Description
Alec has a lot of eggs. One day, he want to sort them in a ascending sequence by weight. But he only can switch two eggs which are adjoining by each other because he has two hands only. Now he ask for your help, and you are enthusiastic. You decide help him calculate the total numbers of switch he need at least.
Attention: the weight of each egg less than 100,000,000 units.
Input
There are multiply test case.The first line describe the number of test case?. For each test case, the first line describe the number of eggs that Alec has,?. The second line describe the weight of each eggs splited by a space.
Output
Output the total number of switch that Alec need at least. One line for each test case. The total number may be very very large but fited in 64-bit integer.
Sample Input
2
2
2 1
2
2 3
Sample Output
1
0
題意：
題意：將相鄰的兩個元素進行排序，用到二路歸并算法，
貼上代碼：
?
1
#include <stdio.h>
2
#include <stdlib.h>
3
4
#define
MAX 100001
5
6
int
n, a[MAX], t[MAX];
7
long
long
sum;
8
9
/*
歸并
*/
10
void
Merge(
int
l,
int
m,
int
r)
11
{
12
/*
p指向輸出區間
*/
13
int
p =
0
;
14
/*
i、j指向2個輸入區間
*/
15
int
i = l, j = m +
1
;
16
/*
2個輸入區間都不為空時
*/
17
while
(i <= m && j <=
r)
18
{
19
/*
取關鍵字小的記錄轉移至輸出區間
*/
20
if
(a[i] >
a[j])
21
{
22
t[p++] = a[j++
];
23
/*
a[i]后面的數字對于a[j]都是逆序的
*/
24
sum += m - i +
1
;
25
}
26
else
27
{
28
t[p++] = a[i++
];
29
}
30
}
31
/*
將非空的輸入區間轉移至輸出區間
*/
32
while
(i <= m) t[p++] = a[i++
];
33
while
(j <= r) t[p++] = a[j++
];
34
/*
歸并完成后將結果復制到原輸入數組
*/
35
for
(i =
0
; i < p; i++
)
36
{
37
a[l + i] =
t[i];
38
}
39
}
40
41
/*
歸并排序
*/
42
void
MergeSort(
int
l,
int
r)
43
{
44
int
m;
45
if
(l <
r)
46
{
47
/*
將長度為n的輸入序列分成兩個長度為n/2的子序列
*/
48
m = (l + r) /
2
;
49
/*
對兩個子序列分別進行歸并排序
*/
50
MergeSort(l, m);
51
MergeSort(m +
1
, r);
52
/*
將2個排好的子序列合并成最終有序序列
*/
53
Merge(l, m, r);
54
}
55
}
56
57
int
main()
58
{
59
int
i;
60
int
t;
61
scanf(
"
%d
"
, &
t);
62
while
(t--
)
63
{
64
scanf(
"
%d
"
, &
n);
65
if
(n ==
0
)
break
;
66
sum =
0
;
67
for
(i =
0
; i < n; i++
)
68
{
69
scanf(
"
%d
"
, &
a[i]);
70
}
71
MergeSort(
0
, n -
1
);
72
printf(
"
%lld\n
"
, sum);
73
}
74
return
0
;
75
}
?
?
上面代碼速度沒有優化，
下面的是網友的代碼：
1
#include <stdio.h>
2
#define
MAX 100000
3
int
n,flag,a[MAX],b[MAX];
4
long
long
cnt;
5
void
Merge(
int
*res,
int
*cop,
int
l,
int
lt,
int
r,
int
rt)
6
{
7
int
base
=
l;
8
while
(
base
<=
rt)
9
{
10
while
((l<=lt && cop[l]<=cop[r]) || (l<=lt && r>rt)) res[
base
++]=cop[l++
];
11
while
((r<=rt && cop[l]>cop[r]) || (r<=rt && l>
lt))
12
{
13
cnt+=lt-l+
1
;
14
res[
base
++]=cop[r++
];
15
}
16
}
17
}
18
void
MergeSort()
19
{
20
int
step=
2
,semi,c,s,i,l,r,lt,rt;
21
int
*res=NULL,*cop=
NULL;
22
while
(step<
2
*
n)
23
{
24
c=n/
step;
25
s=n%
step;
26
semi=step/
2
;
27
if
(s>semi) c++
;
28
flag++
;
29
if
(flag%
2
)
30
{
31
res=
b;
32
cop=
a;
33
}
34
else
35
{
36
res=
a;
37
cop=
b;
38
}
39
for
(i=
0
;i<c;i++
)
40
{
41
l=i*
step;
42
r=l+
semi;
43
lt=r-
1
;
44
rt=(n-
1
)<(l+step-
1
)?(n-
1
):(l+step-
1
);
//
邊界非整段處理
45
Merge(res,cop,l,lt,r,rt);
46
}
47
if
(rt<n-
1
){
48
for
(i=rt+
1
;i<n;i++) res[i]=cop[i];
//
非常關鍵，尾部剩余有序要記得拷貝
49
}
50
step*=
2
;
51
}
52
}
53
54
int
main()
55
{
56
int
i;
57
int
t;
58
scanf(
"
%d
"
, &
t);
59
while
(t--
){
60
scanf(
"
%d
"
,&
n);
61
62
flag=cnt=
0
;
63
for
(i=
0
;i<n;i++) scanf(
"
%d
"
,&
a[i]);
64
MergeSort();
65
printf(
"
%lld\n
"
,cnt);
66
67
}
68
return
0
;
69
}
?
?另一個版本：
1
long
long
a[
100004
],b[
100004
];
int
N;
2
3
long
long
total,i,j;
4
5
void
Mergesort(
int
start,
int
end)
6
{
7
if
(start<
end){
8
int
i,j,mid=(start+end)/
2
;
9
Mergesort(start,mid);
10
Mergesort(mid+
1
,end);
11
int
t=start;j=mid+
1
;i=
start;
12
while
(i<=mid && j<=
end){
13
if
(a[i]<=
a[j])
14
b[t++]=a[i++
];
15
else
{
16
b[t++]=a[j++
];
17
total+=mid-i+
1
;
18
}
19
}
20
while
(i<=
mid)
21
b[t++]=a[i++
];
22
while
(j<=
end)
23
b[t++]=a[j++
];
24
for
(i=start;i<=end;i++
)
25
a[i]=
b[i];
26
}
27
}
28
29
30
int
main()
31
{
32
int
t;
33
double
ac;
34
scanf(
"
%d
"
, &
t);
35
while
(t--
) {
36
scanf(
"
%d
"
,&
N);
37
total=
0
;
38
for
(i=
0
;i<N;i++
)
39
scanf(
"
%lld
"
,&
a[i]);
40
Mergesort(
0
,N-
1
);
41
ac = (total+N)/(N*
1.0
);
42
printf(
"
%.2f\n
"
,ac);
43
44
}
45
}
?
?
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