s，
Internationalorganizations (same as international organizations)
Low,
Internationalorganizations (same as international organizations)
High)
{
Internationalorganizations (same as international organizations)
Pivot;
//Central perspective
if
(low
& lt
High)
{
pivot
=
Partition,
Low,
High);
Quick sort,
Low,
pivot
-
1);
//Recursively sort the low-order sub-table.
Quick sort,
pivot
+
1,
High);
//Recursively sort the high child tables.
}
}
Internationalorganizations (same as international organizations)
Master ()
{
Internationalorganizations (same as international organizations)
t，
n；
Internationalorganizations (same as international organizations)
Me;
Internationalorganizations (same as international organizations)
a[ 1000]；
cin
& gt& gt
t；
During ...
(T -)
{
cin
& gt& gt
n；
for
(I
=
0;
I
& lt
n；
i++)
cin
& gt& gt
a[I]；
Quick sort (a,
0,
ordinary
-
1);
for
(I
=
0;
I
& lt
n；
i++)
Standard output
& lt& lt
a[i]
& lt& lt
endl
}
return
0;
}
Bubble sorting algorithm: where t is the number of tests and n is the number of elements.
# Including
& ltiostream & gt
use
Namespace
std
Internationalorganizations (same as international organizations)
Master ()
{
Internationalorganizations (same as international organizations)
t，
n；
Internationalorganizations (same as international organizations)
I
j，
k，
t；
Internationalorganizations (same as international organizations)
a[ 1000]；
cin
& gt& gt
t；
During ...
(T -)
{
cin
& gt& gt
n；
for
(I
=
0;
I
& lt
n；
i++)
cin
& gt& gt
a[I]；
for
(j
=
0;
j
& lt
ordinary
-
1;
j++)
{
for
(k
=
0;
k
& lt
ordinary
-
j
-
1;
k++)
{
if
(a[k]
& gt
a[k]
+
1])
{
t
=
a[k]
+
1];
a[k]
+
1]
=
a[k]；
a[k]
=
t；
}
}
}
for
(I
=
0;
I
& lt
n；
i++)
Standard output
& lt& lt
a[i]
& lt& lt
endl
}
return
0;
}